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I am seeking a proof for the following...

Suppose $p$ and $q$ are distinct primes. Show that $ p^{q-1} + q^{p-1} \equiv 1 \quad (\text{mod } pq)$ $$ $ I gather from Fermat's Little Theorem the following: $q^{p-1} \equiv 1 \quad (\text{mod } p)

and

p^{q-1} \equiv 1 \quad (\text{mod } q)$$

How can I use this knowledge to give a proof? I'm confident I can combine this with the Chinese Remainder Theorem, but I am stuck from here.

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    My apologies if it is, I could not find it.2012-04-19

4 Answers 4

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Write $a=p^{q-1}+q^{p-1}$. Immediately we have

$\begin{cases} a\equiv 0^{\,q-1}+q^{p-1} \equiv 1 \mod p \\ a\equiv p^{q-1}+0^{\,p-1} \equiv 1 \mod q.\end{cases}$

What does CRT tell us $a$ must be $\bmod pq$?

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    You don't even need the full CRT for this proof. The first congruence means $p$ divides $a-1$, while the second means $q$ divides $a-1$. Since $p,q$ are relatively prime, $pq$ must divide $a-1$....2012-04-20
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Hint $\rm\ mod\ (p,q)\!:\ q^{p-1}\equiv (1,0),\ \ p^{q-1}\equiv (0,1)\:$ so their sum $\rm\:x \equiv (1,1)$

So $\rm\:p,q\ |\ x\!-\!1\ \Rightarrow\ pq = lcm(p,q)\ |\ x\!-\!1.\:$ (CRT isn't needed for this simple constant case)

More generally, if $\rm\:b,c\:$ are coprime then $\rm\: e' = b^{\phi(c)}$ satisfies $\rm\: e'\equiv 0\pmod{b},\ e'\equiv 1\pmod{c}\:.\ $ Hence, $\ $ using $\rm\:e'\:$ and its complement $\rm\:\ e = 1-e'\:,\ $ where $\rm\ \ e\: \equiv 1\pmod{b},\ e\:\equiv 0\pmod{c}\:,\:$ yields the following closed-form for solutions of congruences by the Chinese Remainder Theorem

$\begin{eqnarray}\rm x\ \equiv\ a\pmod{b} \\ \rm x\ \equiv\ d\pmod{c}\end{eqnarray}\rm\ \ \iff\ \ x\ \equiv\ a\ e + d\ e'\pmod{b\:c}$

The relationship between CRT and the orthogonal idempotents $\rm\:(1,0),\ (0,1)\:$ will become clearer when you study the Peirce decomposition induced by such idempotents.

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first using Fermat's theorem,

$q^{p-1}\equiv 1 \pmod p$

Also,$\ \ \ p^{q-1}\equiv0 \pmod p$

Second using Fermat's theorem,

$p^{q-1}\equiv 1 \pmod q$

Also,$q^{p-1}\equiv0 \pmod q$

Using first part we get,

$q^{p-1}+p^{q-1}\equiv1 \pmod p$

using second part we get,

$p^{q-1}+q^{p-1}\equiv1 \pmod q$

Since $p$ and $q$ are distinct prime, therefore $\gcd (p,q)=1$

$q^{p-1}+p^{q-1}\equiv1 \pmod {pq}$

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Since $\gcd(p,q)=1$, by Fermat little theorem, $p^{q-1}\equiv1 \pmod q$.

Now, q^{p-1}\equiv 0 \pmod q$ $(\because q\mid q^{p-1}).

Thus we have, p^{q-1}+q^{p-1}\equiv 1 \pmod q\tag{1} Again by Fermat little theorem,

q^{p-1}\equiv 1 \pmod p

And p^{q-1}\equiv 0 \pmod q$ $(\because p\mid p^{q-1})

From this we have, p^{q-1}+q^{p-1}\equiv 1 \pmod p\tag{2}$ From $(1)$ and $(2), we have,

p^{q-1}+q^{p-1}\equiv 1 \pmod {pq} (\because (p,q)=1)$