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Let $a,b$ be distinct real numbers such that $a^n -b^n$ is integer for every positive integer $n$ , then is it true that $a,b$ are integers ?

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As many others have noticed, it readily follows that $a,b\in\mathbb Q$. Let $a=\frac cd$ and $b=\frac ef$ such that $\gcd(c,d)=\gcd(e,f)=1$ and $d,f>0$. Writing $\frac cd=z+\frac ef=\frac{fz+e}f$ with $z\in\mathbb Z$, we see that $d=f$.

Note that we have $d^n\mid c^n-e^n$ for every $n$.

I will use the following result, where $\nu_p(n)$ denotes the exponent (possibly $0$) of the prime $p$ in the prime factorization of $n$:

**Lemma.** Let $x$ and $y$ be (not necessarily positive) distinct integers and let $n$ be a positive integer. If $p$ is a prime such that $p\mid x-y$ but $p\nmid n,x,y$, then $\nu_p(x^n-y^n)=\nu_p(x-y)$.

*Proof.* See here, Lemma 1 or by using the result from this earlier question. As you can see, the proof is elementary and very short.

Now suppose $p$ is a prime divisor of $d$, and let $k=\nu_p(c-e)$.

Consider the integers $n$ that are not divisible by $p$, for example $n=pm+1$. From $d^n\mid c^n-e^n$ and the above lemma it follows that $n\leqslant k$ for all such $n$, which is clearly a contradiction because $n$ can be arbitrarily large.

We conclude that no such $p$ exists. Hence $d=1$, and $a,b\in\mathbb Z$.

This answer is really just the same as barto’s, but written up in a more self contained fashion. We may write, as barto has already shown, $a = \frac{c}{d}, b = \frac{e}{d}$ with $ce$ relatively prime to $d$ and $c,d,e$ integers with $d >0.$ Now for every $d,$ we have $c^{n} \equiv e^{n}$ (mod $d^{n}$). Suppose that $d > 1.$ Then for every integer $n>0,$ we have $c^{n}-e^{n} = (c-e) \sum_{j=0}^{n-1}c^{j}e^{n-1-j}.$ Hence $\frac{c^{n}-e^{n}}{c-e} \equiv nc^{n-1}$ (mod $d$) But $c$ is relatively prime to $d,$ so whenever $n$ is relatively prime to $d,$ we conclude that $\frac{c^{n}-e^{n}}{c-e}$ is relatively prime to $d$. Hence $c-e$ is divisible by $d^{n},$ as $c^{n}-e^{n}$ is, a contradiction, as $d >1$ and $n$ can be arbitrarily large.

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