Another Pythagorean Triple Problem

If $c$ is a value that satisfies the conditions, then $-c$ also satisfies them. Therefore, the sum of all possible values for $c$ is invariant, is $0$ .

Applying Euclid's generating triples formula, $37=k({ m }^{ 2 }-{ n }^{ 2 })=k(m+n)(m-n)$ , for some integers $k, m,n$ , and as $37$ is prime, then we have two cases:

• $k=1$ , thus $(m+n)=1$ or $-1$ , and $m-n=37$ or $-37$ ; also, $(m+n)=37$ or $-37$ , and $m-n=1$ or -1.

Solving for $m$ and $n$ in each case, we have the following pairs $(m,n)$ : $(19,18),(19,-18),(-19,-18),(-19,18)$

Each pair is just a signs permutation. As $b=k(2mn)$ , therefore the possible values for $b$ are $684$ and $-684$ .

• The second case, $k=-1$ will generate the same values for $b$ but another permutations for $m$ and $n$ that don't affect the triple.

$c=685$ or $-685$ , so, in total, we have $4$ triples: $(37,684,685),(37,684,-685),(37,-684,685),(37,-684,-685)$ .

Consequently, $A=4$ and $B=0$ , hence, $A+B=4$ .

Note that ${ 37 }^{ 2 }=1369=684+685$ . Note that this demonstration works for every prime number.