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1. What the problem asks

Let

[ f_{p}(a)= (a+1)^{p}-a^{p}-1 ,\qquad a\in \mathbb Z ,\; p>2\text{ prime}. ]

Because (f_{p}(a)\equiv0\pmod p) for every integer (a) (reduce the binomial expansion modulo (p)), the congruence

[ f_{p}(a)\equiv0\pmod {p^{2}} \tag{1} ]

asks for an extra factor (p).

Define

[ \mathcal P:={p\text{ prime }:\; \exists a\in\mathbb Z,\; p\nmid a(a+1)\text{ and }(1)\text{ holds}}. ]

The exam problem proved that

  • (x^{p}+y^{p}=z^{p}) has a solution in the group of (p)-adic units (\mathbb Z_{p}^{\times}) iff (p\in\mathcal P).

The question we have to answer is:

  • For which primes does (1) have a solution?
    In particular we want to know what happens for primes (p\equiv2\pmod3) and to prove that both the “good’’ primes (those not in (\mathcal P)) and the “bad’’ primes (those in (\mathcal P)) occur infinitely often.

2. A convenient reformulation

Write

[ u\equiv a\pmod{p^{2}},\qquad u\in(\mathbb Z/p^{2}\mathbb Z)^{\times } . ]

Using the binomial theorem and the fact that (\binom{p}{k}=p\frac{(p-1)!}{k!(p-k)!}) is divisible by (p) for (1\le k\le p-1), we obtain

[ \begin{aligned} (a+1)^{p} &= a^{p}+p\,a^{p-1} +\frac{p(p-1)}{2}\,a^{p-2}+\cdots +p\,a+1\[2mm] &\equiv a^{p}+1+p\sum_{k=1}^{p-1}\frac{1}{k}\binom{p-1}{k-1}a^{p-k} \pmod{p^{2}} . \end{aligned} ]

Hence (1) is equivalent to

[ \sum_{k=1}^{p-1}\frac{1}{k}\binom{p-1}{k-1}a^{p-k}\equiv0\pmod p . \tag{2} ]

The sum in (2) is the Fermat quotient of the unit (a+1) minus the Fermat quotient of the unit (a):

[ q_{p}(b):=\frac{b^{p-1}-1}{p}\pmod p . ]

Indeed, a short calculation gives

[ q_{p}(a+1)-q_{p}(a)\equiv\sum_{k=1}^{p-1}\frac{1}{k}\binom{p-1}{k-1}a^{p-k} \pmod p . ]

Consequently

[ \boxed{\; (a+1)^{p}\equiv a^{p}+1\pmod{p^{2}} \iff q_{p}(a+1)\equiv q_{p}(a)\pmod p\

Original question: Finding primes so that $x^p+y^p=z^p$ is unsolvable in the $p$-adic units on Mathematics Stack Exchange, licensed CC BY-SA.