Primes and squares

$\text{We can see that,} (p+q+r+s) \text{being a prime has to be odd .} \text{ Thus,at least one of } p,q,r,s \text{ has to be even.}\\\text{But since } p,q,r,s\text{ are distinct primes exactly one of them has to be 2.} \\$

$\text {Let, }$

$\normalsize\color{#333333}\begin{aligned} p^2+qs&=n^2\\\ p^2+qr&=m^2\\\\ \text{Assume q=2}&,\\ \implies p^2+2s&=n^2\\ (n^2-p^2)\pmod{4}&\equiv2s\pmod{4}\\ (n^2-p^2) \pmod{4}&\equiv0,1,-1 \hspace{16mm} \small\color{#3D99F6}\text{As, }n^2,p^2 \pmod{4}\equiv0,1\\ 2s\pmod{4}&\equiv2 \hspace{16mm} \small\color{#3D99F6}\text{As, s is an odd prime }\\ \implies(n^2-p^2)\pmod{4}&\not\equiv 2s\pmod{4}\\ \implies n^2-p^2&\neq2s \\ \text{ or } q&\neq2\\\\ \text{Similary we can prove r, s} &\neq2\\ \implies p&=2\\\\ \text{As p=2 ,we can write,}\\ 4+qs&=n^2\\ 4+qr&=m^2\\ n^2-4&=qs\\ (n+2)(n-2)&=qs\\ \text{Similarly }(m+2)(m-2)&=qr\\ \text{we have } n^2&\neq m^2 \hspace {7mm}\small\color{#3D99F6}\text{As, }s\neq r\\ \text{Let, }n&>m\\ \text{Since q,r,s are primes we}& \text{ have},\\ q&=(n-2)=(m+2)\\ r&=(n+2)\\ s&=(m-2)\\ \implies n-m&=4\hspace{16mm}\small\color{#3D99F6}\text{As } q=(n-2)=(m+2)\\ \text{Rewriting in terms of m,}\\ r&=m+6\\ q&=m+2\\ s&=m-2\\ r\pmod{3}&\equiv m\pmod{3}\\ s\pmod{3}&\equiv m+1\pmod{3}\\ q\pmod{3}&\equiv m+2\pmod{3}\\ \implies\text{ At least one of q,r,s}&\text{ is divisible by } 3 \\\\ \text{Since q,r,s are distinct}&\text{ primes exactly one of q,r,s=3}\\\\ \text{However, if r,q=3,s will}&\text{ be negative}\\\\ \implies s&=3\\ q&=7\\ r&=11\\ \text{Also if we had assumed}&\text{ m>n ,r and s would just interchange their values }\\\\ \text{Thus the possible }&\text{solutions are ,}\\ (2,7,3,11) \text{ and } &(2,7,11,3)\\\\ n+\left(\sum_{i=1}^n(p_{i}+q_{i}+r_{i}+s_{i})\right)-23&=2+2\times(2+3+7+11)-23\\ &=\color{#EC7300}\boxed{\color{#333333}25}\\\\\\\\\\\\ \text{P.S. : Sorry for the lengthy solution.} \end{aligned}$