Verify: $a,b$ be positive rational numbers such that for infinitely many integers $n,$ $a^n-b^n$ is an integer, Prove that $a,b$ are also integers. 
Let $a,b$ be two positive rational numbers such that for infinitely many integers $n,$ $a^n-b^n$ is an integer, Prove that $a,b$ are also integers.

Let $a=\frac{x}{z}$ and $b=\frac yz .$
Hence $z^n\mid x^n-y^n$ for infinite $n.$
Note that

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*if $2|z\implies 2|x-y\implies n\nu_2(z)= \nu_2 (z^n)=\nu_2(x^n-y^n)= \nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1.$
Now $z$ is fixed and so in $x,y.$ But the LHS is increasing quite fast.

*

*if $p>2\mid z \implies  n\nu_p(z)= \nu_p (z^n)= \nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n).$
But again bounding gives that for large $z$ this won't be true.
So $z=1.$
Is the method correct? I am not sure of the bounding though.. Any other solution would be helpful too.
 A: You have the correct general idea, but there are several issues. First, $a \neq b$ is required since, otherwise, any positive rational $a = b$ works. Second, you expressed $a$ and $b$ with a common denominator of $z$, but using the Lifting-the-exponent lemma (LTE) with any $p \mid z$ specifically requires that $p \not\mid x$ and $p \not\mid y$. To show this can always be done, first note we have for some positive integer $n$ and $k$ that
$$\left(\frac{x}{z}\right)^n - \left(\frac{y}{z}\right)^n = k \; \to \; x^n - y^n = k(z^n) \tag{1}\label{eq1A}$$
Now, consider if $\gcd(x,z) = d \gt 1$. Then $d^n \mid k(z^n)$ and $d^n \mid x^n$, so $d^n \mid y^n \; \to \; d \mid y$. Thus, $x$, $y$ and $z$ can be divided by $d$ to form new values, with $\gcd(x,z) = 1$ now, plus $\gcd(y,z) = 1$ as well. Alternatively, we could have started with the $\gcd$ of $y$ and $z$ first. In either case, we get coprime $x$, $y$ and $z$. Next, using \eqref{eq1A}, we have the requirement that you already provided, i.e., that
$$z^n \mid x^n - y^n \tag{2}\label{eq2A}$$
is true for infinitely many integer $n$. As you stated, if $2 \mid z$, then the co-prime requirements mean that $x$ and $y$ are both odd, so $2 \mid x - y$. However, there are $2$ cases for LTE. If $4 \mid x - y$, then for any $n$ where \eqref{eq2A} holds,
$$n(\nu_2(z)) = \nu_2(z^n) \le \nu_2(x^n - y^n) = \nu_2(x - y) + \nu_2(n) \tag{3}\label{eq3A}$$
For all primes $p$, note that $\nu_p(n) \le \frac{n}{2}$, so along with $1 \le \nu_2(z) \; \to \; n \le n(\nu_2(z))$, then \eqref{eq3A} shows $n \le \nu_2(x - y) + \frac{n}{2} \; \to \; \frac{n}{2} \le \nu_2(x - y)$. Since the right side is a fixed value, this obviously can't be true for infinitely many $n$.
If $\nu_2(x - y) = 1$, then for even $n$ we basically have what you showed, i.e.,
$$n(\nu_2(z)) = \nu_2(z^n) \le \nu_2(x^n - y^n)= \nu_2(x - y) + \nu_2(x + y) + \nu_2(n) - 1 \tag{4}\label{eq4A}$$
Similar to what was done with \eqref{eq3A}, this cannot be true for infinitely many $n$. Since
$$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}(y) + \ldots + x(y^{n-2}) + y^{n-1}) \tag{5}\label{eq5A}$$
then, for odd $n$, there are $n$ odd terms being summed inside the second main set of brackets. This means this sum is odd, so $\nu_2(x^n - y^n) = \nu_2(x - y) = 1$. Thus, \eqref{eq2A} cannot be true for infinitely many $n$ in this case either.
Note that, for any odd prime $p \mid z$, having $p \mid x^n - y^n$ does not necessarily mean $p \mid x - y$, e.g., $3 \mid 2^2 - 1^2$ but $3 \not\mid 2 - 1$. Nonetheless, there's always at least one positive $n$ where $p \mid x^n - y^n \; \to \; x^n \equiv y^n \pmod{p}$ (e.g., Fermat's little theorem shows that $n = p - 1$ works), so let $m$ be the smallest such positive integer. Next, for any $n$ where $p \mid x^n - y^n$, Euclidean division states there are integers $g$ and $h$ where
$$n = gm + h, \; \; 0 \le h \lt m \tag{6}\label{eq6A}$$
Therefore,
$$\begin{equation}\begin{aligned}
x^{gm + h} & \equiv y^{gm + h} \pmod{p} \\
(x^{gm})(x^h) & \equiv (y^m)^g(y^h) \pmod{p} \\
(x^{m})^g(x^h) & \equiv (x^m)^g(y^h) \pmod{p} \\
x^h & \equiv y^h \pmod{p}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Since $q = m$ is the smallest positive integer where $x^q \equiv y^q \pmod{p}$, then $0 \le h \lt m$ means $h = 0$, i.e., \eqref{eq6A} gives that $n = gm$. Using the LTE with this gives
$$n(\nu_p(z)) = \nu_p(z^n) \le \nu_p((x^m)^g - (y^m)^g) = \nu_p(x^m - y^m) + \nu_p(g) \tag{8}\label{eq8A}$$
Similar to what was done before, $\nu_p(g) \le \frac{g}{2} \le \frac{n}{2}$, which results in $\frac{n}{2} \le \nu_p(x^m - y^m)$. The right side being a fixed value means there can't be infinitely many $n$ where \eqref{eq8A} is true.
The above shows $z$ can't have a factor of $2$ or any odd prime, which means $z = 1$, i.e., $a$ and $b$ must be integers.
