(a): Note that for all $n = p$ for $p$ prime, we have that by FLT, $$x \equiv x^p \equiv y^p \equiv y\pmod{p},$$so $$p\mid x-y$$for all primes $p$, implying $x-y = 0$, so $x = y$. $\blacksquare$ (b): The answer is all integers $x, y$ for which $x-y\not \in \{-1, 1\}$. To show that $x-y\not \in \{-1, 1\}$, note that there exists an $n > 1$ for which $x^n \equiv y^n \pmod{n}$, so if $p$ is the smallest prime divisor of $n$, then $$p \mid n\mid x^n - y^n.$$If $p\mid x$, then $p\mid y$, so then $p\mid x-y$. Else, $\gcd(x, p) = \gcd(y, p) = 1$. In this case, note that $\gcd(n, p-1) = 1$, since the smallest prime divisor of $n$ is $p$ and all prime divisors of $p-1$ (if any) are less than $p$. Therefore, we have that since $$\left(\frac{x}{y}\right)^n \equiv 1\pmod{p}$$and $\gcd(n, p-1) = 1$, we must have $$\frac{x}{y} \equiv 1\pmod{p},$$so $x\equiv y\pmod{p}$ and so $p\mid x-y$. Thus, in both cases this implies that there exists a prime dividing $x-y$, so $x-y \not \in \{-1, 1\}$. To show that all $x, y$ for which $x-y \not \in \{-1, 1\}$ work, note that clearly $x=y$ works, and if $x\neq y$, then consider a prime $p$ dividing $x-y$. If $p$ divides $x$, then $p$ divides $y$, and so $$x^{p^{\ell}}\equiv y^{p^{\ell}} \equiv 0 \pmod {p^{\ell}}$$for every $\ell \geq 1$, since $$\nu_p(x^{p^{\ell}}), \nu_p(y^{p^{\ell}}) \geq p^{\ell} \geq \ell.$$Else, $p\nmid x, y$ and $p\mid x-y$. If $p\neq 2$, then note that by LTE, we have that for all $\ell \geq 1$, $$\nu_p(x^{p^{\ell}} - y^{p^{\ell}}) = \nu_p(x-y) + \nu_p(p^{\ell}) \geq \ell,$$so $p^{\ell} \mid x^{p^{\ell}} - y^{p^{\ell}}$ for all $\ell \geq 1$. If $p = 2$, then note that $4\mid x^2-y^2$ since $x^2\equiv y^2\equiv 1\pmod{4}$. Therefore, since we can apply LTE for $p=2$ when the difference of the two numbers is divisible by $4$, we have that for all $\ell \geq 1$, $$\nu_2(x^{2^{\ell}} - y^{2^{\ell}}) = \nu_2((x^2)^{2^{\ell-1}} - (y^2)^{2^{\ell-1}}) = \nu_2(x^2-y^2) + \nu_2(2^{\ell-1}) \geq 1 + (\ell-1) = \ell,$$so $2^{\ell} \mid x^{2^{\ell}} - y^{2^{\ell}}$ for all $\ell \geq 1$. Thus, in all cases, we have that if $x-y \not \in \{-1, 1\}$, there exists infinitely many positive integers $n$ for which $x^n \equiv y^n \pmod{n}$ and if $x-y \in \{-1, 1\}$, then there does not, so the only pairs of integers $(x, y)$ that work are those with $x-y \not \in \{-1, 1\}$, as claimed. $\blacksquare$