Diophantine equation $x^2+y^2=z^2+t^2$?

Question

I would like to find some source books or articles which discuss the Diophantine equation $$ x^2+y^2=z^2+t^2,\qquad |y-z|=1 $$ for which $x,z$ are odd positive and $y,t$ are even positive integers. Any brief explanation is also welcome and appreciated.

Thanks!

Rewrite it as $y^2 - z^2 = t^2 - x^2$ and factor both sides. I don't know why you expect such a specific question to be discussed in an article or a book, though. — Qiaochu Yuan, Dec 07 '15 at 09:32
http://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 Use any formula. Substitute in the 2nd equation and then we solve the Pell equation. Its solution and then substitute in the formula. — individ, Dec 07 '15 at 10:18

score 2 · Answer 1 · edited Mar 21 '16 at 14:08

2

Is not so hard. $x^2-t^2=z^2-y^2 $, so $(x-t)*(x+t)=(z-y)*(z+y)$. Now suppose that$ z=y+1$.So $ (x-t)*(x+t)=y+z=2y+1$. Now it's so easy, you must to keep account by initial conditions.

edited Mar 21 '16 at 14:08

Martin Sleziak

53,687

answered Dec 07 '15 at 09:35

ale

1,744

Without loss of generality you can let $z=y+1$. – Ian Miller Dec 07 '15 at 09:37

score 1 · Answer 2 · answered Dec 09 '15 at 12:26

What to do is just substitute your conditions on your equations. Since $x,z$ are odd and $y,t$ are even and $|y-z|=1$, then,

$$(2m-1)^2+(2u)^2 = (2u-1)^2+(2n)^2\tag1$$

Expanding, you can see that $2$nd powers of $u$ will cancel, so you'll only have a linear equation in $u$. Letting Walpha do the work, we get the condition,

$$u = -m^2+m+n^2$$

which makes $(1)$ true for any $m,n$.

Diophantine equation $x^2+y^2=z^2+t^2$?

2 Answers2