b) Denote \(2n=a^2+b^2\left(a,b\in Z\right)\), then a², b² are both odd or even. So, a,b are both odd or even. Hence, \(a\pm b⋮2\) or \(\dfrac{a\pm b}{2}\in Z\).

We have :

\(n=\dfrac{a^2}{2}+\dfrac{b^2}{2}=\left(\dfrac{a^2}{4}+2.\dfrac{a}{2}.\dfrac{b}{2}+\dfrac{b^2}{4}\right)+\left(\dfrac{a^2}{4}-2.\dfrac{a}{2}.\dfrac{b}{2}+\dfrac{b^2}{4}\right)\)

\(=\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{a-b}{2}\right)^2\) is the sum of 2 perfect squares

We have : n = a² + b² (a ; b e z) then:

2n = 2a² + 2b² = a² - 2ab + b² + a² + 2ab + b² = (a - b)² + (a + b)²

a) Denote \(n=a^2+b^2\left(a,b\in Z\right)\), then :

\(2n=2a^2+2b^2=a^2-2ab+b^2+a^2+2ab+b^2=\left(a-b\right)^2+\left(a+b\right)^2\)