How many groups of 4 primes exist such that their sum is a prime and that $p^2+qs$ and $p^2+qr$ are squares? How many groups of 4 primes of the form $(p, q, r, s)$ exist such that their sum is a prime and that $p^2+qs$ and $p^2+qr$ are both square numbers?
I'm having trouble finding a quadruple satisfying all conditions. I've found a few satisfying the last condition, but they've always failed the first condition. Any help? Thanks!
Can someone give me a start or solution?
 A: The only solution with $\langle p,q,r,s\rangle$ distinct is $\langle 2, 7, 11, 3\rangle$.
First of all, note that by parity considerations we must have one of the four primes equal to $2$ (otherwise the sum $p+q+r+s$ is even and thus not prime).  What's more, that prime must be $p$; if $p$ were odd, then $p^2\equiv 1\pmod 4$, but then (at least) one of $p^2+qr$ or $p^2+qs$ must be $\equiv 3\pmod 4$ (since mod $4$, it's $1$ plus $2$ times an odd number) and thus can't be a square.  Therefore, $p=2$; I'll omit $p$ in favor of $2$ in what follows.
Next, we know that $4+qr$ and $4+qs$ are both square.  Since they're larger than $4$, they can be written as $(2+a)^2=4+4a+a^2$ and $(2+b)^2=4+4b+b^2$.  Thus we have $qr=a(4+a)$ and $qs=b(4+b)$.  But assuming $a$ and $b$ are different (i.e., $r$ and $s$ are different) then we can take $q=a$, $r=4+a$, $s=b$, $q=4+b$; that is, we have $s$, $q=4+s$, and $r=4+q = 8+s$ all prime. 
Now, $s$, $q$, $r$ are three consecutive members of an arithmetic progression with common difference not divisible by $3$, so one of them is a multiple of $3$ and therefore (since it's prime) must be exactly $3$; of necessity this is $s$, the smallest.  This gives $q=7$ and $r=11$, giving the listed solution.
A: Suppose $p\le q\le r\le s$. Take
$$
p=2, \qquad q=4k-1, \qquad r=s=4k+3,  \tag{1}
$$
such that $$2+q+r+s \quad \mbox{ is prime. }  \tag{2}$$  
Then all given conditions are satisfied because
$$
4 + qr = 4+qs = 4+(4k-1)(4k+3) = (4k+1)^2.
$$
For $p=2$, here are a few $q$ and $r$ satisfying $(1)$ and $(2)$: 
$$q=3, \quad r=s=7, $$
$$q=7, \quad r=s=11, $$
$$q=19, \quad r=s=23, $$
$$q=43, \quad r=s=47, $$
$$q=67, \quad r=s=71. $$
A computer search gives $74$ solutions of the form $(1)$, $(2)$ with $q=r-4$, $r=s<10000$. It is very likely that there are infinitely many solutions of this form. By the Hardy-Littlewood $k$-tuple conjecture, the numbers $4k−1$ and $4k+3$ are both prime infinitely often.
