Legendre 3 Square problem

Question

I got a little help on the way here yesterday but it seems like my question is dead.

Problem:

If $ n \in \mathbb{N} $ can be represented as $ n = n_1^2 + n_2^2 + n_3^2 ,\quad n_1, n_2, n_3 \in \mathbb{Z}, $ show that then $ n \neq 4^a(8m+7)$ for $a,m \in \mathbb{Z}.$

Solution:

$ 4^a(8m+7) \equiv 0 \mod 4, \forall a \geq 1.$ If $ n_i $ is an odd square, $ (n_i)^2 = (2k +1)^2 = 4(k^2+1) + 1 \equiv 1 \mod 4.$

So $n_1, n_2, n_3 $ must be even squares. So we can write $ (\frac{n_1}{2})^2 +(\frac{n_2}{2})^2 + (\frac{n_3}{2})^2 = \frac{n}{4} $, which yields to

$$ \frac{n}{4} = 4^{a-1}(8m+7). $$

Now I want to apply $ \mod 8 $ somehow I believe.

We got the easier parts: $ (8m+7) \equiv 7 \mod 8 ,\quad 4^{a-1} \equiv \begin{cases} 4 \quad a =2\\ 0 \quad a > 2\end{cases} $.

So $ 4^{a-1}(8m+7) \equiv 4 \quad \text{or} \quad 0 \mod 8. $

How do I proceed? $n$ is even, $ \frac{n}{4} \equiv \quad ? \mod8$. Any help is muy appriciated! <3

Edit;

The squares in $ \mathbb{Z_8} $:

• $ 0^2 = 0$
• $ 1^2 = 1 $
• $ 2^2 = 4 $
• $3^2 = 1 $
• $4^2 = 0 $
• $5^2 = 1 $
• $ 6^2 = 4 $
• $7^2 = 1 $

With the sum of three squares, we can never reach $ 7 \mod 8$?

Answer 1

What you have done so far is to prove that:

Let $a \geq 1$. If $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}4 = 4^{a-1}(8m+7)$.

Applying this observation $a$ times, one notices that if $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}{4^a} = 8m+7$. Thus, if we prove that no number of the form $8m+7$ can be written as a sum of three squares, then no number of the form $4^a(8m+7)$ can be written as a sum of three squares either.

So consider a number of the form $8m+7$. Modulo $8$, it is congruent to $7$.

As you've noticed, the squares modulo $8$ are $0$, $1$ and $4$. So if $8m+7 = x_1^2+x_2^2+x_3^2$ with $x_j \in \mathbb Z$, then

$$8m + 7 \equiv 7 \equiv x_1^2 + x_2^2 + x_3^2 \pmod 8,$$

which is not a solvable equation, since $7$ cannot be written as the sum of $0$, $1$ and $4$, using three summands (repetitions allowed).