Prove that there exists $s$ such that $s(ab-1)^n +1$ is composite I find this interesting question in a number theory book.
Given two positive integers $a, b$ such that $a>1, b>1, \gcd(a, b)=1$. Prove that there exists a positive integer $s$ such that $s(ab-1)^n + 1$ is a composite number $\forall n \in \mathbb{N}.$
Its solution is short but beautiful :)
 A: Hint $\,\ c\mid 1\!+\!(\overbrace{c\!-\!1}^{\Large s})(\overbrace{c\!+\!1}^{\large ab-1})^n$
Derivation $\ $ Suppose $\ c\mid 1\!+\!st^n\,$ for all $\ n\ge 0.\,$  $\ n=0\,\Rightarrow\, c\mid 1\!+\!s,\,$ so $\,s = jc\!-\!1,\,$ some $\,j\in\Bbb Z.\,$   $\,{\rm mod}\ c\!:\,\ {\rm for\ \ all}\ \ n\!:\,\ 0\,\equiv\, 1\!+\!st^n\equiv 1\!-t^n\!\! \iff\! t^n\equiv 1\!\!\!\overset{\  \ \large n=1}\iff\!t\equiv 1\!\iff\! t= kc\!+\!1,\,$ some $\,k\in\Bbb Z.\,$ Therefore for all $\,n\!:\ c\mid 1+st^n\! \iff\! 1\!+\!st^n = 1\!+\!(jc\!-\!1)(kc\!+\!1)^n\ $ for integers $\,j,k.\phantom{I^{I^{I^I}}}$
A: Bill's hint is more than enough and kills it beautifully, but as another solution, let $p$ be a prime divisor of $ab-2$ and take $s = (p-1)!.$ By Wilson's theorem, we'd have
$$(p-1)!(ab-1)^n + 1 \equiv (-1)(1)^n + 1 \equiv 0 \;\; (\text{mod} \; p)$$
P.S. Coprime condition is redundant.
A: If $s=ab-3$ then 
$$s(ab-1)^n+1\equiv -1(1)^n+1\equiv 0\pmod{ab-2}$$
Hence $(ab-2)|s(ab-1)^n+1$.
Since $a>1$ and $b>1$ and $\gcd(a,b)=1$, we have $ab-3\ge 3>1$. Hence $$ab-2<(ab-3)(ab-1)+1$$
Noting that $ab-2>1$ we have that $s(ab-1)^n+1$ is composite for all $n\in\mathbb{N}$ since it is divisible by but not equal to $ab-2$.
