Problem 3

Suppose for contradiction that $t<\frac{h-1}2$, i.e.\ $2t+1<h$. Then $0<(2t+1)^2+4<h^2+4$. Since $h=\lceil\sqrt p\rceil$ and $p\ge2013$, we have $h^2+4<2p$, so $(2t+1)^2+4$ is a positive multiple of $p$ (by the previous step) that is less than $2p$, forcing $(2t+1)^2+4=p$. But $2t+1$ is odd, and every odd square is $\equiv1\pmod8$, so $(2t+1)^2+4\equiv5\pmod8$; since $p\equiv1\pmod8$, this is a contradiction. Hence $t\ge\frac{h-1}2$, i.e.\ $x=\frac{p+1}2+t\ge\frac{p+h}2$.