April wrote: Let $ n$ be a positive integer and let $ p$ be a prime number. Prove that if $ a$, $ b$, $ c$ are integers (not necessarily positive) satisfying the equations \[ a^n + pb = b^n + pc = c^n + pa\] then $ a = b = c$. Let $ a = max (a;b;c)$. Case I: $ a > c > b$. First $ if p=2$. If $ a$ or $ b$ or $ c$ is equal to $ 0$ . It is trivial to see $ a=b=c=0$. Now $ |a|;|b|;|c|\ge 1$. We have $ a^n - b^n = (a - b)(a^{n - 1} + .. + b^{n - 1}) = p(c - b)$. Because $ a-b>c-b>0$ thus $ 2>a^{n-1}+..+b^{n-1}$. So that $ n=1$ or $ n=2$. Both cases imply $ a=b=c$. If $ p>3$. It is obvious to see that if $ x = y$ ($ x;y$ are chosen in $ (a;b;c)$ ) then $ a = b = c$. Thus $ a^n - b^n = (a - b)(a^{n - 1} + .. + b^{n - 1}) = p(c - b)$;$ c^n - b^n = p(a - c) = (c - b)(c^{n - 1} + .. + b^{n - 1})$;$ a^n - c^n = (a - c)(a^{n - 1} + .. + c^{n - 1}) = p(a - b)$. It means $ (a^{n - 1} + .. + b^{n - 1})(c^{n - 1} + .. + b^{n - 1})(a^{n - 1} + .. + c^{n - 1}) = p^3 = ABC$ where $ C = (a^{n - 1} + .. + b^{n - 1})$ .etc. If there exists one of $ A;B;C$ be equal to 1, Assume that It is C (The same agruments will work when it is either B or A). Thus $ a^n - b^n = a - b$. If $ B$ or C is also equal to $ 1$.It is very easy to see that $ a = b = c$ contradiction. So that $ A = p$; $ B = p^2$ or $ B = p$ ;$ A = p^2$. If $ A = p$; $ B = p^2$ henxe $ c^n - b^n = p(c - b) = p(a - c)$ and $ a^n - c^n = p^2(a - c) = p(a - b)$. The first equation implies $ 2c = a + b$, the second imlpies $ p(a - c) = a - b$ which is equivalent to $ (p - 2)a = (p - 2)b$. Then $ a = b = c$ contradiction. If $ B = p$ ;$ A = p^2$ thus $ a^n - c^n = P(a - c) = p(a - b)$ which implies $ a = b = c$. In conclusion, None of $ A;B;C$ is equal to 1. It means $ A = B = C = p$. Now it shows immediately $ a = b = c$. Case II: $ a > b > c$. The proof is absolutely similar to the above!. We have done! Sorry for my poor English. Warm me if I make any silly mistakes (Grammar or Maths)

Let $ P(a)=a^n-pa$. Note: $ P(a)$ is a polynomial of $ a$, and $ p(a)$ is $ p\cdot a$. Assume that not all of them are equal, and I'll leave out trivial cases (one is equal to zero, etc.). We have that $ |a-b| \mid |P(a)-P(b)|$ for all $ a,b$. Without loss of generality, let $ a\ge b\ge c$ (the case $ a\ge c\ge b$ is similar). $ P(a)=c^n-pb=c^n-pc+p(c-b)=P(c)+p(c-b)$. $ a-c\mid p(b-c)$, similar we get $ a-b\mid p(a-c)$ and $ b-c\mid p(a-b)$. Let $ a-b=x,b-c=y,a-c=x+y,gcd(x,y)=d,x=rd,y=sd$. $ r,s$ are positive integers. $ x+y\mid py,x\mid p(x+y),y\mid px$. $ r+s\mid ps,r\mid pr+ps,s\mid pr$. As $ gcd(r,s)=1$, we get $ r\mid p,s\mid p$. But, as $ p$ is a prime, we must have $ r=1$ or $ s=1$. Let $ r=1$. $ 1+s\mid ps$ so $ 1+s\mid p$, also $ s\mid p$. Again we conclude $ s=1$, and $ p=2$. We also conclude $ a-b=b-c$, or $ 2b=a+c$ $ a^n+2a+2c=b^n+2c=c^n+2a$. $ (a-b)(a^{n-1}+\ldots+b^{n-1})=2(c-b)$, or $ a^{n-1}+\ldots+b^{n-1}=-2$. It's easy to see that we get a contradiction, and so $ a=b=c$.

I've solved a generalisation of this problem, but I think there might be some mistake, please check my solution: Instead of prime number, let $ p$ be a positive odd number (well, we miss the case $ p=2$, so it's not really a generalisation ). Note that $ \frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{a^n-c^n}{a-b}=p$. So $ \frac{(a^n-b^n)(b^n-c^n)(a^n-c^n)}{(a-b)(b-c)(c-a)}=p^3$. If $ n$ is odd, then $ a^n-b^n,b^n-c^n,c^n-a^n$ have the same signs with $ a-b,b-c,c-a$ respectively, which implies that $ p$ is negative, a contradiction. Therefore $ n$ is even. Note that at least two of $ a,b,c$ have the same parity, wlog assume $ a\equiv b\pmod2$. We have $ \frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1}$. If they are both even, clearly this expression is even. If they are both odd, the expression equals the sum of $ n$ odd integers, and hence is even. So $ \frac{a^n-b^n}{a-b}$ is even. But $ \frac{b^n-c^n}{b-c}$ and $ \frac{a^n-c^n}{c-a}$ are integers, so $ \frac{(a^n-b^n)(b^n-c^n)(a^n-c^n)}{(a-b)(b-c)(c-a)}$ is an even integer, contrary to the fact that $ p$ is odd. QED

I think you are right. I have proved the same (for all $ p\equiv 1 \mod 2$) this way: Let $ P(x) = a^n + pb - x^n$ be a polynomial and note that $ P$ has integer coefficients. Now, $ P(a) = pb$, $ P(b) = pc$ and $ P(c) = pa$. Assume wlog $ a \equiv b \mod 2$. Now $ P(a) \equiv P(b) \mod 2$, so $ pb \equiv pc \mod 2$, $ b \equiv c \mod 2$ since $ p$ is odd. So $ a \equiv b \equiv c \mod 2$. We can now prove by induction on $ m$ that $ a \equiv b \equiv c \mod 2^m$ for $ m \in \mathbb{N}$. Now choose $ m$ such that $ a,b,c \in ( - 2^m,2^m)$. wlog $ a = b$, so $ P(a) = P(b)$, so $ b = c$, thus $ a = b = c$ as desired. So $ p$ isn't even required to be positive

x164 wrote: Let $ P(x) = a^n + pb - x^n$ be a polynomial and note that $ P$ has integer coefficients. Now, $ P(a) = pb$, $ P(b) = pc$ and $ P(c) = pa$. I don't understand the definition of $ P(x)$. Why is $ P(b)=pc$? If $ P(x)=a^n+pb-x^n$, then it should be $ P(b)=a^n+pb-b^n$.

You are right, $ P(b) = a^n + pb - b^n$. Now we use the fact that $ a^{n} + pb = b^{n} + pc$: Since $ a^n + pb = b^n + pc$, $ P(b) = (a^n + pb) - b^n = (b^n + pc) - b^n = pc$.

There may be something wrong here, but anyways... $(a-b)(a^{n-1}+\ldots +b^{n-1})=p(b-c)$ $(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$ $(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$. First, it is clear that if any two of $a,b,c$ are equal, then all three are equal as we see that if, WLOG $a=b$, then $a^n=b^n$ so $pb=pc$ or $b=c=a$. Multiplying the above three equations, we get that if $X=(a^{n-1}+\ldots +b^{n-1})$, $Y=(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$, $Z=(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$, then $XYZ=p^3$. Clearly $X,Y,Z>0$. If one of $X,Y,Z$ is one, then n=1, as for example, if $a^n-b^n=a-b$, this must imply n=1. Then $a+pb=b+pc=c+pa$. So we get $a-b=p(b-c)=p^2(c-a)=p^3(a-b)$ so either $p=1$ - impossible, or $a=b=c$. Otherwise, $X=Y=Z=p$ and we get $p(a-b)=p(b-c)=p(c-a)$, so we get $a-b=b-c$ or $2b=a+c$ and similarly $2c=a+b$ which gives $a=2c-b$, and substituting into the earlier equation, we get $3b=3c$, which gives us $b=c=a$, as desired. Hence $a=b=c$.

manifestdestiny wrote: There may be something wrong here, but anyways... $(a-b)(a^{n-1}+\ldots +b^{n-1})=p(b-c)$ $(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$ $(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$. First, it is clear that if any two of $a,b,c$ are equal, then all three are equal as we see that if, WLOG $a=b$, then $a^n=b^n$ so $pb=pc$ or $b=c=a$. Multiplying the above three equations, we get that if $X=(a^{n-1}+\ldots +b^{n-1})$, $Y=(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$, $Z=(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$, then $XYZ=p^3$. Clearly $X,Y,Z>0$. If one of $X,Y,Z$ is one, then n=1, as for example, if $a^n-b^n=a-b$, this must imply n=1. Then $a+pb=b+pc=c+pa$. So we get $a-b=p(b-c)=p^2(c-a)=p^3(a-b)$ so either $p=1$ - impossible, or $a=b=c$. Otherwise, $X=Y=Z=p$ and we get $p(a-b)=p(b-c)=p(c-a)$, so we get $a-b=b-c$ or $2b=a+c$ and similarly $2c=a+b$ which gives $a=2c-b$, and substituting into the earlier equation, we get $3b=3c$, which gives us $b=c=a$, as desired. Hence $a=b=c$. What about $5^2-(-4)^2=5-(-4)=9$, it isn't correct at all, when you say $X=1$ for example. for the solution of Gunardi: if they heve the same signs, $p$ has to be positive, hence that proof isn't good. solution with $P(x)$ isn't correct, you define $P(a)$ with help of $b$ and similar.

assume that $a>b>c$ (if two of are equal,it's easy too check that all of them are equal) $\frac{a^n-b^n}{c-b}=\frac{c^n-b^n}{c-a}=\frac{c^n-a^n}{b-a}=p$ 1)$p>2$: $\left \| c-b \right \|_p=x\Rightarrow \left \| a^n-b^n \right \|_p=x+1$ $\left \| c^n-b^n \right \|_p=x+\left \| n \right \|_p\Rightarrow \left \| c-a \right \|_p=x+\left \| n \right \|_p -1$ $\left \| c^n-a^n \right \|_p=x+2\left \| n \right \|_p-1\Rightarrow \left \| a-b \right \|_p=x+2\left \| n \right \|_p-2$ $\Rightarrow x+1=\left \| a-b \right \|_p+\left \| n \right \|_P=x+3\left \| n \right \|_p-2\Rightarrow \left \| n \right \|_p=1$ $\Rightarrow \left \| a-b \right \|_p=\left \| b-c \right \|_p=\left \| c-a \right \|_p=x$ $a-b=p^x .k_1$ , $a-c=p^x.k_2$ , $b-c=p^x.k_3$ $k_1\mid k_3,k_3\mid k_2,k_2\mid k_1\Rightarrow k_1=k_2=k_3\Rightarrow a=b=c$ contradiction 2)p=2 is not different with the 1

mousavi wrote: assume that $a>b>c$ (if two of are equal,it's easy too check that all of them are equal) $\frac{a^n-b^n}{c-b}=\frac{c^n-b^n}{c-a}=\frac{c^n-a^n}{b-a}=p$ 1)$p>2$: $\left \| c-b \right \|_p=x\Rightarrow \left \| a^n-b^n \right \|_p=x+1$ $\left \| c^n-b^n \right \|_p=x+\left \| n \right \|_p\Rightarrow \left \| c-a \right \|_p=x+\left \| n \right \|_p -1$ $\left \| c^n-a^n \right \|_p=x+2\left \| n \right \|_p-1\Rightarrow \left \| a-b \right \|_p=x+2\left \| n \right \|_p-2$ $\Rightarrow x+1=\left \| a-b \right \|_p+\left \| n \right \|_P=x+3\left \| n \right \|_p-2\Rightarrow \left \| n \right \|_p=1$ $\Rightarrow \left \| a-b \right \|_p=\left \| b-c \right \|_p=\left \| c-a \right \|_p=x$ $a-b=p^x .k_1$ , $a-c=p^x.k_2$ , $b-c=p^x.k_3$ $k_1\mid k_3,k_3\mid k_2,k_2\mid k_1\Rightarrow k_1=k_2=k_3\Rightarrow a=b=c$ contradiction 2)p=2 is not different with the 1 Are you sure, if $p \not| v_p(b-c)$ you aren't correct and where would you conclude that from?

Allnames wrote: It means $ A = B = C = p$. Now it shows immediately $ a = b = c$. I was doing this question, and similarly arrived at $A= B= C= p$, however, I could not reduce this down to $a= b= c$ immediately, could you elaborate?

I shall present my proof.(I have skipped trivial calculations) Note that equality of any two of $a,b,c$ implies equality of all the three,so we assume that no two of them are equal. Subtracting the terms taking two at a time we get $p(c-b)=a^n-b^n$ $p(a-c)=b^n-c^n$ $p(b-a)=c^n-a^n$ Multiplying the three equalities yeilds $\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{a-c}=p^3$ Note that $\frac{a^n-b^n}{a-b}$ never equals $1$ unless $n=1$ or $a,b \in (0,1,-1)$. If $n=1$ then $p=1$,a contradiction.Work out the details in the second case to show impossibility. Since no term can equal $1$,the factors must be $(p,p,p)$ or $(-p,-p,p)$ in some order.In any case atleast one of them equals $p$,say $\frac{a^n-b^n}{a-b}$.Then $a^n-b^n=p(a-b)=a^n-c^n \implies b^n=c^n \implies p=0$ a contradiction. The case when first two terms are $-p$ and the last one is $p$ needs to be addressed seperately.Then if $n$ is odd then $a<b<c$.But this is not possible since then $0<p(c-b)=a^n-b^n<0$.If $n$ is even then $\frac{a^\frac{n}{2}-b^\frac{n}{2}}{a-b} \cdot (a^\frac{n}{2}+b^\frac{n}{2})=-p$ and similarly the LHS of the other equality is factorized.Work out the details to show that this case is also not possible. In conclusion we must have $a=b=c$

April wrote: Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a$, $b$, $c$ are integers (not necessarily positive) satisfying the equations \[ a^n + pb = b^n + pc = c^n + pa\] then $a = b = c$. Proposed by Angelo Di Pasquale, Australia I have a surprisingly easy proof for the required result which might very well be wrong.It would help if you could tell me what you think of this. So here goes: First if $a=b=c$ then we are clearly done. If not,then assume that two of these are equal(without loss of generality, say $a=b$) Then replacing $b$ with $a$,the first equality becomes, $a^n+pa=a^n+pc$ which just simplifies to $a=c$,implying the required result. Now consider the case where all three are distinct. WLOG let $a>b>c$ Then if $n$ is odd,we would have $a^n>b^n$ and $pb>pc$,whose addition would contradict the first equality.(Note that this holds even when $a$ and $b$ are negative because still $a^n>b^n$ when $n$ is odd) 1.Therefore $n$ is even Now assume that all three of these are positive.Then $a^n>b^n$ and $pb>pc$ and addition would give us the previous contradiction. So one of $a,b,c$ must be negative. Since $a>b>c$,then $c<0$ Now consider three cases: The first- $b>0$ Then $a>b$ gives $a>0$ Once more,$a^n>b^n$ and $pb>pc$ would be a contradiction. The second- $b<0$ Then since by 1, $n$ is even, $b^n<c^n$ and $pc<pa$ whose addition would contradict the second equality. The third- $b=0$ Then the entire equality becomes $a^n=pc=c^n+pa$ Divisibility conditions could be used to simplify this trivially(eg: since $p|a$ put $a=pk$ and simplify) Doing so we arrive at another contradiction. Therefore,our original assumption must have been wrong so $a=b=c$ $QED$ This is a huge generalization of the original problem so there must be something wrong with my proof.Looking forward to any comments

I think that this is not a generalization. Quote: This is a huge generalization of the original problem Becouse every thing fails when b=0

samithayohan wrote: I think that this is not a generalization. Quote: This is a huge generalization of the original problem Becouse every thing fails when b=0 Why is that?

In the case b=0 Korapus5732 have used some results about divisibility.So still a,b,c must be integers

samithayohan wrote: In the case b=0 Korapus5732 have used some results about divisibility.So still a,b,c must be integers Solvable without divisibility. Since $n$ is even, $c^n > 0$, and since $p$ is positive $pa>pc$. Adding the two inequalities, we get a contradiction.

Dear Korapus5732 I think that you are completely wrong. In your proof you have assumed that $a$$>$$b$$>$$c$. But observe that this equation is not symmetric. It is just cyclic. So your proof is wrong even from the begining. . You cant use WLOG in a cyclic statement.

samithayohan wrote: Dear Korapus5732 I think that you are completely wrong. In your proof you have assumed that $a$$>$$b$$>$$c$. But observe that this equation is not symmetric. It is just cyclic. So your proof is wrong even from the begining. . You cant use WLOG in a cyclic statement. Samitha Yohan is right. Korapus5732's assumption is wrong from the beginning.

Let: $$(x,y,z)=(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+ab^{n-2}+b^{n-1},b^{n-1}+b^{n-2}c+b^{n-3}c^2+\ldots+bc^{n-2}+c^{n-1},c^{n-1}+c^{n-2}a+c^{n-3}a^2+\ldots+ca^{n-2}+a^{n-1}).$$We rearrange the equations into: \begin{align*} x(a-b)&=p(c-b)\\ y(b-c)&=p(a-c)\\ z(c-a)&=p(b-a). \end{align*}If some two of $a,b,c$ are equal, WLOG $a=b$. Then from the first equation, $b=c$, so the conclusion holds. Otherwise, assume FTSOC that $a\ne b\ne c\ne a$. Multiplying these three equations gives: $$xyz=p^3.$$ $(\text a)$ If $p\in\{x,y,z\}$, then WLOG $x=p$. Then $p(a-b)=p(c-b)$, so $a=c$, contradiction. $(\text b)$ If $\{-1,1\}\subseteq\{x,y,z\}$, then WLOG $|x|=1$. Suppose $x=1$ and that $n$ is odd, then $a^n-b^n=a-b$, or $a^n-a=b^n-b$. If $ab=0$, WLOG $b=0$, then $a=\pm1$ and we reach a contradiction upon concluding that $a^n=pc$. If $n$ is odd then $a,b$ must have the same sign. WLOG $a,b$ are both positive, then since $ab\ne0$ we have $a,b\ge1$. Now let $f(x)=x^n-x$. By expansion, $(x+1)^n>x^n+1$. Then $f(x+1)>f(x)$, so since $f(a)=f(b)$ we need $a=b$, contradiction. Suppose $x=1$ and that $n$ is even. If $a,b$ have the same sign then, as before, $a=b$. So $a$ and $b$ must have different signs, then WLOG $a<0$ and $b>0$, and $a+b\ge0$. Let $d=-a$, we have $d^n+d=(-d)^n+d=b^n-b$. Note that $b\ge d$. If $b=d$ then $b=0$ which is a contradiction, so $b\ge d+1$. Using the result we got above, $d^n+d\ge(d+1)^n-d-1$. But $(d+1)^n>d^n+2d+1$ by expansion, so we reach a contradiction. The same approach can be used for the case $x=-1$; it is impossible. Since one of the cases $(\text a)$ or $(\text b)$ must hold, we're done.

After trying to derive relations that allows applying LTE, I found this solution which do not use it at all. Assume for contradiction that $a,b,c$ are distinct. We can easily see that if $a,b,c$ are not pairwise distinct, then they must be all equal. Then, we must have $a \ne b, b \ne c, c\ne a.$ If $n=1,$ we will have $$ a-b = -p(b-c), b-c = -p(c-a), c-a= -p(a-b)$$and multiplying everityhing and using the pairwise distinc assumption yelds $p^3 = -1,$ contradiction since $p$ is prime. Then $n \ge 2$ and we can write, $$ (a-b)T(a,b) = -p(b-c) ~(1A)$$$$ (b-c)T(b,c) = -p(c-a)~(1B)$$$$ (c-a) T(c,a) = -p(a-b)~(1C)$$where $T(x,y) = x^{n-1} + x^{n-2}y + \dots +y^{n-1}.$ Multiplying everything again yelds $$T(a,b)T(b,c)T(c,a) = -p^3~(*)$$Claim: Neither of $T(a,b),T(b,c), T(c,a)$ can be equal to $p$ or $-p.$

The Claim and the fact that $p$ is prime now implies that $(*)$ can only be satisfied if one of $|T(a,b)|, |T(a,c)|, |T(b,c)|$ be equal to $p^3$ and the remaining two be equal to $ 1.$ Then assume without of generality that $|T(a,b)| = p^3$ and $|T(b,c)| = |T(c,a)| =1.$ Thus, $(1A), (1B)$ and $(1C)$ will implie that $$ v_p(a-b) + 3 = 1 + v_p(b-c) $$$$ v_p(b-c) = 1 + v_p(c-a) $$$$ v_p(c-a) = 1 + v_p(a-b) $$That is $v_p(b-c) = 2 + v_p(a-b)$ and $v_p(c-a) = 1 + v_p(a-b);$ Then we will have $$ a-b = p^{\alpha}k$$$$ b-c = p^{\alpha + 2}s $$$$ c-a = p^{\alpha+1}r $$where $\alpha$ is an non-negative integer and $k,s,r$ are integers not divisible by $p.$ (Note that this can be done since, $a-b, b-c, c-a$ are assumed to be non-zero integers.) Then, adding up this equations gives $$ 0 = p^{\alpha}(k + ps + p^2r) \Rightarrow k = -ps -p^2r,$$contradiciton since $k$ is an integer not divisible by $p.$ So done. $\square$

Solved with pyramid_fan90. Assume for the sake of contradiction that $a$, $b$, and $c$ are pairwise distinct. If $n$ is odd then suppose that $a>b$. It follows that $c>b$ since otherwise we would have $a^n+pb>b^n+pc$. We also obtain $c<a$ because otherwise $c^n+pa>a^n+pb$. However, this implies $b^n+pc<c^n+pa$, a contradiction. Then $a<b$. Similarly, $b<c$ and $c<a$, a contradiction. Henceforth assume that $n$ is even. Notice that we may obtain \[\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}=p\]which implies that \[\prod\frac{a^n-b^n}{a-b}=\prod(a^{n-1}+a^{n-2}b+\dots+b^{n-1})=-p^3.\]Notice that each of these sums is even when the parity of the two variables is the same, and each one is odd when the parity of the two variables differs. This implies that the product is even, or that $p=2$. Now, notice that the parity of all the variables is the same. This implies that each of the sums was an even number. Now, the three sums can only be $(-2, -2, -2)$ or $(-2, 2, 2)$. FTSOC, suppose that the sums are $(-2, 2, 2)$ and WLOG set $\frac{a^n-b^n}{a-b}=2$. Then we also have that $\frac{a^n-b^n}{c-b}=2$ which implies that $a=c$, a contradiction. Finally, suppose the sums are $(-2, -2, -2)$. Then clearly $b-a=c-b$, $c-b=a-c$, and $a-c=b-a$, which implies $a=b=c$. We are done. $\blacksquare$

For the sake of contradiction we will have that $a,b,c$ are not equal to each other. CLAIM: We have that if $n$ is odd then there is not $a,b,c$ that satisfy the problems equation. PROOF: : WLOG $a>b$ and $c>b$ and we have $c<a$ but after plugging our values in we get that $b^n+pc<c^n+pa$ which is a contradiction. Similarly we have that the other cyclic variants are also a contradiction. QED. $N$ is even case For this case after some rearrangements we get that $\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}=p$. Since we know that $a-b|a^n-b^n$ we have that $\prod(a^{n-1}+a^{n-2}b+\dots+b^{n-1})=-p^3.$ We will notice that $(a^{n-1}+a^{n-2}b+\dots+b^{n-1})$ has a divisor of $2$ which implies that $p=2$. Since the product is $-p^3$ we have that the sums could be $(-2,-2,-2)$ or $(-2,2,2)$ we would like to have $(-2,2,2)$ as the sums but we see that there is a immediate contradiction that $a=c$ so we have that $a=b=c$ works . QED $\blacksquare$ REMARKS: This problem was really nice but it took me some time for me to notice that $p=2,$

If two of the variables are equal, say $a=b,$ then we get $pb=pc$ or $a=b=c$. Assume for the sake of contradiction that $a\neq b\neq c$. We find that $p = \frac{a^n-b^n}{c-b} = \frac{b^n-c^n}{a-c} = \frac{c^n-a^n}{b-a}$. It is easy to check that if $n$ were odd then one of the terms is negative, contradiction since $p$ is positive. This means $n$ is even. We must have two of the variables congruent mod 2 by pigeonhole, so WLOG, $a\equiv b \pmod 2.$ But since $\frac{a^n-c^n}{a-b}$ is an integer, $a\equiv c \pmod 2$ as well. Taking the product of the three values, we see that $$p^3 = - \prod_{cyc} \frac{a^n-b^n}{a-b}.$$If $2\mid a,b$, let $\nu_2(a) = k, \nu_2(b)=m$. If $k\neq m$, then $\nu_2(a-b) = \min(m,k)$ and $\nu_2(a^n-b^n) = \min(mn, kn)$ and since $n$ is even, $\min(mn,kn) > \min(m,k)$ implying $2\mid p$ or $p=2.$ If $m=k$, let $a = 2^m \cdot x, b=2^m \cdot y.$ Then we see that $\nu_2(a-b) = m + \nu_2(x-y)$ and $\nu_2(a^n-b^n) = mn + \nu_2(x^n-y^n) = mn + \nu_2(x-y) + \nu_2(x+y) + \nu_2(n) - 1 > m+\nu_2(x-y)$, so $2\mid p$ or $p=2.$ If $p=2$, then if 3 non 1 integers have a product of $-8$, they must be $(-2,-2,-2)$ or $(-2,2,2)$. However it is easy to check that one of the cyclic values $\frac{a^n-b^n}{c-b}$ must be equal to the corresponding cyclic value $\frac{a^n-b^n}{a-b},$ or $a=b$, contradiction. This means $a=b=c.$

Subtracting pairs of equations from each other, \begin{align*} a^n-b^n &= p(c-b) \\ b^n-c^n &= p(a-c) \\ c^n-a^n &= p(b-a). \end{align*}This implies $$-p^3 = \frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a}.$$All three terms must equal $-p$, as they are all integers and otherwise two of them would contradict the third. But now, cross-equating, $$-p(a-b) = a^n-b^n = p(c-b) \implies a+c=2b.$$Thus, each of $a, b, c$ is the average of the other two, so $a=b=c$.

Took me $5$ hours to type the LaTeX along with fixing all the fakesolves I had made . Don't someone dare say this was easy. Might be easy for some omegaorsmax guy like you but I being a nub felt that this problem was very very instructive as it connects all the different ways of algebraic manipulations and joins all of them together. We case bash this . Firstly, if any two of the ones become equal, it's easy to see that all of them become equal. FTSOC assume that none of them are equal. Before we actually begin, notice that the equations are just cyclic and not symmetric. So we cannot take $a>b>c$. What we can actually do is take $\max\{a,b,c\}=a$. Now manipulating the first and the second equalities, we get,\[\dfrac{a^n-b^n}{c-b}=\dfrac{b^n-c^n}{a-c}=\dfrac{c^n-a^n}{b-a}=p.\tag{$\spadesuit$}\]Cross multiplying and simplifying further we get, $a^{n+1}+b^{n+1}+c^{n+1}=ab^n+bc^n+ca^n$. That is,\[a\cdot a^n + b\cdot b^n+c\cdot c^n=c\cdot a^n+a\cdot b^n+b\cdot c^n.\tag{$\clubsuit$}\]Now if $n$ is odd, note that for any ordering of $a$, $b$ and $c$ (like say for $a>b>c$), we are going to have a similar ordering for $a^n$, $b^n$ and $c^n$ (like say for $a^n>b^n>c^n$). This however forces a contradiction to the rearrangement inequality applied to $(\clubsuit)$. Thus we must have that $n$ is even. Now $a$ must be positive, otherwise we arrive at a contradiction for $\dfrac{c^n-a^n}{b-a}=p$ as the LHS becomes -ve. Now we divide the rest into two cases : $a>b>c$. From $(\spadesuit)$ we have $\dfrac{a^n-b^n}{c-b}=p$ which then forces $|a|<|b|$. This then implies that $c<b<-a$ as $a$ is +ve which further gives that $|c|>|a|$ which however is a contradiction to $\dfrac{c^n-a^n}{b-a}=p$ as the LHS again becomes -ve. So we must have: $a>c>b$. Now moving onto the final part of the solution, manipulating $(\spadesuit)$ we get the following identities, \begin{align} \dfrac{(a-b)(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})}{(c-b)}&=p\\ \dfrac{(b-c)(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})}{(a-c)}&=p\\ \dfrac{(c-a)(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})}{(b-a)}&=p .\end{align} Now from $a>c>b$ we are going to have $\dfrac{b-a}{c-a}>1$ from which we can deduce $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=\dfrac{p(b-a)}{(c-a)}>p$. Now multiply $(1)\times(2)\times(3)$ to get,\[(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=-p^3\]which along with our inequality that we arrived at above forces $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})\in\{p^2,p^3\}$. Now again case bash OOOOOOF!!!!! If $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=p^2\implies\dfrac{b-a}{c-a}=p$. Then we multiply $(1)\times(2)$ to get, \[(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-p.\]Now note that $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})$ can't be $=p$ otherwise this forces $a=c$. Thus we must have $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})=1$ and $(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-p$. Thus our final equations change to, \begin{align} \dfrac{(a-b)}{(c-b)}&=p\\ \dfrac{(b-c)}{(a-c)}&=-1\\ \dfrac{(c-a)}{(b-a)}&=\dfrac{1}{p} .\end{align} Now from $(3)$ and $(1)$ we get $\dfrac{1}{p}=\dfrac{c-a}{b-a}=\dfrac{b-c}{b-a}$ which upon adding gives $\dfrac{1}{p}=\dfrac{1}{2}$ that is $p=2$ and $a+b=2c$. So now our problem statment becomes \[(2c-b)^n+2b=b^n+2c=c^n+2(2c-b).\]From this we get $2(c-b)=(2c-b)^n-b^n$ which gives $|2c-b|> |b|$. Now note that the RHS is an increasing function (can be easily proved by differentiating and checking off the rest of the cases manually) and the equality clearly holds at $n=1$ which thus forces $n=1$ but we had that $n$ is even which forces a contradiction. Now for the other case, if $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=p^3\implies\dfrac{b-a}{c-a}=p^2$. Again multiply $(1)\times(2)$ to get $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})=1$ and $(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-1$ and then we have our final equations as follows, \begin{align} \dfrac{(a-b)}{(c-b)}&=p\\ \dfrac{(b-c)}{(a-c)}&=-p\\ \dfrac{(c-a)}{(b-a)}&=\dfrac{1}{p^2} .\end{align} So from $(1)$ and $(2)$, we get $p-\dfrac{1}{p}=\dfrac{a-b}{c-b}+\dfrac{a-c}{b-c}=1$ which upon solving gives a clear contradiction.

lol what is this proof Define \[f(a,b)=a^{n-1}+a^{n-2}b+\dots+b^{n-1}.\] If two of these are equal then obviously all three are equal. So assume FTSOC that $a$, $b$, and $c$ are pairwise distinct. Then \[p=\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}\]so \[-p^3=f(a,b)\cdot f(b,c)\cdot f(c,a).\] Lemma 1. If $|f(a,b)|=1$, then $p\mid a-b$ and $p\nmid n$. Proof. \[|a^n-b^n|=|a-b|\rightarrow p=\frac{|a-b|}{|c-b|}\rightarrow p\mid a-b.\]So by LTE, $p\nmid n$. Claim. The absolute values of the factors being $1$, $1$, and $p^3$ is impossible. Proof. If $|f(a,b)|=1$, then we can use lemma 1 to get $p\mid a-b$ and $p\nmid n$. Similarly, WLOG $|f(b,c)|=1$, so $p\mid b-c$. So $p\mid c-a$. We have $|f(c,a)|=p^3$, so \[\nu_p\left(|c^n-a^n|\right)>\nu_p\left(|c-a|\right).\]But $c-a$ is a multiple of $p$, contradiction by LTE. Therefore, at least one of the absolute values must be $p$. Claim. None of the factors can be $p$. Proof. Assume WLOG that this is $f(a,b)$. Then \[p=\frac{a^n-b^n}{a-b}=\frac{a^n-b^n}{c-b}\rightarrow a=c,\]contradiction. Lemma 2. If $f(a,b)=-p$, then $a+c=2b$. Proof. Assume WLOG that this is $f(a,b)$. Then \[p=\frac{a^n-b^n}{b-a}=\frac{a^n-b^n}{c-b}\rightarrow a+c=2b.\] Claim. The factors can't be all $-p$. Proof. Otherwise, by lemma 2, we have $a+c=2b$ and cyclic variations so $a=b=c$, contradiction. Claim. The factors can't be $1$, $-p$, and $p^2$ nor $-1$, $-p$, and $-p^2$. Proof. Let $|f(a,b)|=1$. By lemma 1, $p\mid a-b$ and $p\nmid n$. So with an argument similar to the first claim, $p\nmid b-c$ and $p\nmid c-a$. But we know that $a+b=2c$ or $b+c=2a$, both of which are enough to show that $a$, $b$, and $c$ are all equivalent mod $p$, contradiction. We are done!

Storage from a while ago We have that $a^n-b^n=p(c-b)$ and cyclically. If n=1 it's evident that a=b=c, if WLOG $a=b$ it's evident that a=b=c. Assume henceforth $n>1$ and $a,b,c$ distinct. Then $$\frac{a^n-b^n}{a-b}\frac{b^n-c^n}{b-c}\frac{c^n-a^n}{c-a}=-p^3.$$Hence (noting $|a^n-b^n|>|a-b|$ $a^n-b^n=\pm(a-b)p$ (otherwise if it has $p^2$ or smth then some other term will have same magnitude). If it's equal to the positive version, then $a-b=c-b$, contradiction, hence all of them must be negative multiplied by p, implying $a+c=2b$ and cyclically, so $2b-2c=c-b$ or $c=b$ again, contradiction! $\blacksquare$

Note that $n=1$ iff $a=b=c$. Then, assume that $n \neq 1$. We have $$p=\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{a^n-c^n}{a-b} \Rightarrow -p^3=\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{a^n-c^n}{a-c}.$$Note that each of these terms must be integers. $\textbf{Claim:}$ Each of the terms contain a factor of $p$. $\emph{Proof.}$ Assume for the sake of contradiction that instead at least one of the terms is $\pm 1$. But since $n \neq 1$, we always have $|x^n-y^n| > x-y$ for integers $x$ and $y$, contradiction. From here, it immediately follows that each term has exactly one factor of $p$. There are now $2$ cases to check: either $1$ term is equal to $-p$ or all $3$ terms are equal to $-p$. Considering the first case, assume $$a^n-b^n=-p(a-b),$$$$b^n-c^n=p(b-c),$$$$a^n-c^n=p(a-c).$$But from above, we also see that $b^n-c^n=p(a-c)$, so $a=b$, violating the condition that $n \neq 1$. Now, consider the second case in which all terms are equal to $-p$. We have $$a^n-b^n=-p(a-b),$$$$b^n-c^n=-p(b-c),$$$$a^n-c^n=-p(a-c).$$Again from above, we have $b^n-c^n=p(a-c)$, so $a+b=2c$. Similarly, $a+c=2b$ and $b+c=2a$. Solving, we get $a=b=c$ again. We have covered all cases, so we are done.

This is more alg than NT (as usual bruh) By rearranging we have that \begin{align*} a^n-b^n &=p(c-b)\\ b^n-c^n &= p(a-c)\\ a^n - c^n &= p(a-b)\\ \end{align*}Multiplying these together we get that, \[\frac{a^n-b^n}{a-b}\frac{b^n-c^n}{b-c}\frac{a^n-c^n}{a-c}=-p^3\]First note that $n$ is clearly even, as if $n$ was odd, then the denominator and numerator of each fraction would have the same sign, making the product positive. Thus, $n$ is in fact even. Note that one of these fractions can be $1$ or $-1$ if and only if $a,b,c\in{0,1,-1}$. We can check the cases when $a=0,-1,1$ by substituting back into the equations, to see that these cases in fact do not work. Thus, each fraction is either $p$ or $-p$. Now, there are two cases $(-p,-p,-p)$ or $(p,p,-p)$. $Case \ 1:$ In this case we first consider \begin{align*} \frac{b^n-c^n}{b-c} &= -p\\ b^n-c^n = -p(&b-c) = b^n - a^n \end{align*}Thus, in-fact $a^n=b^n \implies a=b$ and in turn $a=b=c$. $Case \ 2:$ We know that always, atleast one of $\frac{a^n-b^n}{a-b}$ and $\frac{a^n-c^n}{a-c}$ is equal to $p$. Assume WLOG, that $\frac{a^n-b^n}{a-b} = p$. Then, we have that \begin{align*} \frac{a^n-b^n}{a-b} &= p\\ a^n-b^n = p(&a-b) = a^n - c^n \end{align*}Thus, we must have that $b^n = c^n \implies b=c$, and thus $a=b=c$. Now, substituting $a=b=c$, we see that all three equations are satisfied without contradictions. Therefore, $a,b,c\in \mathbb{Z}$ satisfy the given equations if and only is $a=b=c$.

First, for completeness, we note $a=b=c$ can hold. Assume by contradiction that $a$, $b$, and $c$ are not all the same. Suppose two of the three are equal; WLOG let $a=b$. Equating the first two expressions, we get \[pb = pc \implies b=c,\] contradiction. Next, suppose all three are distinct. Then \[(a^n-b^n)(b^n-c^n)(c^n-a^n) = p(c-b) \cdot p(a-c) \cdot p(b-a)\]\[\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{a^n-c^n}{a-c} = -p^3,\]implying that all three factors must be $\pm p$, and it's easy to verify all cases lead to $a=b=c$. $\blacksquare$

We will rewrite the equations as follows \begin{align*} a^n-b^n &= p(c-b) \\ b^n-c^n &= p(a-c) \\ c^n-a^n &= p(b-a) \end{align*} Multiplying and simplifying, we get \[\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a}=-p^3\] Then, $a^n-b^n=\pm p(a-b)$ since $a-b \mid a^n-b^n$ and they cannot be equal. (This is the same for cyclic variations) If $a^n-b^n=p(a-b)$, we have \[a^n-b^n=p(a-b)=p(c-b)\] so $a=c$, implying through some simple calculations that $a=b=c$. If $a^n-b^n=-p(a-b)$, we have \[a^n-b^n=-p(a-b)=p(c-b)\] Then, $2b=a+c$ and same with cyclic variations, meaning $a=b=c$. This covers all cases, so our proof is complete. $\square$

We have $a^n - b^n = p(c - b)$ and cyclic variants, so we have $(a - b) \mid p(b - c), (b -c) \mid p(c - a), (c - a) \mid p(a - b) $. If one of the differences is zero, WLOG $a - b$, we have $p(c - b) = 0$, so $b = c$, then we have $a = b = c$ as desired. So assume that all of them are nonzero, we force a contradiction. Treat each difference its absolute value, then clearly two of them sum to the other, and one is maximal. WLOG let the maximal one be $a- b$. Then clearly $\frac{b - c}{a - b}$ is not an integer, so it must have denominator exactly $p$ when simplified. Thus we can write $\frac{b -c }{a - b} = \frac xp$ for some $x$. Then we also know that $p\frac{c - a}{b - c} = p\frac{p - x}{x}$, which must be an integer, since $\gcd(x,p) = 1$, we have $x = 1$. We then have $\frac{p^2}{p - x}$ is an integer, impossible for $x = 1$, contradiction.

Note that by manipulation we eventually get the equations \begin{align*} a^n-b^n &= -p(b-c) \\ b^n-c^n &= -p(c-a) \\ c^n-a^n &= -p(a-b). \end{align*} If there are two variables that are equal, say WLOG $a=b,$ then it is clear that $b=c,$ and we would be done. So, for the sake of a contradiction assume otherwise. Then multiplying the three equations yields $$\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a} = -p^3.$$Note that if $n$ is odd, $a^n-b^n$ will have the same sign as $a-b$ and the same holds for cyclic variations so the LHS is positive while the RHS is negative, a contradiction. Therefore $n \geq 2$ is even. Notice that there are two variables with the same parity, say WLOG $a, b.$ Then $$\frac{a^n-b^n}{a-b} = a^{n-1}+a^{n-2}b+a^{n-3}b^2 + \cdots + ab^{n-2}+b^{n-1}.$$If $a, b$ are both even then this value is even. If $a, b$ are both odd then the value is the sum of $n$ odd numbers, which is even. Therefore the LHS of our equation is even so $p=2,$ which gives $$\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a} = -8.$$ Now we show that $a, b, c$ all have the same parity. If $a, b$ are both odd, then consider the equation $a^n-b^n=-2(b-c).$ By LTE $v_2(a^n-b^n) =v_2(a+b)+v_2(a-b)+v_2(n)-1 \geq 2,$ so $2|(b-c),$ and thus $c$ is odd. Meanwhile, if $a, b$ are both even, then considering the equation $a^n-b^n=-2(b-c)$ again, we see that $2^n$ divides the LHS, but $n \geq 2$ so $2 | (b-c),$ and thus $c$ is even. Therefore by the same logic as above each of $\frac{a^n-b^n}{a-b}$ and its cyclic variations are even, and thus each of them equals $\pm 2.$ We either have $(-2, 2, 2)$ or $(-2, -2, -2).$ In the former case, WLOG $a^n-b^n=-2(a-b), b^n-c^n=2(b-c), c^n-a^n=2(c-a).$ The last equation implies $c-a=b-a \implies b=c,$ a contradiction. The latter implies $$a^n-b^n=-2(a-b) \implies a-b=b-c \implies a+c=2b$$and cyclic variations, which by simple algebra yields $a=b=c,$ contradiction. Therefore we have proven by contradiction that $a=b=c,$ and it is clear that this solution works.