Show that $23a^2$ is not the sum of 3 squares.

Question

I know that Legendre's theorem states that a number is expressible as a sum of 3 squares iff. it's not of the form $4^x (8m+7)$, so I need to show that $23a^2$ is of this form, how could I go about doing this?

Answer 1

Note that $4^x(8m + 7)$ is a product of two terms:

• a power of $4$, and
• the remaining odd part.

This motivates assuming $a$ to be of the form $2^xr$ where $r \ge 1$ is odd and $x \ge 0$. (Both $r$ and $x$ are integers.)
Note that every integer can indeed be written in the above form (in a unique manner).

Now, we get that $a^2 = 4^xr$. This is promising because we have gotten a $4^x$ term.
This shows that $$23a^2 = 4^x(23r^2).$$

Now, we need to show that $23r^2$ is of the form $8m + 7$. Note that $23 = 8\cdot2 + 7$.
So, if we can show that $r^2$ is of the form $8k + 1$, then we would be done.

This can be done easily by exhaustion.
Since $r$ is odd, there are only the following possibilities for $r$:
$r$ is of one of the following forms:

1. $8k + 1$
2. $8k + 3$
3. $8k + 5$
4. $8k + 7$

You can square each and verify that $r^2$ is always of the form $8k + 1$. Thus, $23a^2$ further simplifies as $$\begin{align}23a^2 &= 4^x(23r^2)\\ &=4^x(23(8k+1))\\ &=4^x(23\cdot8k + 16 + 7)\\ &=4^x((23k + 2)\cdot8 + 7)\\ &= 4^x(8m + 7),\end{align}$$ as desired.

Answer 2

If a number $n$ is of this form, then:

$$n = 4^x \left(8m + 7\right) = 4^x \cdot 8m + 7 \cdot 4^x = 23a^2$$

Therefore

$$n \equiv 7 \cdot 4^x \equiv \begin{cases}0 & \text{if } x > 1 \\4 & \text{if } x = 1\\7 & \text{if } x = 0\end{cases}\pmod{8}$$

and

$$n \equiv 23 a^2 \equiv \begin{cases}0 & \text{if } a \equiv 0 \pmod{4}\\4 & \text{if } a \equiv 2 \pmod{4}\\7 & \text{if } a \text{ is odd}\end{cases}\pmod{8}$$

Lets bust the case where $n \equiv 7 \pmod{8}$. It means that $a$ is odd. So $23a^2$ is odd, too. However $4^x \left(8m + 7\right)$ is even. Because $n$ can't be even and odd at the same time, it is a contradiction.

Next case: $n \equiv 4 \pmod{8}$

$$4 \left(8m + 7\right) = 23a^2 \quad a \in \{2,6,10,12,...\}$$

Because $\left(8m + 7\right)$ is not divisible by $4$, $a$ needs to be $2$, then:

$$8m + 7 = 23$$

Therefore $m = 2$.

So actually it is of the form!

$$4^1 \left( 8 \cdot 2 + 7 \right) = 23 \cdot 2^2$$

One possible solution: $x = 1$, $m = 2$, $a = 2$