The sum of two numbers is a square. The sum of their squares is also a square. Find a method to get all such numbers.

Answer

let a + b = x^2

and a^2 + b^2 = y^2.

the second equation is the characteristic Pythagorean triplet. so we can write a and b as p^2-1 and 2p where p is a positive integer > 1 .

therefore, (p^2-1) + (2p) has to be a square number. [which is not true except for p=1/2]

thus there is no base Pythagorean triplet which satisfies this relation, but multiples of them may easily do so. like: 3,4,5. 3+4 = 7 not a square. but for 21,28,35. 21 + 28 =7^2.

thus (p^2+2p-1)(p^2-1) and (p^2+2p-1)(2p)

or better still, q^2*(p^2+2p-1)(p^2-1) and q^2*(p^2+2p-1)(2p) satisfy the given condition, where p,q are both positive integers.

Answer

let a + b = x^2

and a^2 + b^2 = y^2.

the second equation is the characteristic Pythagorean triplet. so we can write a and b as p^2-1 and 2p where p is a positive integer > 1 .

therefore, (p^2-1) + (2p) has to be a square number. [which is not true except for p=1/2]

thus there is no base Pythagorean triplet which satisfies this relation, but multiples of them may easily do so. like: 3,4,5. 3+4 = 7 not a square. but for 21,28,35. 21 + 28 =7^2.

thus (p^2+2p-1)(p^2-1) and (p^2+2p-1)(2p)

or better still, q^2*(p^2+2p-1)(p^2-1) and q^2*(p^2+2p-1)(2p) satisfy the given condition, where p,q are both positive integers.