From $p | (2t^2 - 1)$, $t^2 \equiv 1/2 \pmod{p}$ From $p^2 | (2st + 1)$, $st \equiv -1/2 \pmod{p}$ If $p|t$, then $2t^2 - 1 \equiv -1 \pmod{p} \not\equiv 0 \pmod{p}$ which does not satisfy the first condition. So $t \not\equiv 0 \pmod{p}$. $t^2 + st = t(s + t) \equiv 0 \pmod{p}$ But since $t \not\equiv 0 \pmod{p}$, $s + t \equiv 0 \pmod{p}$. Then $p| s + t \Leftrightarrow p^2|s^2 + 2st + t^2 \Leftrightarrow p^2|s^2 + 2st + t^2 - (2st + 1) \Leftrightarrow p^2| s^2 + t^2 - 1$.