How Many Integers?

Lemma: If $p,q$ are primes such that $p \mid \dfrac{x^q-1}{x-1}$ for some $x \in \mathbb{Z},$ either $p=q$ or $p \equiv 1 \pmod{q}$ depending on whether $x \equiv 1 \pmod{p}$ or $x \not \equiv 1 \pmod{p}.$

Proof: Consider two cases.

• Case 1: $x \equiv 1 \pmod{p}$

Then, note that $\dfrac{x^q-1}{x-1} = x^{q-1} + x^{q-2} + \cdots + 1 \equiv q \pmod{p}.$ Since $\dfrac{x^q-1}{x-1} \equiv 0 \pmod{p},$ we have that $p \mid q \implies p=q.$

• Case 2: $x \not \equiv 1 \pmod{p}$

Then, we have that $q \mid x^p - 1,$ implying the order of $x \pmod{q}$ divides $p,$ i.e. is either $1$ or $p.$ However, it cannot be $1$ since otherwise we would be back to the first case. By FLT, $x^{q-1} \equiv 1 \pmod{p},$ so $p \equiv 1 \pmod{q}.$

In both cases, we're done. $\blacksquare$

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Now, we make a quick table of fifth power residues modulo $7.$ $\begin{array}{c|c|c} x & x^5 & x^5 \pmod{7} \\ \hline 1 & 1 & 1 \\ \hline 2 & 32 & 4 \\ \hline 3 & 243 & 5 \\ \hline 4 & 4^5 & (-3)^5 \equiv 2 \\ \hline 5 & 5^5 & (-2)^5 \equiv 3 \\ \hline 6 & 6^5 & (-1)^5 \equiv 6 \end{array}$ By the lemma, $y^5-1 \equiv 0 / 1 \pmod{7} \implies y^5 \equiv 1 / 2 \pmod{7}.$ By the table, we see that either $y \equiv 1 \pmod{7}$ or $y \equiv 4 \pmod{7}.$ However, these yield that $\dfrac{y^5-1}{y-1} = y^4 + y^3 + \cdots + 1$ are congruent to $5,3 \pmod{7}$ respectively, which is a contradiction since all primes in the factorization of $\dfrac{y^5-1}{y-1}$ are either $0$ or $1 \pmod{7}$ (because $\dfrac{y^5-1}{y-1}$ divides $\dfrac{x^7-1}{x-1}$ and all primes in the factorization of the latter are either $0$ or $1 \pmod{7}$ ).

Hence, there are no solutions.

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This problem is taken from ISL 2006 N5.