💡First, I need to determine the time it takes Micah to finish the marathon. Let's denote this time as ( M ) hours.Jake takes ( frac{1}{3} ) times more time than Micah, which means Jake's time is ( M + frac{1}{3}M = frac{4}{3}M ) hours.The total time taken by Micah, Dean, and Jake is 23 hours. Dean takes 9 hours, so the equation becomes:[M + frac{4}{3}M + 9 = 23]Combining the terms:[frac{7}{3}M + 9 = 23]Subtracting 9 from both sides:[frac{7}{3}M = 14]Solving for ( M ):[M = 14 times frac{3}{7} = 6]Micah takes 6 hours to finish the marathon. Since speed is inversely proportional to time, the ratio of Micah's speed to Dean's speed is the inverse of their times:[frac{text{Micah's speed}}{text{Dean's speed}} = frac{9}{6} = frac{3}{2}]Therefore, the ratio of Micah's speed to Dean's speed is ( 3:2 ).

💡Alright, so I've got this problem here about quadratic functions and sets. Let me try to wrap my head around it step by step. The function is given as ( f(x) = ax^2 + bx + c ), where ( a ), ( b ), and ( c ) are real numbers. Then, there are two sets defined: ( M = { f(2n) mid n text{ is an integer} } ) and ( N = { f(2n+1) mid n text{ is an integer} } ). The task is to prove that either ( M = N ) or ( M cap N = emptyset ). Okay, so ( M ) is the set of all values of ( f ) evaluated at even integers, and ( N ) is the set of all values of ( f ) evaluated at odd integers. We need to show that either these two sets are identical or they don't share any common elements. Hmm, interesting. First, maybe I should write out what ( f(2n) ) and ( f(2n+1) ) look like. Let's compute them:For ( f(2n) ):[f(2n) = a(2n)^2 + b(2n) + c = 4an^2 + 2bn + c]For ( f(2n+1) ):[f(2n+1) = a(2n+1)^2 + b(2n+1) + c = a(4n^2 + 4n + 1) + b(2n + 1) + c = 4an^2 + 4an + a + 2bn + b + c]Simplifying that:[f(2n+1) = 4an^2 + (4a + 2b)n + (a + b + c)]So, ( f(2n) ) is a quadratic in ( n ) with coefficients ( 4a ), ( 2b ), and ( c ), while ( f(2n+1) ) is also a quadratic in ( n ) but with coefficients ( 4a ), ( 4a + 2b ), and ( a + b + c ). Now, the question is about the relationship between the sets ( M ) and ( N ). To show that ( M = N ) or ( M cap N = emptyset ), perhaps I should consider whether there's any overlap between the values produced by these two quadratics. If there is an overlap, meaning some value is achieved by both ( f(2n) ) and ( f(2m+1) ) for some integers ( n ) and ( m ), then maybe that leads to ( M = N ). Alternatively, if there's no overlap, then ( M cap N ) is empty.Let me suppose that there exists some integer ( n ) and ( m ) such that ( f(2n) = f(2m+1) ). That would mean:[4an^2 + 2bn + c = 4am^2 + (4a + 2b)m + (a + b + c)]Simplifying this equation:[4an^2 + 2bn + c = 4am^2 + 4am + 2bm + a + b + c]Subtracting ( c ) from both sides:[4an^2 + 2bn = 4am^2 + 4am + 2bm + a + b]Let me rearrange terms to bring everything to one side:[4an^2 - 4am^2 + 2bn - 4am - 2bm - a - b = 0]Factor out common terms:[4a(n^2 - m^2) + 2b(n - m) - 4am - 2bm - a - b = 0]Notice that ( n^2 - m^2 = (n - m)(n + m) ), so:[4a(n - m)(n + m) + 2b(n - m) - 4am - 2bm - a - b = 0]Factor out ( (n - m) ) from the first two terms:[(n - m)(4a(n + m) + 2b) - 4am - 2bm - a - b = 0]Hmm, this is getting a bit messy. Maybe I can factor out some terms differently. Let me see:Looking back at the equation:[4a(n^2 - m^2) + 2b(n - m) = a + b + 4am + 2bm]Wait, perhaps I can factor ( (n - m) ) on the left side:[(n - m)(4a(n + m) + 2b) = a + b + 4am + 2bm]Let me denote ( k = n - m ). Then ( n = m + k ). Substituting back:[k(4a(2m + k) + 2b) = a + b + 4am + 2bm]Expanding the left side:[k(8am + 4ak + 2b) = a + b + 4am + 2bm]Which is:[8amk + 4ak^2 + 2bk = a + b + 4am + 2bm]This seems complicated. Maybe I need a different approach.Alternatively, perhaps I can consider the difference between ( f(2n+1) ) and ( f(2n) ). Let's compute ( f(2n+1) - f(2n) ):[f(2n+1) - f(2n) = [4an^2 + (4a + 2b)n + (a + b + c)] - [4an^2 + 2bn + c] = (4a + 2b - 2b)n + (a + b + c - c) = 4an + a + b]So, the difference between consecutive terms in ( N ) and ( M ) is linear in ( n ). Interesting.If ( f(2n+1) - f(2n) = 4an + a + b ), then for this difference to be zero for some ( n ), we'd have:[4an + a + b = 0 implies n = -frac{a + b}{4a}]But ( n ) has to be an integer. So, unless ( frac{a + b}{4a} ) is an integer, there is no integer ( n ) such that ( f(2n+1) = f(2n) ). Wait, but this is just for consecutive terms. Maybe the overlap isn't necessarily between consecutive terms. So, perhaps I need a more general approach.Let me think about the structure of ( M ) and ( N ). Both are sets of values of quadratic functions evaluated at integers. Quadratic functions are either convex or concave depending on the sign of ( a ). If ( a neq 0 ), the function is a parabola.Now, if ( a = 0 ), the function becomes linear: ( f(x) = bx + c ). In that case, ( f(2n) = 2bn + c ) and ( f(2n+1) = 2bn + b + c ). So, ( M = {2bn + c mid n in mathbb{Z}} ) and ( N = {2bn + b + c mid n in mathbb{Z}} ). If ( b neq 0 ), then ( M ) and ( N ) are arithmetic sequences with the same common difference ( 2b ), but shifted by ( b ). So, unless ( b = 0 ), ( M ) and ( N ) are just two different arithmetic sequences. If ( b = 0 ), then both ( M ) and ( N ) collapse to the single value ( c ), so ( M = N ).But in the case ( a = 0 ), unless ( b = 0 ), ( M ) and ( N ) are different but overlapping? Wait, no. If ( b neq 0 ), then ( M ) and ( N ) are two different arithmetic sequences with the same common difference. So, they are either identical (if the shifts are multiples of the common difference) or disjoint.Wait, actually, if ( M ) is ( c + 2bmathbb{Z} ) and ( N ) is ( c + b + 2bmathbb{Z} ), then ( N = M + b ). So, unless ( b ) is a multiple of ( 2b ), which would require ( b = 0 ), ( M ) and ( N ) are translates of each other and hence disjoint.So, in the linear case (( a = 0 )), either ( M = N ) (if ( b = 0 )) or ( M cap N = emptyset ). That seems to fit the conclusion we need to prove.But in the quadratic case (( a neq 0 )), it's more complicated. Let me consider that.Suppose ( a neq 0 ). Then ( f(x) ) is a parabola. The sets ( M ) and ( N ) are the images of the even and odd integers under ( f ). We need to see if these images can overlap or not.One approach is to consider the function ( f ) modulo some integer. But I'm not sure. Alternatively, maybe I can analyze the equations ( f(2n) = f(2m+1) ) for integers ( n, m ).So, setting ( f(2n) = f(2m+1) ), we have:[4an^2 + 2bn + c = 4am^2 + (4a + 2b)m + (a + b + c)]Simplify:[4a(n^2 - m^2) + 2b(n - m) = a + b + 4am + 2bm]Factor:[4a(n - m)(n + m) + 2b(n - m) = a + b + 4am + 2bm]Let ( k = n - m ), then ( n = m + k ). Substitute back:[4a k (2m + k) + 2b k = a + b + 4am + 2bm]Expand:[8a m k + 4a k^2 + 2b k = a + b + 4a m + 2b m]Rearrange terms:[8a m k + 4a k^2 + 2b k - 4a m - 2b m - a - b = 0]Factor terms with ( m ):[m(8a k - 4a - 2b) + 4a k^2 + 2b k - a - b = 0]This is a linear equation in ( m ). For integer solutions ( m ), the coefficient of ( m ) must divide the constant term. Let's denote:[A = 8a k - 4a - 2b][B = 4a k^2 + 2b k - a - b]So, ( A m + B = 0 implies m = -B / A ). For ( m ) to be integer, ( A ) must divide ( B ).But ( A ) and ( B ) depend on ( k ), which is ( n - m ). This seems recursive. Maybe another approach is needed.Alternatively, perhaps I can consider the function ( f(x) ) and its behavior on even and odd integers. Since ( f(x) ) is quadratic, it's symmetric about its vertex. The vertex occurs at ( x = -b/(2a) ). If the vertex is at an integer, say ( x = t ), then ( f(t) ) is the minimum or maximum value of ( f ). If ( t ) is even, then ( f(t) ) is in ( M ); if ( t ) is odd, it's in ( N ). But I'm not sure if this directly helps. Maybe I need to think about the periodicity or something. Wait, quadratic functions aren't periodic, so that might not be useful.Another idea: suppose that ( M ) and ( N ) are both infinite sets (if ( a neq 0 ), they are, since ( f(x) ) tends to infinity as ( x ) does). If they overlap, then perhaps they coincide entirely. Otherwise, they don't overlap at all.But how to formalize this? Maybe by assuming that there's at least one common value and then showing that all values must coincide.Suppose there exists some ( y ) such that ( y = f(2n) = f(2m+1) ) for some integers ( n, m ). Then, as before, we have:[4an^2 + 2bn + c = 4am^2 + (4a + 2b)m + (a + b + c)]Simplify:[4a(n^2 - m^2) + 2b(n - m) = a + b + 4am + 2bm]Factor:[4a(n - m)(n + m) + 2b(n - m) = a + b + 4am + 2bm]Let ( k = n - m ), so ( n = m + k ). Substitute:[4a k (2m + k) + 2b k = a + b + 4a m + 2b m]Expand:[8a m k + 4a k^2 + 2b k = a + b + 4a m + 2b m]Rearrange:[8a m k + 4a k^2 + 2b k - 4a m - 2b m - a - b = 0]Factor ( m ):[m(8a k - 4a - 2b) + 4a k^2 + 2b k - a - b = 0]Let me denote:[C = 8a k - 4a - 2b][D = 4a k^2 + 2b k - a - b]So, ( C m + D = 0 implies m = -D / C ). For ( m ) to be integer, ( C ) must divide ( D ).But ( C ) and ( D ) are expressions in ( k ), which is ( n - m ). This seems like a loop. Maybe I need to find conditions on ( a ) and ( b ) such that this equation has integer solutions ( m ) and ( k ).Alternatively, perhaps I can think about the function ( f(x) ) modulo 2. Since ( f(2n) ) and ( f(2n+1) ) are evaluated at even and odd integers, maybe their images have different properties modulo something.Compute ( f(2n) ) modulo 2:[f(2n) = 4an^2 + 2bn + c equiv 0 + 0 + c equiv c pmod{2}]Similarly, ( f(2n+1) ):[f(2n+1) = 4an^2 + 4an + a + 2bn + b + c equiv 0 + 0 + a + 0 + b + c equiv a + b + c pmod{2}]So, ( f(2n) equiv c pmod{2} ) and ( f(2n+1) equiv a + b + c pmod{2} ). Therefore, if ( c equiv a + b + c pmod{2} ), then ( a + b equiv 0 pmod{2} ). Otherwise, ( f(2n) ) and ( f(2n+1) ) have different parities, meaning ( M ) and ( N ) consist of numbers with different parities, hence ( M cap N = emptyset ).So, if ( a + b ) is odd, then ( M ) and ( N ) have different parities, so they can't overlap. If ( a + b ) is even, then ( f(2n) ) and ( f(2n+1) ) have the same parity, so it's possible for them to overlap.But this is just a necessary condition, not sufficient. So, if ( a + b ) is even, ( M ) and ( N ) could potentially overlap, but we need to see if they must coincide entirely or not.Wait, in the linear case (( a = 0 )), if ( a + b = b ) is even, then ( M ) and ( N ) are arithmetic sequences with the same parity. If ( b ) is even, then ( M = c + 2bmathbb{Z} ) and ( N = c + b + 2bmathbb{Z} ). If ( b ) is even, say ( b = 2k ), then ( N = c + 2k + 4kmathbb{Z} ), which is a shift of ( M ) by ( 2k ). So, unless ( k = 0 ), ( M ) and ( N ) are different but overlapping? Wait, no. If ( b ) is even, ( M ) and ( N ) are still translates, so they don't overlap unless ( b = 0 ). Wait, no. If ( b ) is even, say ( b = 2k ), then ( M = c + 4kmathbb{Z} ) and ( N = c + 2k + 4kmathbb{Z} ). So, ( N = M + 2k ). If ( 2k ) is a multiple of ( 4k ), which would require ( k = 0 ), then ( M = N ). Otherwise, ( M ) and ( N ) are disjoint.So, in the linear case, even if ( a + b ) is even, unless ( b = 0 ), ( M ) and ( N ) are disjoint. If ( b = 0 ), then ( M = N = {c} ).So, in the quadratic case, maybe a similar logic applies. If ( a + b ) is odd, ( M ) and ( N ) have different parities, so ( M cap N = emptyset ). If ( a + b ) is even, then it's possible for ( M ) and ( N ) to overlap, but we need to see if they must coincide entirely.Alternatively, maybe I can consider the function ( f(x) ) and see if it's symmetric in some way that would make ( M = N ).Wait, another idea: suppose that ( f(2n) = f(2m+1) ) for some ( n, m ). Then, as we saw earlier, this leads to a quadratic equation in ( n ) and ( m ). If this equation has solutions, then perhaps the entire sets ( M ) and ( N ) must coincide.Alternatively, maybe I can consider the function ( f(x) ) and see if it's periodic or something, but quadratic functions aren't periodic.Wait, perhaps I can think about the function ( f(x) ) evaluated at even and odd integers and see if they can be made equal by some transformation.Alternatively, maybe I can consider the function ( f(x) ) and see if it's symmetric about some point, making the images of even and odd integers coincide.Wait, let me think differently. Suppose that ( M = N ). That would mean that for every even integer ( 2n ), there exists an odd integer ( 2m+1 ) such that ( f(2n) = f(2m+1) ), and vice versa. So, the function ( f ) must map even and odd integers onto the same set.Alternatively, if ( M cap N = emptyset ), then no even integer maps to the same value as any odd integer.So, to prove that either ( M = N ) or ( M cap N = emptyset ), I need to show that if there's at least one common value, then all values must coincide.Suppose there exists some ( y ) such that ( y = f(2n) = f(2m+1) ) for some integers ( n, m ). Then, as before, we have:[4an^2 + 2bn + c = 4am^2 + (4a + 2b)m + (a + b + c)]Simplify:[4a(n^2 - m^2) + 2b(n - m) = a + b + 4am + 2bm]Factor:[4a(n - m)(n + m) + 2b(n - m) = a + b + 4am + 2bm]Let ( k = n - m ), so ( n = m + k ). Substitute back:[4a k (2m + k) + 2b k = a + b + 4a m + 2b m]Expand:[8a m k + 4a k^2 + 2b k = a + b + 4a m + 2b m]Rearrange:[8a m k + 4a k^2 + 2b k - 4a m - 2b m - a - b = 0]Factor ( m ):[m(8a k - 4a - 2b) + 4a k^2 + 2b k - a - b = 0]Let me denote:[C = 8a k - 4a - 2b][D = 4a k^2 + 2b k - a - b]So, ( C m + D = 0 implies m = -D / C ). For ( m ) to be integer, ( C ) must divide ( D ).But ( C ) and ( D ) are expressions in ( k ), which is ( n - m ). This seems recursive. Maybe I can find a relationship between ( a ) and ( b ) that makes this possible.Alternatively, perhaps I can consider specific cases. Let's suppose ( a = 1 ), ( b = 0 ), ( c = 0 ). Then ( f(x) = x^2 ). So, ( M = {4n^2} ) and ( N = {(2n+1)^2} ). Clearly, ( M ) and ( N ) are disjoint because squares of even and odd integers are distinct. So, ( M cap N = emptyset ).Another case: ( a = 1 ), ( b = -2 ), ( c = 1 ). Then ( f(x) = x^2 - 2x + 1 = (x - 1)^2 ). So, ( f(2n) = (2n - 1)^2 ) and ( f(2n+1) = (2n)^2 ). So, ( M = {(2n - 1)^2} ) and ( N = {4n^2} ). Again, ( M ) and ( N ) are disjoint.Wait, but in this case, ( f(2n) = (2n - 1)^2 ) and ( f(2n+1) = (2n)^2 ). So, ( M ) is the set of odd squares and ( N ) is the set of even squares. They are disjoint.Another case: ( a = 1 ), ( b = 2 ), ( c = 1 ). Then ( f(x) = x^2 + 2x + 1 = (x + 1)^2 ). So, ( f(2n) = (2n + 1)^2 ) and ( f(2n+1) = (2n + 2)^2 = 4(n + 1)^2 ). So, ( M = {(2n + 1)^2} ) and ( N = {4(n + 1)^2} ). Again, disjoint.Wait, but what if ( a = 0 ), ( b = 0 ), ( c = 0 ). Then ( f(x) = 0 ), so ( M = N = {0} ).Another case: ( a = 0 ), ( b = 1 ), ( c = 0 ). Then ( f(x) = x ). So, ( M = {2n} ) and ( N = {2n + 1} ). These are disjoint.Wait, but if ( a = 0 ), ( b = 2 ), ( c = 0 ). Then ( f(x) = 2x ). So, ( M = {4n} ) and ( N = {4n + 2} ). These are disjoint.But if ( a = 0 ), ( b = 0 ), ( c = 1 ). Then ( f(x) = 1 ). So, ( M = N = {1} ).So, in the linear case, ( M ) and ( N ) are either identical (if ( b = 0 )) or disjoint.In the quadratic case, from the examples above, ( M ) and ( N ) seem to be disjoint. But is that always the case?Wait, let me think of a quadratic function where ( M = N ). Is that possible?Suppose ( f(x) = x^2 - x ). Then, ( f(2n) = 4n^2 - 2n ) and ( f(2n+1) = (2n+1)^2 - (2n+1) = 4n^2 + 4n + 1 - 2n - 1 = 4n^2 + 2n ). So, ( M = {4n^2 - 2n} ) and ( N = {4n^2 + 2n} ). Are these sets equal?Let me see: For ( M ), ( 4n^2 - 2n ). For ( N ), ( 4n^2 + 2n ). If I replace ( n ) with ( -n ) in ( M ), I get ( 4n^2 + 2n ), which is exactly ( N ). So, ( M = N ) because for every ( m in M ), there's an ( n ) such that ( m = f(2n) = f(2(-n)+1) ). Wait, no, because ( f(2n) = 4n^2 - 2n ) and ( f(2(-n)+1) = 4(-n)^2 + 2(-n) = 4n^2 - 2n ). So, yes, ( M = N ).So, in this case, ( M = N ). Interesting. So, it's possible for ( M = N ) in the quadratic case.So, in this example, ( f(x) = x^2 - x ), ( M = N ). So, the conclusion is that either ( M = N ) or ( M cap N = emptyset ).Therefore, to generalize, we need to show that if there exists some overlap between ( M ) and ( N ), then ( M = N ). Otherwise, they are disjoint.So, going back to the earlier equation:[4a(n - m)(n + m) + 2b(n - m) = a + b + 4am + 2bm]Let me factor ( (n - m) ):[(n - m)(4a(n + m) + 2b) = a + b + 4am + 2bm]Let me denote ( k = n - m ), so ( n = m + k ). Substitute back:[k(4a(2m + k) + 2b) = a + b + 4am + 2bm]Expand:[8a m k + 4a k^2 + 2b k = a + b + 4a m + 2b m]Rearrange:[8a m k + 4a k^2 + 2b k - 4a m - 2b m - a - b = 0]Factor ( m ):[m(8a k - 4a - 2b) + 4a k^2 + 2b k - a - b = 0]Let me denote:[C = 8a k - 4a - 2b][D = 4a k^2 + 2b k - a - b]So, ( C m + D = 0 implies m = -D / C ). For ( m ) to be integer, ( C ) must divide ( D ).But ( C ) and ( D ) are expressions in ( k ), which is ( n - m ). This seems recursive. Maybe I can find a relationship between ( a ) and ( b ) that makes this possible.Alternatively, perhaps I can consider the function ( f(x) ) and see if it's symmetric in some way that would make ( M = N ).Wait, in the example where ( M = N ), ( f(x) = x^2 - x ), which can be written as ( f(x) = (x - 0.5)^2 - 0.25 ). So, the vertex is at ( x = 0.5 ), which is halfway between 0 and 1. Maybe this symmetry causes ( M = N ).Indeed, if the vertex is at a half-integer, then ( f(2n) ) and ( f(2n+1) ) might be symmetric around that vertex, leading to ( M = N ).So, perhaps if the vertex is at a half-integer, ( M = N ). Otherwise, ( M ) and ( N ) are disjoint.The vertex of ( f(x) = ax^2 + bx + c ) is at ( x = -b/(2a) ). For this to be a half-integer, ( -b/(2a) = k + 0.5 ) for some integer ( k ). So, ( b = -2a(k + 0.5) ).So, if ( b = -2a(k + 0.5) ) for some integer ( k ), then the vertex is at a half-integer, and ( M = N ). Otherwise, ( M cap N = emptyset ).Therefore, in general, either the vertex is at a half-integer, leading to ( M = N ), or it's not, leading to ( M cap N = emptyset ).So, to formalize this, suppose that ( -b/(2a) = t + 0.5 ) for some integer ( t ). Then, ( b = -2a(t + 0.5) ). Substituting back into ( f(x) ):[f(x) = a x^2 - 2a(t + 0.5) x + c]Complete the square:[f(x) = a left( x^2 - 2(t + 0.5)x right) + c = a left( (x - (t + 0.5))^2 - (t + 0.5)^2 right) + c = a(x - t - 0.5)^2 - a(t + 0.5)^2 + c]So, the function is symmetric around ( x = t + 0.5 ). Therefore, for any integer ( n ), ( f(2n) = f(2(t - n) + 1) ). Hence, ( M = N ).Otherwise, if ( -b/(2a) ) is not a half-integer, then ( M ) and ( N ) are disjoint.Therefore, we've shown that either ( M = N ) or ( M cap N = emptyset ).

💡First, I need to determine the total amount of paper Zoe and her friends recycled to earn 6 points. Since they earn 1 point for every 8 pounds recycled, I can calculate the total pounds by multiplying the number of points by 8.Next, I know that Zoe recycled 25 pounds of paper. To find out how many pounds her friends recycled, I subtract the amount Zoe recycled from the total amount recycled.Finally, by performing this subtraction, I will obtain the number of pounds recycled by Zoe's friends.

💡Alright, so I have this problem about random variables and exponential distributions. Let me try to wrap my head around it step by step. The problem states:"Let ξ be a non-negative random variable with a distribution function F = F(x) such that F(0) < 1 and the right-hand derivative F₊' (0) = λ exists. Show thatthe random variable ξ is exponential ⇨ ξ ∧ η = ξ / 2,where η is an independent copy of ξ."Okay, so I need to show an equivalence: ξ is exponential if and only if the minimum of ξ and an independent copy η is equal in distribution to ξ divided by 2.First, let's parse the notation and terms:- ξ is a non-negative random variable. So, it takes values in [0, ∞).- F(x) is its distribution function, meaning F(x) = P(ξ ≤ x).- F(0) < 1: This tells me that there's a positive probability that ξ is greater than 0. So, P(ξ > 0) = 1 - F(0) > 0.- The right-hand derivative F₊' (0) exists and is equal to λ. That means the slope of the distribution function at 0 from the right is λ.- ξ ∧ η is the minimum of ξ and η, where η is an independent copy of ξ. So, η has the same distribution as ξ, and ξ and η are independent.- The statement to prove is that ξ is exponential if and only if ξ ∧ η is equal in distribution to ξ / 2.So, I need to show both directions:1. If ξ is exponential, then ξ ∧ η = ξ / 2 in distribution.2. If ξ ∧ η = ξ / 2 in distribution, then ξ is exponential.Let me tackle each direction one by one.**First Direction: If ξ is exponential, then ξ ∧ η = ξ / 2 in distribution.**Assume ξ is exponential with parameter λ, so its distribution function is F(x) = 1 - e^{-λx} for x ≥ 0.Since η is an independent copy of ξ, it also has the same exponential distribution.Now, let's find the distribution of ξ ∧ η.For any x ≥ 0, P(ξ ∧ η > x) = P(ξ > x, η > x) = P(ξ > x)P(η > x) because ξ and η are independent.Since both ξ and η are exponential with parameter λ, P(ξ > x) = e^{-λx}, so:P(ξ ∧ η > x) = e^{-λx} * e^{-λx} = e^{-2λx}.But e^{-2λx} is the survival function of an exponential random variable with parameter 2λ. So, ξ ∧ η is exponential with parameter 2λ.On the other hand, ξ / 2 would have a distribution function F(x/2) = 1 - e^{-λ(x/2)} = 1 - e^{-λx/2}.Wait, that doesn't seem to match with the survival function of ξ ∧ η, which is e^{-2λx}.Hmm, that suggests that ξ / 2 is exponential with parameter λ/2, while ξ ∧ η is exponential with parameter 2λ. So, they are not the same unless λ = 0, which isn't possible because F(0) < 1 implies λ > 0.Wait, maybe I made a mistake here. Let me think again.If ξ is exponential with parameter λ, then ξ / 2 would have a distribution function F(x) = P(ξ / 2 ≤ x) = P(ξ ≤ 2x) = 1 - e^{-λ(2x)} = 1 - e^{-2λx}.Ah, okay, so ξ / 2 has the same distribution as ξ ∧ η, which is exponential with parameter 2λ. So, their survival functions are both e^{-2λx}, so they are indeed equal in distribution.So, that takes care of the first direction: if ξ is exponential, then ξ ∧ η = ξ / 2 in distribution.**Second Direction: If ξ ∧ η = ξ / 2 in distribution, then ξ is exponential.**This is the converse, and it's probably more involved. I need to show that if the minimum of ξ and an independent copy η is equal in distribution to ξ / 2, then ξ must be exponential.Let me denote G(x) = P(ξ ∧ η > x). Since ξ ∧ η = ξ / 2 in distribution, we have:G(x) = P(ξ ∧ η > x) = P(ξ / 2 > x) = P(ξ > 2x).But also, since ξ and η are independent, G(x) = P(ξ > x)P(η > x) = [P(ξ > x)]^2.So, we have:[P(ξ > x)]^2 = P(ξ > 2x).Let me denote S(x) = P(ξ > x), the survival function of ξ.So, the equation becomes:[S(x)]^2 = S(2x).This is a functional equation for S(x). I need to solve this equation under the given conditions: F(0) < 1 and F₊' (0) = λ exists.First, let's note that S(0) = P(ξ > 0) = 1 - F(0) > 0, which is given.Also, since F is the distribution function, it is right-continuous. The right-hand derivative at 0 exists and is equal to λ, so F₊' (0) = λ.But F(x) = 1 - S(x), so S(x) = 1 - F(x).Therefore, S₊' (0) = -F₊' (0) = -λ.So, the derivative of S at 0 from the right is -λ.Now, let's try to solve the functional equation [S(x)]^2 = S(2x).This kind of equation often suggests that S(x) is an exponential function. Let me test that.Suppose S(x) = e^{-kx} for some k > 0.Then, [S(x)]^2 = e^{-2kx}, and S(2x) = e^{-2kx}.So, indeed, [S(x)]^2 = S(2x).Therefore, S(x) = e^{-kx} is a solution.But are there other solutions?Suppose S(x) is not exponential. Let's see.Assume S(x) is twice differentiable (which we might be able to argue based on the existence of the right-hand derivative at 0). Then, we can take logarithms on both sides of [S(x)]^2 = S(2x):2 ln S(x) = ln S(2x).Differentiate both sides with respect to x:2 (S'(x)/S(x)) = (S'(2x)/S(2x)) * 2.Simplify:2 (S'(x)/S(x)) = 2 (S'(2x)/S(2x)).Cancel the 2's:S'(x)/S(x) = S'(2x)/S(2x).But from the functional equation, S(2x) = [S(x)]^2, so S'(2x) = 2 [S(x)]^2 S'(x) by the chain rule.Wait, let's compute S'(2x):d/dx S(2x) = 2 S'(2x).But from the functional equation, S(2x) = [S(x)]^2, so differentiating both sides:d/dx S(2x) = 2 S(x) S'(x).Therefore,2 S'(2x) = 2 S(x) S'(x).Divide both sides by 2:S'(2x) = S(x) S'(x).So, we have:S'(x)/S(x) = S'(2x)/S(2x).But S'(2x) = S(x) S'(x), so:S'(x)/S(x) = [S(x) S'(x)] / [S(2x)].But S(2x) = [S(x)]^2, so:S'(x)/S(x) = [S(x) S'(x)] / [S(x)]^2 = S'(x)/S(x).So, this is consistent, but it doesn't give us new information.Alternatively, let's consider taking the logarithm:ln S(2x) = 2 ln S(x).Let me define u(x) = ln S(x). Then, the equation becomes:u(2x) = 2 u(x).This is a well-known functional equation, and its solutions are u(x) = k x, leading to S(x) = e^{k x}.But since S(x) is a survival function, it must be decreasing, so k must be negative. Let me write k = -λ, so S(x) = e^{-λ x}.Therefore, the only solution is S(x) = e^{-λ x}, which is the survival function of an exponential distribution with parameter λ.Therefore, ξ must be exponential.Wait, but I need to ensure that this is the only solution. Are there any other functions besides exponential that satisfy [S(x)]^2 = S(2x)?Suppose S(x) is a power function, say S(x) = (1 + a x)^{-b}. Let's test if this satisfies [S(x)]^2 = S(2x).Then, [S(x)]^2 = (1 + a x)^{-2b}, and S(2x) = (1 + 2 a x)^{-b}.These are equal only if (1 + a x)^{-2b} = (1 + 2 a x)^{-b}.Taking both sides to the power of 1/b:(1 + a x)^{-2} = (1 + 2 a x)^{-1}.Cross-multiplying:(1 + 2 a x) = (1 + a x)^2.Expanding the right side:1 + 2 a x = 1 + 2 a x + a^2 x^2.Subtracting 1 + 2 a x from both sides:0 = a^2 x^2.This must hold for all x ≥ 0, which implies a = 0. But then S(x) = 1, which contradicts F(0) < 1. So, power functions other than the exponential do not satisfy the equation.Therefore, the only solution is the exponential function.Additionally, we can use the given condition that F₊' (0) = λ exists. For the exponential distribution, F(x) = 1 - e^{-λ x}, so F₊' (0) = λ, which matches the given condition.Thus, we've shown both directions:1. If ξ is exponential, then ξ ∧ η = ξ / 2 in distribution.2. If ξ ∧ η = ξ / 2 in distribution, then ξ must be exponential.Therefore, the equivalence holds.

💡Okay, so I have this problem where I need to find all functions ( f: mathbb{Z}^* rightarrow mathbb{N}_0 ) that satisfy two conditions. Let me try to understand each condition and see how I can approach this.First, let me restate the problem to make sure I get it. ( mathbb{Z}^* ) is the set of all nonzero integers, and ( mathbb{N}_0 ) is the set of all nonnegative integers. So, ( f ) takes any nonzero integer and maps it to a nonnegative integer.Condition (1) says that for any two nonzero integers ( a ) and ( b ) such that their sum ( a + b ) is also a nonzero integer, the function value at ( a + b ) is at least the minimum of the function values at ( a ) and ( b ). So, ( f(a + b) geq min{f(a), f(b)} ).Condition (2) is about multiplicativity. It states that for any two nonzero integers ( a ) and ( b ), the function value at their product ( ab ) is equal to the sum of the function values at ( a ) and ( b ). So, ( f(ab) = f(a) + f(b) ).Hmm, okay. So, condition (2) reminds me of logarithms, which turn multiplication into addition. But since we're dealing with integers and nonnegative integers, it's more like a valuation or something related to prime factorization.Let me think about condition (2) first because it seems more restrictive. If ( f(ab) = f(a) + f(b) ), then ( f ) is a homomorphism from the multiplicative monoid of nonzero integers to the additive monoid of nonnegative integers. That makes me think of functions like the number of prime factors, or more specifically, the exponent of a particular prime in the prime factorization of a number.For example, if I fix a prime ( p ), then for any integer ( n ), ( v_p(n) ) is the exponent of ( p ) in the prime factorization of ( n ). This function satisfies ( v_p(ab) = v_p(a) + v_p(b) ), which is exactly condition (2). So, ( v_p ) is a function that satisfies condition (2). But does it satisfy condition (1)?Let me check. Suppose ( f(n) = v_p(n) ). Then, for any ( a ) and ( b ), ( f(a + b) geq min{f(a), f(b)} ). Is this true?Yes, this is a well-known property of valuations. Specifically, in the context of p-adic valuations, it's called the non-Archimedean property. It states that ( v_p(a + b) geq min{v_p(a), v_p(b)} ), which is exactly condition (1). So, ( v_p(n) ) satisfies both conditions.But the problem says "find all functions," so I need to check if these are the only possible functions or if there are others.Wait, maybe we can scale ( v_p(n) ) by some constant. Suppose ( f(n) = k cdot v_p(n) ) for some nonnegative integer ( k ). Then, ( f(ab) = k cdot v_p(ab) = k(v_p(a) + v_p(b)) = f(a) + f(b) ), so condition (2) is satisfied. What about condition (1)?( f(a + b) = k cdot v_p(a + b) geq k cdot min{v_p(a), v_p(b)} = min{k cdot v_p(a), k cdot v_p(b)} = min{f(a), f(b)} ). So, yes, condition (1) is also satisfied.So, scaling by a nonnegative integer ( k ) still gives a function that satisfies both conditions.But are there any other functions besides these scaled p-adic valuations?Let me think. Suppose ( f ) is not of the form ( k cdot v_p(n) ). Maybe it's a combination of different valuations? For example, ( f(n) = v_p(n) + v_q(n) ) for two different primes ( p ) and ( q ). Does this satisfy condition (2)?Let's check. ( f(ab) = v_p(ab) + v_q(ab) = v_p(a) + v_p(b) + v_q(a) + v_q(b) = (v_p(a) + v_q(a)) + (v_p(b) + v_q(b)) = f(a) + f(b) ). So, condition (2) is satisfied.But does it satisfy condition (1)? Let's see. Suppose ( a = p ) and ( b = q ). Then ( a + b = p + q ). What is ( f(a + b) )? It's ( v_p(p + q) + v_q(p + q) ).But ( p + q ) is not divisible by ( p ) or ( q ) unless ( p = q ), which they aren't since they are distinct primes. So, ( v_p(p + q) = 0 ) and ( v_q(p + q) = 0 ). Therefore, ( f(a + b) = 0 ).On the other hand, ( min{f(a), f(b)} = min{1, 1} = 1 ). So, ( f(a + b) = 0 ) which is not greater than or equal to 1. This violates condition (1). So, such a function ( f(n) = v_p(n) + v_q(n) ) doesn't satisfy condition (1).Therefore, combining different valuations doesn't work. So, maybe the only functions are scalar multiples of a single p-adic valuation.Wait, but what if ( f ) is identically zero? That is, ( f(n) = 0 ) for all ( n ). Does that satisfy both conditions?Condition (2): ( f(ab) = 0 = 0 + 0 = f(a) + f(b) ). So, yes.Condition (1): ( f(a + b) = 0 geq min{0, 0} = 0 ). So, yes.So, the zero function is also a solution. But in the earlier case, if ( k = 0 ), then ( f(n) = 0 ) for all ( n ), which is covered by the scalar multiple case.So, the zero function is just a special case when ( k = 0 ).Are there any other functions besides these?Suppose ( f ) is not identically zero and not a multiple of a p-adic valuation. Let me see if such a function can exist.Suppose ( f ) is non-zero and not of the form ( k cdot v_p(n) ). Then, there must be at least two different primes ( p ) and ( q ) such that ( f(p) ) and ( f(q) ) are both positive.But earlier, I saw that if ( f(p) ) and ( f(q) ) are both positive, then taking ( a = p ) and ( b = q ), ( a + b = p + q ), and ( f(a + b) ) would have to be at least the minimum of ( f(p) ) and ( f(q) ), which is positive. But ( f(a + b) = f(p + q) ). If ( p + q ) is not divisible by ( p ) or ( q ), then ( f(p + q) ) would have to be zero, which contradicts the requirement that it's at least the minimum of ( f(p) ) and ( f(q) ), which is positive.Wait, but ( p + q ) might be divisible by another prime, say ( r ). So, ( f(p + q) ) would be ( f(r) ) times something? Hmm, no, because ( f ) is defined on all nonzero integers, not just primes.Wait, actually, ( f ) is defined for all nonzero integers, so ( f(p + q) ) is just some nonnegative integer. But if ( p + q ) is not divisible by ( p ) or ( q ), then ( v_p(p + q) = 0 ) and ( v_q(p + q) = 0 ). So, if ( f ) is a combination of valuations, but as I saw earlier, that doesn't work.Alternatively, maybe ( f ) is a constant function? Let's say ( f(n) = c ) for some constant ( c ).Then, condition (2): ( f(ab) = c = c + c = 2c ). So, ( c = 0 ). So, the only constant function is the zero function, which we've already considered.So, constant functions other than zero don't work.What about functions that are zero except at certain points? For example, suppose ( f(n) = 0 ) unless ( n ) is a power of 2, in which case ( f(n) = k cdot log_2 |n| ). But wait, that might not satisfy condition (2).Wait, let me test it. Suppose ( f(n) = k cdot v_2(n) ). Then, ( f(ab) = k cdot v_2(ab) = k(v_2(a) + v_2(b)) = f(a) + f(b) ). So, that works. And condition (1) is satisfied because of the non-Archimedean property.But if I have a function that's zero except at primes other than 2, say, then it might not satisfy condition (1). For example, if ( f(n) = v_3(n) ), then similar reasoning applies.Wait, but if I mix valuations, as I tried earlier, it doesn't work because of the contradiction with condition (1). So, it seems that the only functions that satisfy both conditions are the scalar multiples of a single p-adic valuation, including the zero function.Let me try to formalize this.Suppose ( f ) is a function satisfying both conditions. Then, for any prime ( p ), ( f(p) ) is some nonnegative integer. If ( f(p) = 0 ) for all primes ( p ), then ( f(n) = 0 ) for all ( n ), since any integer can be factored into primes, and ( f(n) ) would be the sum of ( f ) over the primes, each of which is zero.If ( f(p) ) is positive for some prime ( p ), then let me set ( k = f(p) ). Then, for any integer ( n ), ( f(n) = k cdot v_p(n) ). Because, for any integer ( n ), ( n ) can be written as ( p^{v_p(n)} cdot m ), where ( m ) is coprime to ( p ). Then, ( f(n) = f(p^{v_p(n)}) + f(m) ). But since ( m ) is coprime to ( p ), ( f(m) ) must be zero, otherwise, if ( f(m) ) is positive, then similar to earlier reasoning, we can find a contradiction with condition (1).Wait, let me elaborate. Suppose ( m ) is coprime to ( p ) and ( f(m) > 0 ). Then, take ( a = m ) and ( b = -m ). Then, ( a + b = 0 ), which is not in ( mathbb{Z}^* ), so condition (1) doesn't apply here. Hmm, maybe that's not the right approach.Alternatively, take ( a = m ) and ( b = 1 ). Then, ( a + b = m + 1 ). Then, ( f(m + 1) geq min{f(m), f(1)} ). But ( f(1) ) can be determined.Wait, let's compute ( f(1) ). Using condition (2), ( f(1 cdot 1) = f(1) + f(1) ), so ( f(1) = 2f(1) ), which implies ( f(1) = 0 ).Similarly, ( f(-1) ) can be found by ( f((-1)(-1)) = f(1) = 0 = f(-1) + f(-1) ), so ( f(-1) = 0 ).Therefore, ( f(-n) = f(n) ) for all ( n ), because ( f(-n) = f(-1 cdot n) = f(-1) + f(n) = 0 + f(n) = f(n) ).So, ( f ) is even, meaning it's the same for ( n ) and ( -n ).Therefore, we can restrict our attention to positive integers.Now, for any integer ( n ), ( f(n) = f(p^{v_p(n)}) + f(m) ), where ( m ) is coprime to ( p ). If ( f(m) > 0 ), then let me see if that leads to a contradiction.Take ( a = m ) and ( b = p ). Then, ( a + b = m + p ). Then, ( f(a + b) geq min{f(a), f(b)} = min{f(m), k} ).But ( f(a + b) = f(m + p) ). Now, ( m + p ) is congruent to ( m ) modulo ( p ), since ( p ) divides ( p ). So, ( m + p equiv m mod p ). If ( m ) is coprime to ( p ), then ( m + p ) is also coprime to ( p ), so ( v_p(m + p) = 0 ), hence ( f(m + p) = 0 ) if ( f ) is a multiple of ( v_p ). But wait, if ( f ) is not a multiple of ( v_p ), then ( f(m + p) ) could be something else.Wait, no. If ( f ) is not a multiple of ( v_p ), then ( f(m) ) could be positive even though ( m ) is coprime to ( p ). But then, ( f(m + p) geq min{f(m), f(p)} = min{f(m), k} ). However, ( m + p ) is congruent to ( m ) modulo ( p ), so if ( m ) is not divisible by ( p ), then ( m + p ) is also not divisible by ( p ). Therefore, if ( f ) is a multiple of ( v_p ), ( f(m + p) = 0 ). But if ( f ) is not a multiple of ( v_p ), then ( f(m + p) ) could be positive.Wait, but if ( f(m) ) is positive and ( f(m + p) geq min{f(m), k} ), which is positive, but ( m + p ) is not divisible by ( p ), so if ( f ) is a multiple of ( v_p ), ( f(m + p) = 0 ), which would contradict the requirement that it's at least positive. Therefore, if ( f(m) > 0 ), then ( f(m + p) ) must be at least ( min{f(m), k} ), but since ( m + p ) is not divisible by ( p ), if ( f ) is a multiple of ( v_p ), it would be zero, which is a contradiction. Therefore, ( f(m) ) must be zero.Thus, ( f(n) = k cdot v_p(n) ) for all ( n ).Therefore, the only functions that satisfy both conditions are the zero function and functions of the form ( f(n) = k cdot v_p(n) ) for some prime ( p ) and nonnegative integer ( k ).Wait, but ( k ) is a nonnegative integer, right? Because ( f ) maps to ( mathbb{N}_0 ), so ( k ) must be such that ( k cdot v_p(n) ) is a nonnegative integer for all ( n ). Since ( v_p(n) ) is a nonnegative integer, ( k ) must also be a nonnegative integer.So, putting it all together, the functions are either identically zero or scalar multiples of a p-adic valuation for some prime ( p ).Let me just check if there are any edge cases or exceptions. For example, what if ( k = 0 )? Then, ( f(n) = 0 ) for all ( n ), which we've already considered. What if ( k = 1 )? Then, ( f(n) = v_p(n) ), which we know satisfies both conditions. For ( k > 1 ), ( f(n) = k cdot v_p(n) ) still satisfies both conditions because scaling by a positive integer preserves both the additive property and the non-Archimedean property.Therefore, I think I've covered all possible cases, and the only functions that satisfy both conditions are the zero function and functions of the form ( f(n) = k cdot v_p(n) ) for some prime ( p ) and nonnegative integer ( k ).**Final Answer**The functions satisfying the given conditions are all functions of the form ( f(n) = k cdot v_p(n) ) for some prime ( p ) and nonnegative integer ( k ). Thus, the solution is (boxed{f(n) = k cdot v_p(n)}).

💡Alright, so I have this problem about integer-coefficient polynomials and cycles of integers. Let me try to wrap my head around it. The problem says that for any integer-coefficient polynomial ( p(x) ), there don't exist distinct integers ( x_1, x_2, ldots, x_n ) (where ( n geq 3 )) such that ( p(x_1) = x_2 ), ( p(x_2) = x_3 ), and so on, until ( p(x_n) = x_1 ). So, basically, it's saying there's no cycle of length 3 or more where each integer maps to the next via the polynomial, and the last maps back to the first.Hmm, okay. So, first, I need to understand what this means. If such a cycle existed, then applying the polynomial repeatedly would cycle through these integers. But the claim is that this can't happen for any integer-coefficient polynomial with distinct integers in the cycle.Let me think about how polynomials behave. Since ( p(x) ) has integer coefficients, if I plug in an integer ( x ), I should get an integer ( p(x) ). That makes sense because the coefficients are integers, and integer operations preserve integrality.Now, the key here might be the properties of polynomials with integer coefficients. I remember that for any two integers ( a ) and ( b ), the difference ( p(a) - p(b) ) is divisible by ( a - b ). That is, ( a - b ) divides ( p(a) - p(b) ). This is because of the factor theorem and the fact that polynomials with integer coefficients have this divisibility property.So, if ( a neq b ), then ( |a - b| leq |p(a) - p(b)| ). This is because the absolute value of the difference in outputs is at least as big as the absolute value of the difference in inputs. That seems important.Let me try to apply this to the problem. Suppose, for contradiction, that such a cycle exists. So, we have distinct integers ( x_1, x_2, ldots, x_n ) with ( n geq 3 ) such that ( p(x_i) = x_{i+1} ) for ( i = 1, 2, ldots, n-1 ) and ( p(x_n) = x_1 ).Now, using the property I just recalled, for each ( i ), ( |x_i - x_{i+1}| leq |p(x_i) - p(x_{i+1})| ). But ( p(x_i) = x_{i+1} ) and ( p(x_{i+1}) = x_{i+2} ), so ( |x_i - x_{i+1}| leq |x_{i+1} - x_{i+2}| ).Wait, that's interesting. So, each difference is less than or equal to the next difference. But since this is a cycle, the last difference ( |x_n - x_1| ) must also be less than or equal to ( |x_1 - x_2| ). So, putting it all together, we have:[|x_1 - x_2| leq |x_2 - x_3| leq ldots leq |x_n - x_1| leq |x_1 - x_2|]This implies that all these absolute differences are equal. So, ( |x_1 - x_2| = |x_2 - x_3| = ldots = |x_n - x_1| ).Okay, so all the differences between consecutive terms in the cycle are the same in absolute value. That suggests that the integers ( x_1, x_2, ldots, x_n ) are equally spaced around a circle, but since we're dealing with integers on a number line, this might impose some restrictions.But wait, integers are on a straight line, not a circle. So, if we have equally spaced points on a line, they must form an arithmetic progression. However, in a cycle, the last term connects back to the first, which would mean that the progression wraps around, which isn't possible on a straight line unless all the terms are the same, which contradicts the distinctness.Hmm, maybe I should think about the signs of the differences. If all the differences have the same absolute value, they could either all be positive or alternate in sign. But if they alternate in sign, that would mean that the sequence goes up and down, but since it's a cycle, this could lead to contradictions.Let me consider the case where all differences are positive. Then, ( x_1 < x_2 < x_3 < ldots < x_n < x_1 ), which is impossible because ( x_n < x_1 ) contradicts ( x_1 < x_2 < ldots < x_n ).Alternatively, if the differences alternate in sign, say ( x_1 > x_2 < x_3 > x_4 < ldots ), but since all differences have the same absolute value, this would imply that ( x_1 = x_3 = x_5 = ldots ) and ( x_2 = x_4 = x_6 = ldots ), which again contradicts the distinctness of the integers.Wait, maybe I should formalize this. Suppose that for some ( i ), ( x_{i-1} - x_i ) and ( x_i - x_{i+1} ) have opposite signs. Then, ( x_{i-1} - x_i = - (x_i - x_{i+1}) ), which implies ( x_{i-1} = x_{i+1} ). But this contradicts the distinctness of the ( x_i )'s.Therefore, all the differences must have the same sign. So, either all ( x_i ) are increasing or all are decreasing. But if they're increasing, ( x_n > x_1 ), which contradicts ( p(x_n) = x_1 ). Similarly, if they're decreasing, ( x_n < x_1 ), which also contradicts ( p(x_n) = x_1 ).Wait, no, actually, if all differences are positive, then ( x_1 < x_2 < ldots < x_n ), but ( p(x_n) = x_1 ), which would mean ( x_1 ) is less than ( x_n ), but ( p(x_n) = x_1 ) would require that ( x_1 ) is somehow mapped back, which might not necessarily be a contradiction, but considering the polynomial's behavior, maybe it is.Alternatively, if all differences are negative, then ( x_1 > x_2 > ldots > x_n ), but then ( p(x_n) = x_1 ) would require ( x_1 > x_n ), which is consistent, but again, considering the polynomial's behavior, maybe it's not possible.Wait, perhaps I should consider the polynomial's degree. If the polynomial is linear, say ( p(x) = ax + b ), then the behavior is straightforward. Let's see if a linear polynomial can have such a cycle.Suppose ( p(x) = ax + b ). Then, ( p(x_i) = a x_i + b = x_{i+1} ). So, we have ( x_{i+1} = a x_i + b ). For a cycle of length ( n ), we'd have ( x_1 = a x_n + b ). So, substituting recursively, we get ( x_1 = a^n x_1 + b frac{a^n - 1}{a - 1} ). For this to hold, unless ( a = 1 ), which would make the polynomial ( p(x) = x + b ), leading to ( x_{i+1} = x_i + b ). Then, a cycle would require ( x_1 = x_1 + n b ), implying ( b = 0 ). But then ( p(x) = x ), which trivially maps each ( x_i ) to itself, contradicting the distinctness.If ( a neq 1 ), then solving ( x_1 = a^n x_1 + b frac{a^n - 1}{a - 1} ) would require ( x_1 (1 - a^n) = b frac{a^n - 1}{a - 1} ). This can only hold for specific ( x_1 ) and ( b ), but since we're dealing with integers, it's restrictive. Moreover, for ( n geq 3 ), it's unlikely to have such a cycle with distinct integers.But the problem is about any integer-coefficient polynomial, not just linear ones. So, maybe the argument needs to be more general.Going back to the earlier point, since all differences ( |x_i - x_{i+1}| ) are equal, say to some integer ( d ), and the signs must be consistent to avoid contradictions, but as we saw, consistent signs lead to a contradiction because the cycle can't close on a straight line.Therefore, such a cycle cannot exist, which proves the statement.Wait, but I should make sure I'm not missing something. Maybe there's a way to have a cycle with more than three points where the differences alternate in sign but still maintain distinctness. But earlier, I saw that if the differences alternate, it would force some ( x_i )'s to be equal, which contradicts distinctness.So, in summary, assuming such a cycle exists leads to a contradiction because the differences between consecutive terms must all be equal in absolute value and same in sign, which is impossible on a straight line for a cycle of length ( n geq 3 ) with distinct integers.I think that's the gist of it. So, the conclusion is that no such cycle exists for any integer-coefficient polynomial.