Determine all possible triplet(s) (P, Q, R) of nonnegative integers, with P ≥ Q, such that each of P^{2} + 1 and Q^{2} + 1 is a prime, and: (P^{2} + 1)(Q^{2} + 1) = (R^{2} + 1).
If p & q are primes p^2+1 is even as q^2+1. So their product will take the form 4k (k integer).
If R is even R^2+1 is odd. Is R is odd R^2+1 or is odd or is 2k (k odd).
So no way, except if one of the primes involved is 2 (an special prime, non 4k 1 or 4k3).
In this case there are other solutions, as p=2 q=3 r=7. In general when the last digit of r is 3 or 7 or 2 or 8 q^2 will be integer and perhaps q also.
But as the problem imposes p>q, no way. The only solution is p=2 q=1
Edited on February 6, 2016, 6:23 pm

Posted by armando
on 20160206 18:02:12 