Question

find the integer solutions to A^2 + 2*B^2 = C^2?

Answer 1

the equation have an infinite number of solutions in positive integers (b=0 isn't interesting); we can find them all, using the same method as used to find all Pythagorean triples. Firstly note we can assume a,b,c have no common factor; if they did, we could divide them out. Conversely, any solution admits further solutions of the form ka,kb,kc for any positive integer k. Furthermore, if any two of a,b,c have a common factor, then it's also a factor of the third. So let's assume the values are pairwise coprime.

2b²=c²−a²=(c−a)(c+a)

The two factors on the right-hand side have a difference of 2a so are of the same parity. They cannot be both odd, since their product is even. Also if the two factors had any common factor, it would also be a factor of their common difference 2a. If any common prime factor of (c−a) and (c+a)divided a, then it would also divide (c−a)+a=c, contrary to assumption. So both terms on the right-hand side are even, and they have no common factor other than 2. This means one factor must be the square of an even number, let's call it m, and the other is twice the square of an odd number, let's call that n.

So we have the following solutions for any even m and odd n, with (m,n)=1

a=|m²−2n²|/2
b=mn
c=(m²+2n²)/2

with the absolute value needed in a depending on which term in the right-hand side product is which. These are the primitive solutions; any integer multiple of them will also be a solution.

m=2,n=1 gives the easy solution a=1,b=2,c=3 Coprime values of different parity give infinitely many others. m=4,n=3 gives the solution 1²+2⋅12²=17²

Note you can also generate an infinite number of solutions with a=1, and in fact c/b are the alternate continued fraction convergents of √2, 3/2,17/12,99/70,…3/2,17/12,99/70,… - however, for any fixed b or c, there are only finitely many solutions (perhaps zero for some values).