Prove that a positive integer n is the sum of two perfect squares if and only if 2*n is also the sum of two perfect squares.

Suppose n is the sum of 2 squares. Let n=x^2+y^2. Then, (x+y)^2+(x-y)^2=(x^2+2xy+y^2)+(x^2-2xy+y^2)=2x^2+2y^2=2(x^2+y^2)=2n. Then, 2n is the sum of 2 squares. Suppose 2n is the sum of 2 squares. Let 2n=x^2+y^2. Since 2n is even, x and y are either both even or both odd. Then, (x+y)/2 and (x-y)/2 are both integers. Then, ((x+y)/2)^2+((x-y)/2)^2=((x^2+2xy+y^2)/4)+((x^2-2xy+y^2)/4)=(2x^2+2y^2)/4=(x^2+y^2)/2=2n/2=n. Then, n is the sum of 2 squares. Therefore, n is the sum of 2 squares if and only if 2n is the sum of 2 squares.

Posted by Math Man on 2020-07-20 20:06:46