Exercise 5.4.14* (Equations $x^4 + 4y^4 = z^2$ and $a^4 + b^2 = c^4$)

Proof.

(a)

Assume for the sake of contradiction that the equation $$\begin{array}{lll}\hfill x^4+4y^4=z^2& & \hfill \text{(1)}\end{array}$$

has a solution $(x,y,z)$ in positive integers, so that ${(x^2)}^2+{(2y^2)}^2=z^2$. Then there is a solution $(x,y,z)$ where $z$ is minimal, given by

$$z=\min <!--\; FUNCTION\; APPLICATION\; -->\{\zeta \in {\mathbb{N}}^{\text{∗}}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->x\in {\mathbb{N}}^{\text{∗}},\exists <!--\; FUNCTION\; APPLICATION\; -->y\in {\mathbb{N}}^{\text{∗}},{(x^2)}^2+{(2y^2)}^2={\zeta }^2\}.$$

The equation (1) shows that $x$ and $z$ are both even or both odd. If $x$ and $z$ are both even, then $x=2x^{\prime },z=2z^{\prime }$ for some integers $x^{\prime },z^{\prime }$. Then $16{x{}^{\prime }}^2+4y^4=4{z{}^{\prime }}^2$, thus $4{x{}^{\prime }}^2+y^4={z{}^{\prime }}^2$, so $(y,x^{\prime },z^{\prime })$ is a solution of (1) in positive integers, with $z^{\prime }<z$. This contradicts the minimality of $z$, thus $x$ and $z$ are odd.

If some prime number $p$ divides $x$ and $z$, then $p\ne 2$, and $p\mid z^2-x^4=4y^4$, thus $p\mid y$. Then $p^4\mid x^4+4y^4=z^2$, therefore $p^2\mid z$. This shows that $(x∕p,y∕p,z∕p^2)$ is a solution of the same equation, satisfying $z∕p^2<z$, and this contradicts the minimality of $z$. Therefore $x\wedge z=1$, so $(x^2,2y^2,z)$ is a primitive Pythagorean triple, with $2y^2$ even. By Theorem 5.5, there are integers $r,s$ of opposite parity such that $r\wedge s=1,r>s>0$ and

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

Since $y^2=\mathit{rs}$, where $r,s$ are relatively prime (and $r>0,s>0$), then $r=c^2,s=a^2$ for some positive integer $a,c$ (Lemma 5.4), so

$$\begin{array}{lll}\hfill \{\begin{array}{cc}{x}^2\hfill & =c^4-a^4,\hfill \\ y^2\hfill & =a^2{c}^2,\hfill \\ z\hfill & =c^4+a^4\hfill \end{array}& & \hfill \text{(3)}\end{array}$$

Put $b=x$. Then

$$\begin{array}{lll}\hfill a^4+b^2=c^4,& & \hfill \text{(4)}\end{array}$$

where $c\wedge a=1$, $a$ and $c$ have opposite parity, and $c\le c^4<z$, so $c<z$.

Then $(a^2,b,c^2)$ is a primitive Pythagorean triple with $b$ odd (since $b^2=c^4-a^4$ is odd). The same Theorem 5.5 shows that there are integers $u,v$ of opposite parity such that $u\wedge v=1,u>v>0$ and

$$\begin{array}{lll}\hfill \{\begin{array}{cc}b\hfill & =u^2-v^2,\hfill \\ a^2\hfill & =2\mathit{uv},\hfill \\ c^2\hfill & =u^2+v^2\hfill \end{array}& & \hfill \text{(5)}\end{array}$$

Since $a^2=2\mathit{uv},u\wedge v=1,u>v>0$, by Problem 5, $u=2l^2,v=m^2$ or $u=m^2,v=2l^2$ for suitable positive integers $l,m$. In both cases, the last equation of (4) gives

$$\begin{array}{lll}\hfill c^2=m^4+4l^4,& & \hfill \text{(6)}\end{array}$$

thus $(x^{\prime },y^{\prime },z^{\prime })=(m,l,c)$ is a solution of (1) in positive integers. But $z^{\prime }=c<z$, and this contradicts the minimality of $z$. This contradiction shows that $x^4+4y^4=z^2$ has no solution in positive integers.

(b)

It remains to show that $a^4+b^2=c^4$ has no solution in positive integers. If this equation has a solution, there is a solution $(a,b,c)\in {({\mathbb{N}}^{\text{∗}})}^3$ where $c$ is minimal.

We prove $a\wedge c=1$. Indeed if a prime $p$ divides $a$ and $c$, then $p\mid c^4-a^4=b^2$, thus $p\mid b$. Then $(a∕p,b∕p^2,c∕p)$ is a solution of (4) with $c∕p<c$. This contradicts the minimality of $c$. This contradiction shows that $a\wedge c=1$. Therefore $(a^2,b,c^2)$ is a primitive Pythagorean triple. By Theorem 5.5, there are integers $u,v$ of opposite parity such that $u\wedge v=1,u>v>0$ and

$$\begin{array}{lll}\hfill (7.a)\{\begin{array}{cc}b\hfill & =u^2-v^2,\hfill \\ a^2\hfill & =2\mathit{uv},\hfill \\ c^2\hfill & =u^2+v^2,\hfill \end{array}\text{ or }(7.b)\{\begin{array}{cc}{a}^2\hfill & =u^2-v^2,\hfill \\ b\hfill & =2\mathit{uv},\hfill \\ c^2\hfill & =u^2+v^2\hfill \end{array}& & \hfill \text{(7)}\end{array}$$

• If (7.a) is true, we obtain as previously integers $m,l$ such that $u=2l^2,v=m^2$ or $u=m^2,v=2l^2$, so $c^2=m^4+4l^4$ (see (6)), and this is impossible by the first part of the proof.

• If (7.b) is true, note that $a,c,u,v$ satisfy the system

  $$\begin{array}{lll}\hfill a^2=u^2-v^2,c^2=u^2+v^2.(\ast )& & \hfill \text{(8)}\end{array}$$

From (8) we obtain

$$\begin{array}{lll}\hfill c^2+a^2=2u^2,c^2-a^2=2v^2,& & \hfill \text{(9)}\end{array}$$

where $c$ and $a$ are odd and relatively prime, and $c>a$. Therefore there is some positive integer $\alpha$ such that

$$c=a+2\alpha .$$

By the first equation of (9), this gives

$$\begin{array}{llll}\hfill & {(a+2\alpha )}^2+a^2=2u^2,& \hfill & \\ \hfill & 2a^2+4\mathit{a\alpha }+4{\alpha }^2=2u^2,& \hfill & \\ \hfill & a^2+2\mathit{a\alpha }+2{\alpha }^2=u^2,& \hfill & \end{array}$$

thus

$$\begin{array}{lll}\hfill & {(a+\alpha )}^2+{\alpha }^2=u^2.& \hfill \text{(10)}\end{array}$$

Note that $a\wedge c=1$ and $c^2+a^2=2u^2$ imply $a\wedge u=1$. From $a^2+2\mathit{a\alpha }+2{\alpha }^2=u^2$ and $a\wedge u=1$, we see that $\alpha \wedge u=1$. This shows that $(a+\alpha ,\alpha ,u)$ is a primitive Pythagorean triple. By Theorem 5.5, there are positive integers $r,s$ of opposite parity such that $r\wedge s=1$ and

$$\begin{array}{lll}\hfill (11.a)\{\begin{array}{cc}\hfill a+\alpha & =2\mathit{rs},\hfill \\ \hfill \alpha & =r^2-s^2,\hfill \\ \hfill u& =r^2+s^2,\hfill \end{array}\text{ or }(11.b)\{\begin{array}{cc}\hfill a+\alpha & =r^2-s^2,\hfill \\ \hfill \alpha & =2\mathit{rs},\hfill \\ \hfill u& =r^2+s^2.\hfill \end{array}& & \hfill \text{(11) }\end{array}$$

In either cases we have

$$\begin{array}{lll}\hfill c^2-a^2=(c-a)(c+a)=2\alpha \cdot 2(a+\alpha )=8\mathit{rs}(r^2-s^2).& & \hfill \end{array}$$

Then the second equation of (9) gives, after division by $2$,

$$\begin{array}{lll}\hfill v^2=4\mathit{rs}(r^2-s^2).& & \hfill \text{(12)}\end{array}$$

Since $r,s$ are of opposite parity, $r^2-s^2$ is odd, and $(4\mathit{rs})\wedge (r^2-s^2)=1$. By Lemma 5.4, $r,s$ and $r^2-s^2$ are perfect squares, say

$$\begin{array}{lll}\hfill r=u^2,s=v^2,r^2-s^2=w^2.& & \hfill \text{(13)}\end{array}$$

Then

$$\begin{array}{lll}\hfill s^4+w^2=u^4.& & \hfill \text{(14)}\end{array}$$

This shows that $(s,w,u)$ is a solution of our equation $a^4+b^2=c^4$. Moreover $c>a>0$, thus $2c^2>c^2+a^2=2u^2$, where $u>0$, so $c>u$. This contradicts the minimality of $c$. This contradiction proves that $a^4+b^2=c^4$ has no solution in positive integers.