Question

find a Pythagorean triplet whose one member is 5 ​

Answer 1

Answer:

There are two Pythagorean triples that have 5 as one of their positive integer values:

(3, 4, 5) and (5, 12, 13).

Interestingly, we can combine the two related equations to give 3² + 4² + 12² = 13², which as a rectangular prism has all integer sides and an integer long diagonal. This is true when n is any positive integer for n² + (n+1)² + (n(n+1))² = (n(n+1) + 1)².

1² + 2² + 2² = 3², 2² + 3² + 6² = 7², 3² + 4² + 12² = 13², 4² + 5² + 20² = 21², etc.