Exercise 5.3.14 (General solution of $x^2 + 5y^2 =z^2$, $\mathrm{gcd}(x,y) = 1$)

Proof. Let $(x,y,z)$ be a solution of $x^2+5y^2=z^2$, where $x\wedge y=1$ and $x,y,z$ are positive integers.

Since $x\wedge y=1$, $x$ and $y$ cannot both be even. They cannot both be odd either, for if they were we would have $x^2\equiv 1(mod4),y^2\equiv 1(mod4)$, and therefore $z^2\equiv x^2+5y^2\equiv 2(mod4)$, which is impossible. So $x$ and $y$ have distinct parity, and $z$ is odd.

If a prime $p\ne 5$ divides $x$ and $z$, then $p\mid 5y^2=z^2-x^2$, thus $p\mid y$, which is impossible because $x\wedge y=1$. If $p=5$ divides $x$ and $y$, then $5^2\mid 5y^2$, thus $5\mid y$, which is impossible. This proves that $x\wedge z=1$

• We consider first the case were $x$ is odd and $y$ is even. Then we can write the equation $x^2+5y^2=z^2$ as

  $$\begin{array}{lll}\hfill 5{\left(\frac{y}{2}\right)}^2=\frac{z+x}{2}\frac{z-x}{2}.& & \hfill \text{(1)}\end{array}$$

  We define $u=(z+x)∕2,v=(z-x)∕2,w=y∕2$, so that $u,v,w$ are positive integers satisfying

  $$uv=5w^2.$$

  We have proved that $x\wedge z=1$, hence $u\wedge v=1$.

  Since $5\mid uv$, then $5\mid u$ or $5\mid v$.

  • If $5\mid u$, then $u=5u^{\prime }$ for some positive integer $u^{\prime }$, and $u^{\prime }v=w^2$, where $u^{\prime }$ and $v$ are relatively prime (since $u^{\prime }\mid u$). By Lemma 5.4, $u^{\prime }$ and $v$ are both perfect squares, so $u^{\prime }=s^2,v=r^2$ for suitable positive integers $r,s$ such that $r\wedge s=1$, and

    $$u=5s^2,v=r^2.$$

  • If $5\mid v$, we obtain similarly by exchanging the roles of $u$ and $v$

    $$u=r^2,v=5s^2.$$

  In the first case ( $5\mid u$), we obtain $\frac{z+x}{2}=5s^2,\frac{z-x}{2}=r^2$, and $5{(y∕2)}^2=5s^2{r}^2$, so $y=2rs$ (because $y,r,s$ are positive integers). Therefore

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =5s^2-r^2,\hfill \\ y\hfill & =2rs,\hfill \\ z\hfill & =5s^2+r^2.\hfill \end{array}& & \hfill \end{array}$$

  In the second case ( $5\mid v$), we obtain $\frac{z+x}{2}=r^2,\frac{z-x}{2}=5s^2$, and $5{(y∕2)}^2=5r^2{s}^2$, so $y=2rs$. Therefore

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =r^2-5s^2,\hfill \\ y\hfill & =2rs,\hfill \\ z\hfill & =r^2+5s^2.\hfill \end{array}& & \hfill \end{array}$$

  In both cases, since $x,y,z$ are positive integers, we obtain

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =\pm (r^2-5s^2)=|r^2-5s^2|,\hfill \\ y\hfill & =2rs,\hfill \\ z\hfill & =r^2+5s^2,\hfill \end{array}& & \hfill \text{(2)}\end{array}$$

  (which are the equations of Problem 12).

  Conversely, if (2) is satisfied, then

  $$x^2+5y^2={(r^2-5s^2)}^2+20r^2{s}^2=r^4+25s^4+10r^2{s}^2={(r^2+5s^2)}^2=z^2.$$

  (Since in this case $x$ is even, we can assume that $r$ and $s$ have distinct parity).

• Now we consider the case where $x$ is even and $y$ is odd. Then $z$ is odd. We write the equation $x^2+5y^2=z^2$ as

  $$5y^2=(z+x)(z-x).$$

  Here $z+x$ and $z-x$ are both odd. If $p$ is a prime such that $p\mid z+x$ and $p\mid z-x$, then $p\ne 2$, and $p\mid 2z,p\mid 2x$ thus $p\mid x$ and $p\mid z$. This is impossible, since $x\wedge z=1$. Therefore $(z+x)\wedge (z-x)=1$.

  As in the first case, this implies that there are positive integers $s,t$ such that $z-x=5s^2,z+x=t^2$, or $z-x=t^2,z+x=5s^2$, where $s\wedge t=1$. In both cases, $5y^2=5s^2{t}^2$, thus $y=st$ (here $y,s,t$ are positive integers). Note that $s,t$ are odd, because $z-x$ and $z+x$ are odd.

  After replacing $x$ by $-x$, if necessary, there exist odd integers $s$ and $t$ such that $z-x=5s^2,z+x=t^2$. Hence

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =\pm (\frac{t^2-5s^2}{2}),\hfill \\ y\hfill & =ts,\hfill \\ z\hfill & =\frac{t^2+5s^2}{2}.\hfill \end{array}& & \hfill \end{array}$$

  Put $r=(t-s)∕2$, which is an integer because $t$ and $s$ are odd, so that $t=s+2r$. Since $s\wedge t=1$, we have also $r\wedge s=1$, and

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x\hfill & =\pm (2r^2+2rs-2s^2),\hfill \\ y\hfill & =s^2+2rs,\hfill \\ z\hfill & =2r^2+2rs+3s^2.\hfill \end{array}& & \hfill \text{(3)}\end{array}$$

  Conversely, if (3) is satisfied, then

  $$\begin{array}{llll}\hfill x^2+5y^2-z^2& ={(2r^2+2rs-2s^2)}^2+5{(s^2+2rs)}^2-{(2r^2+2rs+3s^2)}^2& \hfill & \\ \hfill & =-5s^4-20r^2{s}^2-20r{s}^3+5s^4+20r{s}^3+20r^2{s}^2& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

In conclusion, if $x,y,z$ are positive integers such that $x\wedge y=1$, then