Some Squares(2)

The numbers that cannot be written as the sum of three squares are precisely those that can be written in the form $4^a (8b + 7)$ , for $a,b \in \mathbb{N}$ . This is Legendre's Three Square Theorem ; it's proof is, by all indications, extremely complicated.

Consider the excluded numbers for the first few values of $a$ :

• $a = 0$ : $\qquad 7, 15, 23, 31, ....\$ i.e. $\dfrac{1}{8}$ of all natural numbers.
• $a = 1$ : $\qquad 28, 60, 92, 124, ....\$ i.e. $\dfrac{1}{4 \cdot 8} = \dfrac{1}{32}$ of all natural numbers.
• $a = 2$ : $\qquad 112, 240, 368, 496, ....\$ i.e. $\dfrac{1}{4^2 \cdot 8} = \dfrac{1}{128}$ of all natural numbers.

$\qquad \; \; \vdots \\ \qquad \; \; \vdots$

In general, for any $a, \dfrac{1}{4^a \cdot 8}$ of the natural numbers cannot be expressed as the sum of three squares.

Then, over all values of $a$ , the fraction of the natural numbers that are not expressible as the sum of three squares is

$\begin{aligned} \dfrac{1}{4^0 \cdot 8} + \dfrac{1}{4^1 \cdot 8} + \dfrac{1}{4^2 \cdot 8} + .... &= \sum_{a = 0}^{\infty} \dfrac{1}{4^a \cdot 8} \\ \\ &= \dfrac{1}{8} \sum_{a = 0}^{\infty} \dfrac{1}{4^a} \\ \\ &= \dfrac{1}{8} \left( \dfrac{1}{1 - \frac{1}{4}} \right) \\ \\ &= \dfrac{1}{8} \cdot \dfrac{4}{3} \\ \\ &= \dfrac{1}{6} \end{aligned}$

and therefore

$\lim_{x\to \infty} \dfrac{x}{f(x)} = \boxed{6}$