Exercise 3.6.12 (The integers $4^m(8k+7)$ are not sums of three squares)

Proof.

(a)

If $x^2+y^2+z^2=n$ and $4\mid n$, then $x^2+y^2+z^2\equiv 0(\mathit{mod}4)$. Suppose for the sake of contradiction that $x$ is odd. Then $x^2\equiv 1(\mathit{mod}4)$, thus $y^2+z^2\equiv 3(\mathit{mod}4)$. But $y^2\equiv 0$ or $1(\mathit{mod}4)$ and $z^2\equiv 0$ or $1(\mathit{mod}4)$, thus $y^2+z^2\not\equiv 3(\mathit{mod}4)$. This contradiction shows that $x$ is even, and with the same arguments, $y$ and $z$ are even.

(b)

We prove by induction the property $$𝒫(m):\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},\forall <!--\; FUNCTION\; APPLICATION\; -->(x,y,z)\in {\mathbb{Z}}^3,x^2+y^2+z^2\ne 4^m(8k+7).$$

• By Problem 11, we know that every $n$ of the form $8k+7$ is not the sum of three squares. Thus $𝒫(0)$ is true.
• Suppose that $𝒫(m)$ is true, so that $4^m(8k+7)$ is never the sum of three squares, whatever the value of the integer $k$. Let $n$ be an integer of the form $n=4^{m+1}(2k+1)$. Suppose for the sake of contradiction, that $n=x^2+y^2+z^2$ for some integers $x,y,z$. Since $4\mid n$, then $x,y$ and $z$ are even by part (a). Thus $n∕4=4^m(2k+1)={x{}^{\prime }}^2+{y{}^{\prime }}^2+{z{}^{\prime }}^2$, where $x^{\prime }=x∕2,y^{\prime }=y∕2,z^{\prime }=z∕2$ are integers. This contradicts the induction hypothesis $𝒫(m)$. Therefore $n$ is not the sum of three squares, so $𝒫(m+1)$ is true.

• The induction is done, so

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->m\in \mathbb{N},\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},\forall <!--\; FUNCTION\; APPLICATION\; -->(x,y,z)\in {\mathbb{Z}}^3,x^2+y^2+z^2\ne 4^m(8k+7).$$

We conclude that if $n$ is of the form $4^m(8k+7)$ then $n$ is not the sum of three squares.