What remainder does $34!$ leave when divided by $71$? What is the remainder of $34!$ when divided by $71$?
Is there an objective way of solving this?
I came across a solution which straight away starts by stating that
$69!$ mod $71$ equals $1$ and I lost it right there.

From $$69!=1\mod 71\Rightarrow 34!36!=-1\mod 71$$ Multiplying both sides by $4$ and noting that $35\cdot 2=-1\mod 71,\ 36\cdot 2=1\mod 71$, we get $$(34!)^2=4\mod 71\Rightarrow x^2=4\mod 71$$ where $34!=x\mod 71$ So, $$71|(x-2)(x+2)\Rightarrow x+2=71, or \ x=2\Rightarrow x=69\ or\ 2$$ since $1\le x\le 70$.
