Proving the Unsolvable: Lagrange's Theorem and 4 Squares

[RichardCypher]

Hi everybody
I'm currently reading Burton's Elementary Number Theory (almost done!) and in the chapter about Lagrange's Theorem about the sum of four squares, there is a supposedly easy question which I can't solve for some reason . I'd really appreciate a hint or two...

Prove that at least one of any four consecutive natural numbers is not a sum of two squares [that is, can't be represented as the sum of two squares of whole numbers]

Thank you all!

 

[iwin2000]

Well I think this one works?

2 + 3 + 4 + 5 = 14

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16

It has to be the SUM of TWO squares... 9 + 4 = 13. None of the others work.

 

[CRGreathouse]

Consider four consecutive numbers mod 4, then consider squares mod 4. The result follows immediately.

iwin2000: the problem was to show the result for all {n, n + 1, n + 2, n + 3}, not just for one such instance.

 

[RichardCypher]

Consider four consecutive numbers mod 4, then consider squares mod 4. The result follows immediately.

iwin2000: the problem was to show the result for all {n, n + 1, n + 2, n + 3}, not just for one such instance.

The square of any natural number mod 4 has to be 0 or 1. Therefore, the sum of two such squares mod 4 has to be 0, 1, or 2. However, out of four consecutive natural numbers mod 4, one has to be 3. Contradiction. Is that right?

Great hint! Thank-you very much

 

[CRGreathouse]

You got it.

I like minimal hints.