Question

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

Answer 1

sure u did this...

\(p(p-1)=2(y-x)(y+x)\)

Answer 2

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

Answer 3

http://www.wolframalpha.com/input/?i=solve++p^2%2Bp%2B2%3D2%28x^2%2By^2%29%2C+p+%3D+x+%2B+y%2C+y+%3D+3x+-+1

Answer 4

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

Answer 5

how u found out that \(p=x+y\)

Answer 6

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

Answer 7

\(p\) 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 \(p=x+y\)