Find all prime numbers $p$ and $q$ such that $2^2+p^2+q^2$ is also prime. Please remember to hide your solution. (by using the hide tags of course.. I don't literally mean that you should hide it )
Problem
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Tags: quadratics, number theory, prime numbers, algebra
Farenhajt
16.08.2011 07:04
If $p,q$ are both odd or both even, then the expression is even and greater than $2$. Thus WLOG $q=2$ and $p$ is odd. Then $p^2+8$ must be prime. If $p=6k\pm 1$, then $p^2+8=36k^2\pm 12k+9=3(12k^2\pm 4k+3)$. Since the expression in the parentheses can't be equal to $1$, which is easily shown, we get that $p^2+8$ is composite.
Therefore $p$ must be odd and can't be representable as $6k\pm 1$. The only such prime is $p=3$, and $3^2+8$ is prime.
Thus the only solutions are $(p,q)\in\{(2,3),(3,2)\}$.
Matheater1
20.08.2011 07:37
let $s=2^2+p^2+q^2$ if both $p,q$ are odd,$s$ is composite. so let's consider $q=2,$ then$s=p^2+8$ now let's assume $p=3k \pm 1$($k$ is natural) then $s=3(3k^2 \pm 2k +3)$ hence $3k^2 \pm 2k +3$ has to have the value $1$.but by quadratic equation it is not possible for any natural number. so $p$ has to be the form $3k$.the only prime of that form is $3$ so $(p,q)=(2,3),(3,2)$