  Let $n$ be a positive integer, $p$ be an odd prime and integers $a,b \not= 0$
with $gcd(a,b)=1$, $p \nmid ab$, and $p|(a^n \pm b^n)$, we prove the identity
$$\nu_p(a^n \pm b^n)-\nu_p(n)=\nu_p(a^{p-1}-b^{p-1}).$$ An unintended
interesting immediate consequence is the following variant of Wieferich's
criterion for FLT : Let $x^n+y^n=z^n$ with $n$ prime and $x,y,z$ pairwise
relatively prime. Then every odd prime $p|y$ satisfies $\nu_p(z^{p-1}-x^{p-1})
\ge n-1$ and every odd prime $p|x$ satisfies $\nu_p(z^{p-1}-y^{p-1}) \ge n-1$,
and every odd prime $p|z$ satisfies $\nu_p(x^{p-1}-y^{p-1}) \ge n-1$, ie. every
odd prime dividing $xyz$ is a Wieferich prime of order at least $n-1$ to some
base pair. In the "first case" where $n \nmid xyz$, the lower bound for the
Wieferich order can be improved to $n$. This gives us very strong intuition why
there should not be any solution even for moderately large $n$.
