Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 11G / b / Solution

Because $I$ and $A$ commute, the matrix binomial theorem givesFor prime $p$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, proving the prime-power binomial congruence for a matrix

Forone has $(I+A{)}^2=2A$ and $(I+A{)}^4=-4I$. If $p=4k+1$, thenso $(-4{)}^k\equiv 1(\mathrm{mod}p)$. If $p=4k-1$, then $A^p=-A$ andso $(-4{)}^k\equiv 2(\mathrm{mod}p)$.

By Euler criterion, $(\frac{2}{p})\equiv 2^{(p-1)/2}(\mathrm{mod}p)$. In either case the preceding congruence gives $(\frac{2}{p})=(-1{)}^k$. Since $(p^2-1)/8$ has the same parity as $k$ for $p=4k\pm 1$, this is the second supplementary law for quadratic reciprocity