Sum of four squares

karpmage

Lagrange's four-square theorem states that any natural number can be expressed as the sum of four integer squares. I've noticed that the first few values of 8n-1 can all only be expressed as a minimum of the sum of four squares. Is this true for all values of n? What's the proof behind it? Thanks.

willem2

The possible values of squares modulo 8 are 0, 1 and 4. There's only one way to make 7 from 0, 1 and 4 using a maximum of 4 numbers.

karpmage

By what i said i mean

7 = 2^2 + 1^2 + 1^2 + 1^2
15 = 3^2 + 2^2 + 1^2 + 1^2
23 = 3^2 + 3^2 + 2^2 + 1^2
31 = 3^2 + 3^2 + 3^2 + 2^2
etc.

i.e. The first few values of 8n-1 (The first 10 at least) can't be shown as a sum of less than four squares.

Not sure what you mean by modulo 8 and what-not, sorry. You might have to dumb it down a bit for me. What exactly are you trying to say?

karpmage

Sorry, meant to quote you in previous post.

MarneMath

Think about clock work math. If it's 11 a.m. and add two more hours we get 1 p.m. You can view this as 11 mod 12 + 2 mod 12 = 1 mod 12. So what he's saying is that the values in mod 8 that are squared can only be 0, 1, and 4. So now if you consider 7 mod 8, you can only get that by adding 1 + 1 + 1 + 4. If you're not familiar with this method:

http://en.wikipedia.org/wiki/Modular...AAAAAAALgaAAA=

karpmage

Ah, I remember now. This came up on a maths test I once did. Completely forgot about it though. Thanks for the help.