Looking mod p we see $m^2 \equiv n^2$ mod p, so $m \equiv \pm n$ mod p. Also if $m \le p$ then $p^2+m^2 \le 2p^2 \Rightarrow n \le \sqrt{2} p$. So $p \le n \le 2p$. Thus we see that $n = p + m$ or $n = 2p-m$. In first case our equation is: $$p^2 + m^2 = (p + m)^2 \Rightarrow 2mp = 0$$Contradiction. In second case our equation is $$p^2+m^2 = (2p-m)^2 \Rightarrow 3p^2 - 4pm = 0 \Rightarrow p(3p-4m) = 0 \Rightarrow 3p=4m$$so $4 | p$ contradiction. Thus we must have that $m>p$.

$p^2=n^2-m^2=(n-m)(n+m)$. So either $n-m=n+m=p$, which is not possible as $m$ is a positive integer, or $n-m=1$ and $n+m=p^2$. So $p^2+m^2=(m+1)^2$. Since $p<n=m+1$, assume that $p=m$. This implies $2p^2=(p+1)^2$. If $3|p$, then $2p^2\equiv0\pmod3$, but $(p+1)^2\equiv1\pmod3$. Otherwise, $2p^2\equiv2\pmod3$, but $(p+1)^2\equiv0$ or $1\pmod3$. So $p\ne m$. Since $p<m+1$, $p<m$, so $m>p$.

Very good

$p^2=n^2-m^2=(m-n)(m+n)$, n and m are positive integer from there we get that $n<m$ , also $n-m=1$ then n=m+1 Now we get thjs equation ,$p^2+m^2=(m+1)^2$, And ,$p^2=2m+1$ ,the little m integer is 4 $m^2>2m+1$ for $m>3$ , from there we get $m>p$

oki

$(n-m)(n+m)=p^2$ $n+m=p^2\hspace{5mm}n-m=1\Rightarrow m=\frac{p^2-1}{2}$ We are trying to prove $\frac{p^2-1}{2}>p$ $$p^2-1>2p$$$$(p-1)^2>2$$Satisfies for $p\geq3$. But what about $p=2$? $$4+m^2=n^2$$$$(n-m)(n+m)=4$$no solution in positive integers

m = p^2-1/2>p