If $p$ and $q$ are primes such that $p>q$, Why is the congruence $$(-q^{p-1}+p^{q-1}-1)a \equiv p^{q-1} \pmod{pq}$$ where $a$ is some integer, considered false? I can't see which gcds I have to compare for this statement to be false. Any help?
2026-04-19 21:47:57.1776635277
Congruence modulo a composite number
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Since $\gcd(p,q)=1$, by Fermat's little theorem, $p^{q-1}\equiv 1 \pmod{q}$ and therefore the given congruence implies $$0\equiv (-q^{p-1}+p^{q-1}-1)a \equiv p^{q-1}\equiv 1\pmod{q}$$ which does not hold.