Sum of four squares

I was looking for numbers who can be expressed as sum of exactly four squares and not less. And I think I have found them. They are all the integers of the form

$$4^{n}\,(7+8k);\;k,\,n\in\mathbb{N}$$

I have no idea how to prove this statement and I wonder if ALL the numbers which need four squares are this kind of numbers.

Edit

Thanks to the comments and the answer the statement can be more precise

A number can be expressed as the sum of four squares and not less if and only if it has the form

$4^n\,(7+8k);\;k,\,n\in\mathbb{N}$

• Make that $4^n(7+8k)$. This is a well-known, but very deep, theorem of Gauss. – Lord Shark the Unknown Jul 14 '17 at 12:15
• See Legendre's three square theorem and Lagrange's four square theorem together, keeping in mind $0^2$ is allowed. And if anyone is still wondering about $178$ (it was an issue raised in at least two separate comments before they were deleted), we have $12^2 + 5^2 + 3^2$ or $13^2 + 3^2$. – Arthur Jul 14 '17 at 12:17
• Thank you for the correction. I reviewed my notes and it was just a typo. But I did not know that the property was known. I am glad anyway to have rediscover it by myself :) – Raffaele Jul 14 '17 at 18:08
• @Arthur Legendre's theorem page looks quite wrong on wikipedia since the OEIS page and my result are about FOUR squares and not three. What do you think about? – Raffaele Sep 5 '17 at 8:59
• @Raffaele No, I stand by what I said. Lagrange's four square theorem says that any number can be written as the sum of four squares (allowing for $0^2$). Legendre says that the number can be written as the sum of three squares iff it is not of the form $4^n(7+8k)$. Thus the numbers that can be written as a sum of four squares, but not as a sum of three are exactly the ones excluded by Legendre. – Arthur Sep 5 '17 at 9:04

At least it is not hard to see that $4^n(7+8k)$ cannot be written as sum of three squares: $a^2+b^2+c^2\equiv m\pmod 4$ where $m$ is the number of odd squares on the left. Hence for $n\ge 1$, any representation of $4^n(7+8k)$ must use three even squares. But then this corresponds to the representation $4^{n-1}(7+8k)=(a/2)^2+(b/2)^2+(c/2)^2$ with smaller $n$. We are thus reduced to the case $n=0$, i.e., $7+8k=a^2+b^2+c^2$. By the above argument, $a,b,c$ must be odd. As odd squares are $\equiv 1\pmod 8$, we obtain $7+8k\equiv 3\pmod 8$, contradiction.