Determine all primes $p$ for which the system
$p + 1 = 2x^2$
$p^2 + 1 = 2y^2$
has a solution in integers $x$; $y$.

Question

Determine all primes $p$ for which the system
$p + 1 = 2x^2$
$p^2 + 1 = 2y^2$
has a solution in integers $x$; $y$.

anonymous · Answer

sure u did this...

$p(p-1)=2(y-x)(y+x)$

anonymous · Answer

and use this  $1<x<y<p$  since without loss of generality we can assume $x$ , $y$ are positive integers

experimentx · Answer

experimentx · Answer

since -1 is not prime ... i guess the answer would only be 7

anonymous · Answer

how u found out that $p=x+y$

experimentx · Answer

p(p-1) = 2(x+y)(x-y)
since p ! = 2*anything and p > p - 1
p-1 = 2(x-y) and p = x+y

anonymous · Answer

$p$ is odd so $p | (y-x)(y+x) $ we know that $p>y-x$ hence $p | x+y$....on the other hand $x+y< 2p$ so $p=x+y$