Since $r$ must be odd and $p=r-2q$, it must be $p>2$. Since $r=p+2q$, thus $s^2 = 2p+3q$, it follows $p>3$, and moreover $p\equiv 2\pmod{3}$; this in turn forces $q\equiv 1\pmod{3}$. So now there is extremely little casework, leading to $(p,q,r,s,n)=(5,13,31,7,2015)$ as sole answer.

mavropnevma wrote: Since $r$ must be odd and $p=r-2q$, it must be $p>2$. Since $r=p+2q$, thus $s^2 = 2p+3q$, it follows $p>3$, and moreover $p\equiv 2\pmod{3}$; this in turn forces $q\equiv 1\pmod{3}$. So now there is extremely little casework, leading to $(p,q,r,s,n)=(5,13,31,7,2015)$ as sole answer. Why?

Vietnam06727 wrote: mavropnevma wrote: Since $r$ must be odd and $p=r-2q$, it must be $p>2$. Since $r=p+2q$, thus $s^2 = 2p+3q$, it follows $p>3$, and moreover $p\equiv 2\pmod{3}$; this in turn forces $q\equiv 1\pmod{3}$. So now there is extremely little casework, leading to $(p,q,r,s,n)=(5,13,31,7,2015)$ as sole answer. Why? Everything is obvious read again

Hypernova wrote: Vietnam06727 wrote: mavropnevma wrote: Since $r$ must be odd and $p=r-2q$, it must be $p>2$. Since $r=p+2q$, thus $s^2 = 2p+3q$, it follows $p>3$, and moreover $p\equiv 2\pmod{3}$; this in turn forces $q\equiv 1\pmod{3}$. So now there is extremely little casework, leading to $(p,q,r,s,n)=(5,13,31,7,2015)$ as sole answer. Why? Everything is obvious read again I did not understand.please tell me once

If q was 0 mod 3, it wouldn’t be prime. If q was 2 mod 3, r = p + 2q = 0 mod 3 so r wouldn’t be prime. q = 1 mod 3 is the only option.

I understood that but i didn't understand the case work,please provide a solution to the case work

$9999 \geq n=pqr \geq p(p+2)(3p+4) \to p \leq 13$ and from $p\equiv 2\pmod{3}$ follows that $p=5,11$ If $p=11$ then $9999 \geq 11*q(2q+11) \to q \leq 17$ and so $q=13$ But $s^2=2p+3q=61$ is not square. So $p=5$ $9999 \geq 5q(2q+5) \to q \leq 29 \to q=7,13,19$ but only for $q=13$ we get $s^2=2p+3q=49$ as square of prime.

RagvaloD wrote: $9999 \geq n=pqr \geq p(p+2)(3p+4) \to p \leq 13$ and from $p\equiv 2\pmod{3}$ follows that $p=5,11$ If $p=11$ then $9999 \geq 11*q(2q+11) \to q \leq 17$ and so $q=13$ But $s^2=2p+3q=61$ is not square. So $p=5$ $9999 \geq 5q(2q+5) \to q \leq 29 \to q=7,13,19$ but only for $q=13$ we get $s^2=2p+3q=49$ as square of prime. thanks