PROBLEM 1.

If the equation y² + a p² = 2 x² (where a is a positive integer, p is an odd prime number) admits a solution (y,x) of integers with gcd(y,x)=1, how can I prove that the equation y² + p² = 2 x² also admits a solution (y,x) of integers with gcd(y,x)=1 ?

PROBLEM 2.

If the two equations y² + p² = 2 x² , y² + q² = 2 x² (where p and q are odd prime numbers) respectively admit the integer solutions (y',x') and (y '', x '') with gcd(y',x')=1 and gcd(y'',x'')=1, how can I prove that the equation y² + (pq)² = 2 x² also admits at least an integer solution (y*,x *) with gcd(y*,x*)=1 ? Is it possible to find a formula that allows to obtain (y*,x *) from the knowledge of (y',x') and (y '', x '') ?

PROBLEM 3.

If the equation y² + n² = 2 x² (where n is a positive integer greater than 3) admits integer solutions con gcd = 1, how can I prove that the equations

y² + p² = 2 x², y² + q² = 2 x², y² + r² = 2 x², y² + s² = 2 x²,......, (where p, q, r, s,....are the prime factors of n) also admit integer solutions con gcd=1?

Certain of your courtesy, I thank you very much.

Angelo.