Question

Find p such that p+1=2x^2, p^2+1=2y^2. Know that p is a prime, x and y are positive integers.​

Answer 1

Answer:

Step-by-step explanation:

Given p + 1 = 2$x^{2}$       → 1 = 2$x^{2}$ - p

         $p^{2} + 1 = 2y^{2}$

        $p^{2} + (2x^{2} - p) = 2y^{2}$

     $p^{2} - p + (2x^{2} - 2y^{2} ) = 0$

The above equation is quadratic in p.

p = [ -(-1) ± $\sqrt{(-1)^{2} - 4* 1*(2x^{2} - 2y^{2} )$ ]  ÷ 2

p =  [1 ± $\sqrt{1 - 8(x^{2}-y^{2} )}$]  ÷ 2