5. Find all integer solutions to the equation $\frac{x^{7}-1}{x-1}=y^{5}-1$.

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5. No integer solutions.

Lemma If $x$ is an integer, and $p$ is a prime factor of $\frac{x^{7}-1}{x-1}$, then either $p \equiv 1(\bmod 7)$, or $p=7$.
Proof of the lemma: Since $p \left\lvert\, \frac{x^{7}-1}{x-1}\right.$, therefore, $p \mid\left(x^{7}-1\right)$.
Thus, $(x, p)=1$.
By Fermat's Little Theorem, we have $x^{p-1} \equiv 1(\bmod p)$.
Assume $7 \mathcal{Y}(p-1)$, then $(p-1,7)=1$.
By Bézout's Theorem, there exist integers $k, m$, such that
$7 k+(p-1) m=1$.
Hence, $x \equiv x^{7 k+(p-1) m} \equiv\left(x^{7}\right)^{k}\left(x^{p-1}\right)^{m} \equiv 1(\bmod p)$.
Therefore, $\frac{x^{7}-1}{x-1}=x^{6}+x^{5}+\cdots+1 \equiv 7(\bmod p)$.
Thus, $p \mid 7$. Hence, $p=7$.
Back to the original problem.
By the lemma, every positive divisor $d$ of $\frac{x^{7}-1}{x-1}$, is either $d=0(\bmod 7)$ or $d \equiv 1(\bmod 7)$.

Assume $x, y$ is a pair of integer solutions. Since for all $x \neq 1$, $\frac{x^{7}-1}{x-1}>0$, thus, $y-1>0$.
Also, $(y-1) \left\lvert\, \frac{x^{7}-1}{x-1}=y^{5}-1\right.$, hence,
$y \equiv 1(\bmod 7)$ or $y \equiv 2(\bmod 7)$.
If $y \equiv 1(\bmod 7)$, then $y^{4}+y^{3}+y^{2}+y+1 \equiv 5(\bmod 7)$;
If $y \equiv 2(\bmod 7)$, then $y^{4}+y^{3}+y^{2}+y+1 \equiv 3(\bmod 7)$.
Since $y^{4}+y^{3}+y^{2}+y+1$ is also a divisor of $\frac{x^{7}-1}{x-1}$, but its remainder modulo 7 is neither 0 nor 1, which is a contradiction.
Therefore, the equation has no integer solutions.