Prove that if $a(a^{pq-p-q+2} -b) $ is not divisible by $pq,$ then $a-b$ is not divisible by $pq$.

Question

Let $a$ and $b$ be positive integers and p and q be two distinct primes. Prove that if $a(a^{pq-p-q+2} -b) $ is not divisible by $pq,$ then $a-b$ is not divisible by $pq$.

I was trying to do a proof by contradiction:

Suppose that $pq\mid a-b$. We have that $p\mid a-b$ and $q\mid a-b$ and so:

$a \equiv b \pmod{pq} \\ a \equiv b \pmod{p} \\a \equiv b \pmod{q}$

I notice that I can refactor it as $a((a^{q-1})^{p}(a^{-q})(a^2) -b))$ or $a((a^{p-1})^{q}(a^{-p})(a^2) - b))$, so possibly Fermat's little theorem could be helpful.

Then I am trying to show that

$a(a^{pq-p-q+2} -b) \equiv a_1 \pmod{p} \\ a(a^{pq-p-q+2} -b) \equiv a_2 \pmod{q}.$

Then I use the Chinese Remainder Theorem to show that $a(a^{pq-p-q+2} -b) \equiv 0 \pmod{pq}$

but I have not had any success with doing this, any help is appreciated.

Answer 1

Contrapositive is: $\bmod m\!=\!pq\!:\,\ a\equiv\color{#c09} b\,\Rightarrow\, 0\equiv a(a^{\large \phi(m)+1}\!-\color{#c00}b)\equiv \color{#0a0}{a^{\large 2}}(a^{\large \phi(m)}\!-1),\, $ which is true more generally for all integers $m$ whose prime factors occur to power at most $\rm\color{#0a0}{two},$ by

Theorem $ $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{\large e_{1}}\cdots\:p_k^{\large e_k}\ $ and suppose also that for all $\,i,\,$ $\ \color{#0a0}{e_i\le e}\ $ and $\ \phi(p_i^{\large e_{i}})\mid f.\ $ Then $\ m\mid \color{#0a0}{a^{\large e}}(a^{\large f}-1)\ $ for all $\: a\in \mathbb Z.$

Proof $\ $ Notice that if $\ p_i\mid a\ $ then $\:p_i^{\large e_{i}}\mid \color{#0a0}{a^{\large e}}\ $ by $\ \color{#0a0}{e_i \le e}.\: $ Else $\:a\:$ is coprime to $\: p_i\:$ so by Euler's phi theorem, $\!\bmod q = p_i^{\large e_{i}}:\, \ a^{\large \phi(q)}\equiv 1 \Rightarrow\ a^{\large f}\equiv 1\, $ by $\: \phi(q)\mid f\, $ and modular order reduction. Thus since all prime powers $\ p_i^{\large e_{i}}$ divide $\, a^{\large e} (a^{\large f} - 1)\ $ so too does their lcm = product = $m$.

Examples $\ $ You can find many illuminating examples in prior questions, e.g. below

$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$

$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$

$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$

$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$

Answer 2

$pq-q-p+2=p(q-1)-q+2=(q-1)(p-1)+1$

If $a=b$ mod $p$ and $a=b$ mod $q$

Little fermat applied to $p$ implies that $a^{pq-q-p+2}=a^{(p-1)(q-1)+1}=a^{q-1+1}=a$ $mod$ $p$, this implies that $a^{pq-p-q+2}-b=a^{pq-p-q+2}-a=0$ $mod$ $p$,

a similar argument shows that $a^{pq-p-q+2}-b=0$ $mod$ $q$.