I have been trying to figure out this problem for hours with no luck.
There exist infinitely many primitive Pythagorean triples in which one of the legs is 8 (which means the difference between the other, longer, leg and hypotenuse is 8). Give the proof for this theorem and the three first triples, those with the shortest hypotenuse.
naturally i was curious, just how many such triples could we find? it was clear to me that b+a, being the larger of the 2 factors of 64, had to be 16 or 32.
b+a = 16 --> a = 6, b = 10. this is not primitive.
b+a = 32 --> a = 15, b = 17.
so it would appear there is but one such primitive triple: (8,15,17).
since the original question asks for the first 3 such triples, i suspect we are not answering the same question the OP is trying to ask.