Past exam of the short options course of the University of Oxford / 2021 / 1 / b

The supplementary law for the Legendre symbol is $(-1/p)=(-1{)}^{(p-1)/2}$ for odd prime numbers $p$. For odd integers $a,b$,and the last term is even. Induction over the prime factors of $Q$ therefore gives ${\sum }_i(p_i-1)/2\equiv (Q-1)/2(\mathrm{mod}2)$. ConsequentlyFor $Q=1$ the exponent is zero.

The second supplementary law for the Legendre symbol is $(2/p)=(-1{)}^{(p^2-1)/8}$. To multiply these signs, observe that for odd $a,b$,Both factors in the numerator are divisible by $8$, so the right side is even. Applying this identity repeatedly to the prime factorization givesAgain the convention at $Q=1$ agrees.