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Kissing Cousins (Posted on 2017-03-02)

Let P be prime, such that
(i) P is a sum of two consecutive squares, and
(ii) Q = (P+4) is also prime.

Prove that Q is a sum of two squares of numbers differing by 3.

Claim: the number of such 'Kissing Cousin' prime pairs is finite - true or false?

See The Solution Submitted by broll

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Part One Proof (spoiler)

Comment 1 of 1

Part one is true even if P and/or Q are not prime.

If P = n^2 + (n+1)^2 = 2n^2 + 2n + 1

Then

P+4 = 2n^2 + 2n + 5 = (n-1)^2 + (n+2)^2

Posted by Steve Herman on 2017-03-02 09:55:35

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