Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$

Question

It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed up. And I found that the rational solutions of $w^2+x^2+y^2=1$ is $$w = \frac{2s}{1+s^2+t^2}$$ $$x = \frac{2t}{1+s^2+t^2}$$ $$y = \frac{1-s^2 - t^2}{1+s^2+t^2}$$ where $s$ and $t$ are both rational numbers. From this I found out that $$(2mnq^2)^2 + (2pqn^2)^2 + (n^2q^2 - m^2q^2-n^2p^2)^2 = (n^2q^2 + m^2q^2+n^2p^2)^2.$$ But it helps little because I cannot prove that the solutions have to be in such form.

Can any one help?

Answer 1

Aspects of your question have been covered here about Lebesgue's Identity,

lebesgue's identity

and a general approach to $x_1^2+x_2^2+\dots+x_n^2 = z^2$,

Diophatine equation $x^2+y^2+z^2=t^2$

Answer 2

In General it is possible in various ways to write the solution to this equation:

$$x^2+y^2+z^2=q^2$$

I like such kind.

$$x=2a^2s^2-2abs^2\pm{2apbs}$$

$$y=2a^2s^2+2abs^2\pm2apbs$$

$$z=p^2b^2-a^2s^2+s^2b^2\pm2apbs$$

$$q=p^2b^2+3a^2s^2+s^2b^2\pm2apbs$$

If you want you can write infinitely many formulas are not the problem. You can even choose a particular form. It is of interest how to solve equations such as this:

$$ax^2+bxy+cxz+dy^2+jyz+rz^2=wq^2$$

$a,b,c,d,j,r,w - $ any specified coefficients.

Answer 3

Here I have a solution just using Pythagorean triples. we know that any solution of $$x^2+y^2=z^2$$ can be written as $$x=2ab,\,\,\,\,\, y=a^2-b^2,\,\,\,\,\ z=a^2+b^2.$$ Therefore for your equation, we can choose $$w=4abc,\,\,\,\,\, x=2(a^2-b^2)c,\,\,\,\,\ y=(a^2+b^2)^2-c^2,\,\,\,\,\,z=(a^2+b^2)^2+c^2.$$

Isn't it interesting? :)