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Question: Let $p$ and $q$ be distinct primes and let $n=pq$. Prove that if $a=p+n\mathbb{Z}$ and $b=q+n\mathbb{Z}$, then $a^{q-1}+b^{p-1}=1.$

I'm not entirely convinced that the above is true. I have applied Fermat's little theorem. Since $p$ and $q$ are distinct primes - $a^{q-1}\equiv 1 \pmod q$ and $b^{p-1}\equiv 1 \pmod p.$ Where I am getting stuck is how to apply that second given: $a=p+n\mathbb{Z}$ and $b=q+n\mathbb{Z}.$

Added due to player3236:
$a^{q-1}+b^{p-1}\equiv 1 \pmod q$ since $b^{q-1}\equiv 0\pmod q$. Similarly for $a$ and $p$. This shows that $a^{q-1}+b^{p-1}-1\equiv 0 \pmod {pq}$,or that $a^{q-1}+b^{p-1}-1$ is divisible by $pq$ since $p$ and $q$ are distinct primes.

Arturo Magidin
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emka
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  • What are $a^{q-1} \pmod p$ and $b^{p-1} \pmod q$? And by extension, what are $a^{q-1}+b^{p-1} \pmod {p } \text{ and }\pmod q$? – player3236 Nov 12 '20 at 16:15
  • $a$ is a coset in $\Bbb Z/n\Bbb Z$. How is $a^{q-1}$ then meant? – Dietrich Burde Nov 12 '20 at 16:15
  • @player3236 I edited my question to address what you mentioned. I didn't feel comfortable typing that amount of LaTeX without seeing it in real time. – emka Nov 12 '20 at 16:27
  • @Dietrich Burde, I'm not sure I follow. – emka Nov 12 '20 at 16:27
  • Please edit your post for clarity. Did you intend to claim that $p^{q-1}+q^{p-1}\equiv 1 \pmod {pq}$? You wrote that it should equal $1$ which, as has been remarked, does not make sense. – lulu Nov 12 '20 at 16:39
  • Well, since you have shown that $a^{q-1} + b^{p-1} \equiv 1 \pmod {pq}$, you proved what the question asked for. – player3236 Nov 12 '20 at 16:52
  • Emka, I said that $a=p+n\Bbb Z$ is a set of numbers, namely ${\cdots ,-2n+p,-n+p.p,p+n,p+2n,\cdots}$. What is a power of a set? – Dietrich Burde Nov 12 '20 at 17:49

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Well $a \equiv p\pmod q$ and $b\equiv q\equiv 0 \pmod q$ so $a^{q-1}+b^{p-1}\equiv p^{q-1}\equiv 1 \pmod q$.

Likewise $a\equiv p\equiv 0\pmod p$ and $b\equiv q\pmod p$ so $a^{q-1}+b^{p-1}\equiv q^{p-1}\equiv 1 \pmod p$.

Now by Chinese Remainder theorem we know any mutual solution $a^{q-1}+b^{p-1}\equiv 1 \pmod p$ and $a^{q-1}+b^{p-1}\equiv 1 \pmod q$ is unique $\mod pq$. And $a^{q-1} + b^{p-1}\equiv 1 \pmod {pq}$ is a solution. So it is the only solution.

fleablood
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