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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.

Bill Dubuque
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Dominick Gerard
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4 Answers4

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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 $\bmod pq$?

anon
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  • From here, could we say $a \equiv a_0 \quad \text{mod } pq$? Would that be enough to complete the proof? – Dominick Gerard Apr 19 '12 at 21:40
  • @Dominick: What's $a_0$ to you? – anon Apr 19 '12 at 21:41
  • can we say that it is 1? we know it must be in the set of all integers right? – Dominick Gerard Apr 19 '12 at 21:43
  • @Dominick: Given $a\equiv1~(p)$ and $a\equiv1~(q)$, at the very least CRT says there is a unique solution modulo $pq$. Obviously $1$ is a solution, so $a\equiv1~(pq)$ by uniqueness. – anon Apr 19 '12 at 21:47
  • thanks, also can you elaborate on why we can go from $a=p^{q-1} + q^{p-1}$ to $a \equiv 0^{q-1} + q^{p-1} \equiv p^{q-1} + 0^{p-1}$? – Dominick Gerard Apr 19 '12 at 22:02
  • @Dominick: You're mixing two different congruences together, can't do that because the moduli (respectively $p$ and $q$) are different. For the first, $p\equiv0~(p)$ implies $p^n\equiv0^n~(p)$ for any $n\ne0$. In other words, any nonzero power of $p$ is a multiple of $p$ (this should be obvious!) and thus can be taken out. – anon Apr 19 '12 at 22:05
  • 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 $a-1$.... – N. S. Apr 20 '12 at 04:24
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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 $\rm\:p,q\ |\ x\!-\!1\ \Rightarrow\ pq = lcm(p,q)\ |\ x\!-\!1\:$ (or use CCRT = constant case CRT)

Generally if $\rm\:p,q\:$ are coprime then $\rm\: e' = p^{\phi(q)}$ satisfies $\rm\: e'\equiv 0\pmod{\!p},\ e'\equiv 1\pmod{\!q}.\ $ Hence using $\rm\,e'$ and its complement $\rm\:\ e = 1-e',\ $ where $\rm\:\! \ e\: \equiv 1\pmod{\!p},\:\!\ e\:\equiv 0\pmod{\!q}\:,\:$ yields this closed-form for solutions of congruences via CRT = Chinese Remainder Theorem

$$\bbox[1px,border:1px solid #c00]{\bbox[8px,border:1px solid #c00]{\begin{eqnarray}\rm x\ \equiv\ a\!\!\pmod{\!p} \\ \rm x\ \equiv\ b\!\!\pmod{\!q}\end{eqnarray}\rm \iff \ x\ \equiv\ a\, e + b\, e'\!\!\!\pmod{\!pq}}}\qquad$$

The relationship between CRT and the orthogonal idempotents $\rm\:(1,0),\ (0,1)\:$ is clarified when onle learns the Peirce product ring decomposition $\,\Bbb Z_{pq}\cong \Bbb Z_p\times \Bbb Z_q\,$ induced by such idempotents, e.g. see this answer for how the common CRT formula follows immediately.

Bill Dubuque
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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 $\gcd (p,q)=1$

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

Mathew Mahindaratne
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Siddhant Trivedi
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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,

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

Mathew Mahindaratne
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Kns
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