Because $a-b\in\mathbb{Z}$ and $a^2-b^2\in\mathbb{Z}$, we must have $a+b\in\mathbb{Q}$, leading to $a=\frac{1}{2}(a+b)+\frac{1}{2}(a-b)$ and $b=\frac{1}{2}(a+b)-\frac{1}{2}(a-b)$ are both rational numbers as well. Write $a=\frac{x}{z}$ and $b=\frac{y}{z}$ as quotients of positive inters, with a smallest possible common denominator $z$. Then the given conditions implies $z^n\mid(x^n-y^n)$ for all $n$.

Suppose a prime $p$ divides $z$. We are going to show that no such $p$ exists. If so, $z$ must be $1$, implying both $a$ and $b$ are positive integers.

First, because $z\mid(x-y)$ so $p$ must divides $(x-y)$ too. If $p\mid x$, then $p\mid y$. This is impossible, because otherwise we could write $a=\frac{\frac{x}{p}}{\frac{z}{p}}$ and $b=\frac{\frac{y}{p}}{\frac{z}{p}}$ with a smaller common denomiator than $z$. Hence $p\not\mid x$ and $p\not\mid y$. This sets up to apply LTE (# 4802).

If $p$ is odd, then we have $$n\le v_p(z^n)=v_p(x^n-y^n)=v_p(x-y)+v_p(n)$$

Taking $p$ to both sides gives $$p^n\le (x-y)n \implies \frac{p^n}{n} \le (x-y)$$

However, this cannot hold for all $n$ because as $n\rightarrow\infty$, the left sides goes infinite while the right side is a constant.

If $p$ is even, i.e. $p=2$, then $$n\le v_2(z^n)=v_2(x^n-y^n)=v_2(x-y)+v_2(n)+v_2(x+y)-1 \implies \frac{2^{n+1}}{n}\le(x-y)(x+y)$$

This relation cannot hold for all $n$ either due to a similar reasoning.

Hence, we find no such primitive exists that divides $z$, leading to $z=1$ and both $a$ and $b$ are positive integers.