Sum Of Two Squares

[MM]

Prove that if a natural number could be represented as sum of two different squares every of its natural power could be represented as sum of two squares.

[broniran]

Use induction and [tex](a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2[/tex]

[amit28it]

Bronian has suggested you the best way to solve it . He has given a good formula ....

[Guest]

Use induction and [tex](a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2[/tex]

This is called Langrange's identity.
Anyway, my solution would be:
Let's say that "a" is the number which can be written as a sum of two squares. Now, you have to prove that for any nonegative n, [tex]a^{n}[/tex] can be written as a sum of two squares.[tex]a= x^{2}+y^{2}[/tex]
As far as "n" is concerned, it can be an odd or an even number.
So, we'll take 2 cases:
Case no. 1: n is an odd number
We can say that n is 2k+1, k being a POSITIVE whole number too.
[tex]a^{n} = a^{2k+1}=a^{2k}Xa= a^{2k}X(x^{2}+y^{2})= (a^{k}Xx)^{2}+(a^{k}Xy)^{2}[/tex]
And the problem is solved.
Case no. 2:n is an even number
We can say that n is 2k+2, k being a POSITIVE whole number too.
[tex]a^{n}=a^{2k+2}[/tex]
But you didn't mention that the squares must be different from 0, so...:
[tex]a^{2k+2}=a^{2k+2}+0^{2}[/tex]
And the problem is solved.

Hope I helped you.