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Primes and Squares (Posted on 2009-08-16)

Determine all possible triplet(s) (P, Q, R) of nonnegative integers, with P ≥ Q, such that each of P² + 1 and Q² + 1 is a prime, and: (P² + 1)(Q² + 1) = (R² + 1).

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Thoughts

| Comment 2 of 3 |

Without the prime restriction, there are an infinite number of solutions with P=Q+1, R=Q^2+Q+1. But the only consecutive primes of the form X^2+1 are 2 and 5, from Q=1.

If there are other solutions in general then PQ+P/Q > R > PQ+Q/P. This implies if P and Q are not consecutive integers then P > 2Q. Also PQ+1 <= R <= PQ+P-Q.

Posted by Brian Smith on 2016-02-06 13:46:57

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