Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / C / C2 / b / i

For odd $m,n$,The last term is even, so induction over the prime factors, including repetitions, gives $\frac{b-1}{2}\equiv {\sum }_i\frac{p_i-1}{2}(\mathrm{mod}2)$. Using the first supplementary law in each Legendre factor,For $b=1$, the product is empty and both sides equal one.