Prove $2(x^2 + y^2)$ is the sum of two squares

Question

Okay so, first time using maths stack exchange (that's why the layout may be off), but I had two questions (both related):

a) How would I prove (by direct argument) that $2(x^2 + y^2)$ is the sum of two squares.

I tried and well all I could reach was the conclusion that $2x^2 + 2y^2 = m^2 + n^2$ (basically I got nowhere).

If you could help with the method of approaching this question it would be very appreciated.

b) If a question is worded as prove or disprove, when do you know when not to disprove, i.e. I kept attempting to disprove a statement (by counter example) and wasted a lot of time in doing so; is there trick?

I could do most of the other proofs fine but this one, unfortunately,got me stuck.

Thank you :)

Answer 1

A neat way to express any multiple $n(x^2+y^2)$ as a sum of two squares is to first express $n$ as the sum of two squares and then use the identity $$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(ay-bx)^2.$$

So, since $2=1^2+1^2$, you are looking for $$(1^2+1^2)(x^2+y^2)=(x+y)^2+(y-x)^2.$$

Answer 2

You could write $((\sqrt 2)x)^2 +((\sqrt 2)y)^2$.

Answer 3

The expression can be rewrited as: $$(x+y)^2+(x-y)^2=x^2+y^2+2xy+x^2+y^2-2xy=2(x^2+y^2)$$

Answer 4

Using the identity $\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2$, $2\left(x^2+y^2\right)=\left(1^2+1^2\right)\left(x^2+y^2\right)=\left(x+y\right)^2+\left(x-y\right)^2$