Let a be some natural number. What are the solutions in integers of the equation ${\displaystyle x^{2}+4a=y^{2}}$? עברית (talk) 16:32, 26 July 2017 (UTC)[reply]

Any even difference of 2 squares is divisible by 4, so there are many solutions to that equation! Georgia guy (talk) 16:38, 26 July 2017 (UTC)[reply]

Hint: Try rewriting as ${\displaystyle y^{2}-x^{2}=4a}$, and then factor and solve in terms of the factors of 4a. Considering whether the factors are even or odd should let you simplify further. --Deacon Vorbis (talk) 16:47, 26 July 2017 (UTC)[reply]

• (edit conflict) Some random thoughts:

Consider a prime number p≠2 that divides both x and y, then p² must divide a since it divides 4a and not 4, and then (x/p, y/p) is solution for a/p². Hence, the general solution should be easily generated from the solutions where ${\displaystyle GCD(x,y)=2^{k}}$by stripping a from its square divisors.

For the divisibility by 2: Obviously x and y are both odd or both even. If they are even, the equation reduces to ${\displaystyle a=(y'-x')(y'+x')}$where ${\displaystyle 2x'=x,2y'=y}$. If they are odd, ${\displaystyle 4a=y^{2}-x^{2}=(2y'+1)^{2}-(2x'+1)^{2}=4y'^{2}-4x'^{2}+4y'-4x'}$which reduces to ${\displaystyle a=(y'-x')(y'+x'+1)}$. Both equations look quite attackable by looking at the pairs m,n such that a=m*n, and retrieving x,y as a function of m,n. Tigraan^{Click here to contact me} 17:24, 26 July 2017 (UTC)[reply]

A curious fact is that the equation always has at least two solutions ${\displaystyle x=a-1}$, ${\displaystyle y=a+1}$and ${\displaystyle x=-a+1}$, ${\displaystyle y=-a-1}$. These are the only solutions if ${\displaystyle a}$is simple. Ruslik_Zero 17:32, 26 July 2017 (UTC)[reply]

Thank you! עברית (talk) 19:40, 26 July 2017 (UTC)[reply]