Solving Diophantine equation $1/x^2+1/y^2=1/z^2$

Question

How can we find positive integers solutions $(x,y,z)$, where $\gcd(x,y,z)=1$ for the equation: $$1/x^2+1/y^2=1/z^2$$

Can we conclude that $x$ and $y$ are not coprimes for it to have solution?

Does this answer your question? Integer solutions for $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$? — Bumblebee, May 26 '22 at 08:58

N. S. · Answer 1 · 2015-01-18T20:03:01.737

7

Multiply both sides by $x^2y^2z^2$.

Then you get $$y^2z^2+x^2z^2=x^2y^2$$

Now use that each of $x^2, y^2,z^2$ divide two of the terms hence the third.

Added Here is the rest of the solution. Let $a=gcd(x,y), b=gcd(x,z), c=gcd(y,z)$.

Then, $gcd(a,b)=1$ and hence $ab|x$. We claim $ab=x$.

Indeed write $x=abd$. Assume by contradiction that $d \neq 1$ and let $p|d$, $p$ prime.

As $x | yz$ we have $abd | yz \Rightarrow d | \frac{y}{a}\frac{z}{b}$.

Then $p$ divides either $\frac{y}{a}$ or $\frac{z}{b}$.

But then, in the first case $pa |x,y$ while in the second $pb | x,z$ contradicting the $gcd$.

Therefore $x=ab$. The same way you can prove that $y=ac, z=bc$.

Replacing in the above equation you get $$a^2b^2c^4+a^2b^4c^2=a^4b^2c^2$$ or $$c^2+b^2=a^2$$

this shows that $(c,b,a)$ is a primitive Pytagoreal triple and $$x=ab \\ y=ac \\ z=bc$$

edited Jan 18 '15 at 20:03

answered Jan 18 '15 at 19:41

N. S.

We get from this that $x\mid yz, y\mid zx, z\mid xy$. What does it imply? – Wojowu Jan 18 '15 at 19:45
@Wojowu If $x|yz$ and $gcd(x,y,z)=1$ then you can show that you can write uniquely $x=x_1x_2$ so that $x_1|y, x_2|z$ and so on. – N. S. Jan 18 '15 at 19:49
How does it answer any of OP's questions? – Wojowu Jan 18 '15 at 19:54
@Wojowu Ok, added more details, clear now? – N. S. Jan 18 '15 at 20:03
Yes, thanks. Really nice solution, I must admit! – Wojowu Jan 18 '15 at 20:04
@Wojowu Ty. The solution is pretty standard, the only thing which is not obvious (but clear to people which are very familiar with Number Theory) is the part $x|yz, gcd(x,y,z)=1$ implies $x=gcd(x,y) \cdot \gcd(x,z)$. I am pretty sure this can be proven in a simpler way than I did. – N. S. Jan 18 '15 at 20:07

score 0 · Answer 2 · answered Jun 15 '16 at 12:58

Although the question's old, I'll leave a solution which I believe is a correct for someone who comes along.

Obviously, $x \ge y > z$.

You should see why $\gcd(x,y)= 1$.

So, now:

$$\frac{1}{y^2} = \frac{x^2 - z^2}{x^2z^2}$$

Hence, $y^2|x^2z^2 \implies y|xz \implies y|z \text{ since } x \text{ and } y \text{ are relatively prime.}$

$y \le z$.

Contradiction.

2 Answers2