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.​

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.​

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Autumn 7 hours 2021-11-25T23:14:01+00:00 1 Answer 0 views 0

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    2021-11-25T23:15:39+00:00

    Answer:

    Step-by-step explanation:

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

           

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