Exercise 5.3.13 (Solutions of $x^2 + 2y^2 = z^2$)

Proof. Let $(x,y,z)$ be a solution of $x^2+2y^2=z^2$ in positive integers with $x\wedge y\wedge z=1$.

If $y=2k+1$ is odd, then $2y^2=2{(2k+1)}^2=8k^2+8k+2\equiv 2(\mathit{mod}8)$, thus $z^2-x^2\equiv 2(\mathit{mod}8)$. But the squares modulo $8$ are $0,1,4$, so the congruence $z^2-x^2\equiv 2(\mathit{mod}8)$ is impossible. Therefore $y$ is even. Moreover $x$ and $z$ have same parity, but $x$ and $y$ cannot be both even, otherwise $2\mid x\wedge y\wedge z=1$. This contradiction shows that $x$ and $z$ are odd (and $y$ is even).

Then $z-x$ and $z+x$ are both even, so we write the equation $x^2+2y^2=z^2$ as

Any common divisor of the two factors $u=\frac{z+x}{2},v=\frac{z-x}{2}$ divides both their sum, $z$ and their difference, $x$. Since $x\wedge z=1$, it follows that these two factors $u,v$ have no common factor. Since $2\mathit{uv}=y^2$ and $u>0,v>0$, by Problem 5, either (a) $u=2s^2,v=r^2$, or (b) $u=r^2,v=2s^2$, for suitable integers $r>0,s>0$ (we intentionally changed the names of the variables in the case (a)). Note that $r\wedge s=1$, because $u\wedge v=1$.

(a)

If $\frac{z+x}{2}=2s^2,\frac{z-x}{2}=r^2$, then $4s^2{r}^2=y^2$, so $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =2s^2-r^2\hfill \\ y\hfill & =\pm 2\mathit{sr}\hfill \\ z\hfill & =2s^2+r^2.\hfill \end{array}& & \hfill \end{array}$$

(b)

If $\frac{z+x}{2}=r^2,\frac{z-x}{2}=2s^2$, then $4r^2{s}^2=y^2$, so $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =r^2-2s^2\hfill \\ y\hfill & =\pm 2\mathit{rs}\hfill \\ z\hfill & =r^2+2s^2.\hfill \end{array}& & \hfill \end{array}$$

In both cases

$$x=\pm (r^2-2s^2),y=\pm 2\mathit{rs},z=r^2+2s^2.$$

Note that $r$ is odd, because $x=\pm (r^2-2s^2)$ and $x$ is odd.

Since $x>0,y>0,z>0$ by hypothesis (and $r>0,s0$), we obtain

$$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =|r^2-2s^2|\hfill \\ y\hfill & =2\mathit{rs}\hfill \\ z\hfill & =r^2+2s^2,\hfill \end{array}& & \hfill \text{(1)}\end{array}$$

where $r>0,s>0$, $r$ odd, and $r\wedge s=1$.

Conversely, if (1) is satisfied, then

$$\begin{array}{lll}\hfill x^2+2y^2={(r^2-2s^2)}^2+8r^2{s}^2=r^4+4s^4+4r^2{s}^2=z^2.& & \hfill \end{array}$$

In conclusion, the solutions of $x^2+2y^2=z^2$ in positive integers with $x\wedge y\wedge z=1$ are given by

$$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =|r^2-2s^2|\hfill \\ y\hfill & =2\mathit{rs}\hfill \\ z\hfill & =r^2+2s^2,\hfill \end{array}& & \hfill \text{(2)}\end{array}$$

where $r>0,s>0$, $r$ odd and $r\wedge s=1$.