HongKong 2022-23 IMO Selection Tests

Clearly, $p\ne 2,5$. Let $m=5^{4n}-5^{3n}+5^{2n}-5^n+1$. Then $$(2\cdot 5^{2n}-5^n+2{)}^2-5\cdot 5^{2n}=4m\equiv 0(\mathrm{mod}p).$$ This gives $5\equiv (5^{-n}(2\cdot 5^{2n}-5^n+2){)}^2(\mathrm{mod}p)$. Using the Legendre symbol, we have $\left(\frac{5}{p}\right)=1$. On the other hand, we have $$(5^{2n}-5^n+1{)}^2+5^n(5^n-1{)}^2=m\equiv 0(\mathrm{mod}p).$$ As above, this implies $\left(\frac{-5^n}{p}\right)=1$. It follows that $$\left(\frac{-1}{p}\right)=\left(\frac{-5^n}{p}\right)\left(\frac{5^n}{p}\right)=1\cdot 1^n=1.$$ Therefore, $p\equiv 1(\mathrm{mod}4)$.