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I have come across this question, which reminded me of Fermats little theorem, i dont know if the Fermats theorem is actually in use in the following mathematical statements

an integer a is a coprime with p and a coprime with q (p and q are different prime numbers )then prove that

$$a^{(p-1)(q-1)} \equiv 1(\operatorname{mod} pq) $$

$$p^{q-1}+q^{p-1} \equiv 1(\operatorname{mod} pq)$$

any help would be appreciated, thanks in advance.

Siong Thye Goh
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New dragon
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2 Answers2

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These are really Chinese Remainder Theorem problems. First look mod $p$, then mod $q$. IF the expression is congruent to the same thing mod $p$ and $q$, then CRT says they're also congruent to that "same thing" mod $pq$.

For the second problem $p^{q-1}+q^{p-1} \equiv p^{q-1} +0 \equiv 1 (\bmod{q})$ by Fermat's little theorem. Likewise $p^{q-1}+q^{p-1} \equiv 1 (\bmod{p})$. So by CRT $p^{q-1}+q^{p-1} \equiv 1 (\bmod{pq}).$

The first problem has the same flavor.

B. Goddard
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  • I like your approach with the CRT. ill definitely look into that, thanks a lot! – New dragon Dec 03 '16 at 20:42
  • would you have any idea how i can find some helpful links proving this statement 2^(()−1) ≡ (+1)/2 (mod n) (if n is odd) – New dragon Dec 03 '16 at 20:51
  • Euler's theorem says $2^{\phi(n)} \equiv 1 \equiv n+1 (\bmod{n})$. Since $n$ is odd, $2$ has an inverse modulo $n$, so multiply through by $2^{-1}$. Wiki should have a decent page about Euler's theorem. – B. Goddard Dec 03 '16 at 20:57
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Hint $\ $ By lil Fermat, both satisfy $\,x\equiv 1\,$ mod $\,p\ \&\ q\,$ thus $\,x\equiv 1\pmod{pq}\ $ by CCRT. Or, more elementarily, we can prove it directly: $\ p,q\mid x\!-1\!\,$ $\,\Rightarrow\,$ $\, pq\mid x\!-\!1,\,$ by $\,p,q\,$ coprime, by Euclid's Lemma or unique factorization. In fact the uniqueness of the CRT solution is equivalent to the universal property of lcm, i.e. $\ p,q\mid n\iff {\rm lcm}(p,q)\mid n$

Bill Dubuque
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