First, we prove that a,b are rational numbers. $a-b \in \mathbb{Q}$ $a+b = \frac{a^2-b^2}{a-b} \in \mathbb{Q}$ $a = \frac{(a+b)+(a-b)}2 \in \mathbb{Q}$ $b = \frac{(a+b)-(a-b)}2 \in \mathbb{Q}$ Then we can find three integers a',b',n such that: $a = \frac{a'}n$ $b = \frac{b'}n$ $d \mid a' \land d \mid b' \land d \mid n \Rightarrow |d| = 1$ The hypotesis translates to $n^k \mid a'^k - b'^k \quad \forall k \in \mathbb{N}$. Now, $n \mid a'-b'$ and we can suppose $(a',n) = 1$. Therefore $(b',n) = 1$. Let d be the integer such that $n^d \mid \mid a'-b'$. First, we suppose d>1. We have $n^{d+1} \mid a'^{d+1} - b'^{d+1} = (a'-b')(a'^d + \ldots + b'^d)$. We get $n \mid (a'^d + \ldots + b'^d)$, but $\gcd(a'-b', a'^d + \ldots + b'^d) \mid n$, therefore $n \mid \mid (a'^d + \ldots + b'^d)$ and $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$. (in this step we must remember that d>1). Now we consider 2(d+1). We know that $n^{2d+2} \mid a'^{2d+2} - b'^{2d+2} = (a'^{d+1} + b'^{d+1})(a'^{d+1}-b'^{d+1})$. But $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$, then $n^{d+1} \mid a'^{d+1} + b'^{d+1}$ and $n^{d+1} \mid (a'^{d+1} + b'^{d+1}) + (a'^{d+1} - b'^{d+1}) = 2a'^{d+1}$. But $(a',n) = 1$, then finally $n^{d+1} \mid 2$. But therefore, n=1 and a,b are integers. If $n \mid \mid a'-b'$, then $n^2 \mid (a'+b')(a'-b')$ and $n \mid a'+b'$ and $n \mid 2a'$, as before. Then n=2 and a',b' are odd integers such that a'-b' is a multiple of 2 but not of 4. Let us take d such that $2^d \mid \mid a'+b'$. Let us take a positive integer k such that $2^k - k > d$ (this obviously exists). But then $2^{2^k} \mid a'^{2^k} - b'^{2^k} = (a'^{2^{k-1}} - b'^{2^{k-1}}) \cdots (a'^2+b'^2)(a'+b')(a'-b')$. We must find $2^k$ factors 2 in this product. But if x,y are odd integers, $2 \mid \mid x^2+y^2$. And $2 \mid \mid a'-b'$ by hypotesis, and $2^d \mid \mid a'+b'$ by hypotesis. Therefore we find exactly $d + k$ factors 2 in this product, which is too low. I hope it is correct! I hope it is correct!

Lets say we showed that a,b are rational (like edriv did). Let: $ a = \frac {x}{y}, b = \frac {s}{t}, (s,t) = (x,y) = 1$ $ a^n - b^n \in Z, n \in N \leftrightarrow (yt)^n|(xt)^n - (ys)^n$. If we show that $ yt|xt, ys$, we get: $ y|x$, $ t|s$, or: $ a,b$ are integers. It follows from the following lemma: Lemma: If $ a^n|b^n - c^n$ for all $ n \in N$, where $ a$,$ b \neq c$ are integers, then: $ a|b,c$. Proof: Let p be a prime divisor of a, $ g = gcd(b,c)$, ($ b' = b/g, c = c'/g$). We have, by "Lifting the Exponent" lemma: $ ord_p(a^n) \le ord_p(b^n - c^n) =$ $ ord_p(g^n) + ord_p(b'^n - c'^n) =$ $ ord_p(g^n) + ord_p(b' - c') + ord_p(n) =$ $ ord_p(n(b - c)g^{n - 1})$ or (in the case p=2): $ ord_p(a^n) \le ord_p(b^n - c^n) = ord_p(g^{n - 2}(b^2 - c^2)n/2)$. We can put that together in the following way - $ ord_p(a^n) \le ord_p(g^{n - 1}(b^2 - c^2)n)$. It implies that: $ a^n | g^{n}(b^2 - c^2)n$ for all n>1. Let $ gcd(a,g) = g'$. We have: $ (\frac {a}{g'})^n|(b^2 - c^2)n$, which implies that: $ |(\frac {a}{g'})^n| \le |b^2 - c^2|n$ for all n>1. It can't happen for large n unless $ b^2 - c^2 = 0$ or $ a = \pm g'$. First case - $ b^2 = c^2$. It means that $ b = - c$, or: $ a^{2n - 1}|2b^{2n - 1}$ for all n, which gives $ a|b$, and then also $ a|c$. Second case - $ a = \pm g'$, or: $ a|b,c$. Q.E.D. "Lifting the Exponent": If p is an odd prime, and a,b are integers, and: $ p|a - b, (a,b,p) = 1$, we have: $ ord_p(a^n - n^n) = ord_p(n(a - b))$ If p=2, $ p|a - b, (a,b,p) = 1$, we have: $ ord_p(a^n - n^n) = ord_p(n(a^2 - b^2)/2)$ I have dropped out some minor details, ask a question or point an error if you find one.

Sorry for spam.But what "GML" mean ?

I assume it's George T. Gilbert, Mark I. Krusemeyer, Loren C. Larson, The Wohascum County Problem Book, MAA.

Here is a proof a little bit different from edriv. It is the same up to $ n|a'-b'$, $ (n,a')=(n,b')=1$. We can also assume $ n>0$ wlog. Let $ a'=An+b'$, we know $ n^k|a'^k-b'^k=\sum_{i=2}^k c_i(An)^ib'^{k-i}+kAnb'^{k-1}$, where $ c_i$ is binomial coefficients. We can always find big $ k$ with $ (k,n)=1$. Note the first term on RHS is multiple of $ (An)^2$. So we have $ n^2|kAnb'^{k-1}$, or $ n|A$. Then the first term on RHS is multiple of $ (An)^2$, or $ n^4$. So we have $ n^4|kAnb'^{k-1}$, or $ n^3|A$. Then the first term on RHS is multiple of $ (An)^2$, or $ n^8$. So we have $ n^8|kAnb'^{k-1}$, or $ n^7|A$. .... We have $ n^m|A$ for any big integer $ m$, $ A$ being nonzero finite number implies $ n=1$. Hence $ a=a'$ and $ b=b'$

$ a = \frac {x}{z}, b = \frac {y}{z}, (x,y,z) = 1$ $p|z$ $p^n|x^n-y^n$ and $p|x-y$ by $LTE lemma$ $n \le v_p(x^n-y^n)=v_p(x-y)+v_p(n)$ $n=p^k$ $p^k-k \le v_p(x-y)$ contradiction