Distinct integer solutions to $a^2+b^2=c^2+d^2$

Question

I'm trying to find integer solutions to $$a^2+b^2=c^2+d^2$$

with values $a> c > d > b>0$

Or in other words, two triangles with integer legs and equal hypotenuse lengths, not necessarily integer. Seems like a Diophantine equation to me, but I only learned how to solve Diophantine equations in the form of Pell's equations. I couldn't find anything on this equation when I checked Wikipedia. It is similar to a Pythagorean quadruple, although not quite, so that's not helpful either. How do I find integer solutions to this?

@A---B oops forgot to mention all values are are unique. I'll add that in — Ryan, Sep 14 '17 at 17:06
Subtract $b^2$ and $d^2$ from each side to get $a^2-d^2=c^2-b^2$, or $$(a+d)(a-d)=(c+b)(c-b).$$ — Théophile, Sep 14 '17 at 17:07
Clearly you would count $25^2+60^2=39^2+52^2$. What about $25^2+60^2=60^2+25^2$? Or $25^2+60^2=0^2+65^2$? Or $25^2+60^2=(-25)^2+(-60)^2$ ? — Henry, Sep 14 '17 at 17:09
$7^2 + 24^2 = 15^2 + 20^2$. So, $(7k, 24k, 15k, 20k)$ should be the solution, — , Sep 14 '17 at 17:10
https://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 — individ, Sep 14 '17 at 17:11
You should clarify the restriction: if $a\neq b$ and $b\neq c$, it can still be true that $a=c$. — gen-ℤ ready to perish, Sep 14 '17 at 17:22
Instead of writing "$a\neq b\neq c\neq d$" which is ambiguous whether or not $a$ must be different from $c$, I would recommend instead just using words "with all of $a,b,c,d$ distinct from one another." Alternatively, if you were to rename these with subscripts as $a_1,a_2,a_3,a_4$ you can say $a_i\neq a_j$ for all $i\neq j$. — JMoravitz, Sep 14 '17 at 17:29

score 8 · Answer 1 · answered Sep 14 '17 at 17:07

One way to find some solutions is to rewrite as $a^2-c^2=d^2-b^2$, which in turn becomes $(a-c)(a+c)=(d-b)(d+b)$.

Then choose a number that factors in more than one way and see what you get.

Example: $15=3\cdot 5=1\cdot 15$

Then $15=(4-1)(4+1)=(8-7)(8+7)$.

So take $a=4, b=7, c=1, d=8$

score 4 · Answer 2 · answered Sep 14 '17 at 17:45

4

HINT.-The general solution of the equation $x^2+y^2=z^2+w^2$ is given by the known enough parametrization with four parameters $$x=tX+sY\\y=tY-sX\\z=tX-sY\\w=tY+sX$$.

answered Sep 14 '17 at 17:45

Piquito

29,594

MaximusFastidiousIrreverence · Answer 3 · 2017-09-15T04:41:35.083

2

$$\implies a^2-c^2=b^2-d^2=N$$ For some $N\in \mathbb{Z}$

If $$d_1 d_2=d_3 d_4=N$$

Then $$\begin{cases} \begin{align} a&=\frac{d_1 +d_2}{2} \\ c&=\frac{d_1 -d_2}{2} \\ b&=\frac{d_3 +d_4}{2} \\ d&=\frac{d_3 -d_4}{2} \\ \end{align} \end{cases}$$

What's needed is that $(d_1,d_2)$ need to be of the same parity and $(d_3,d_4)$ need to be of the same parity.

This answer is of the same spirit as paw88789

edited Sep 15 '17 at 04:41

answered Sep 14 '17 at 20:42

MaximusFastidiousIrreverence

1,735

score 1 · Answer 4 · edited Sep 15 '17 at 02:37

Strong Hint: $$\begin{align} a^2 + b^2 &= c^2 + d^2 \\ \implies a^2 + b^2 - c^2 &= d^2 \\ \implies a^2 + (b + c)(b - c) &= d^2 \\ \implies (b + c)(b - c) &= d^2 - a^2 \\ \implies (b + c)(b - c) &= (d + a)(d - a) \end{align}$$. $$\therefore a^2 + b^2 \neq \{p : p = \text{prime number}\} \iff a^2 + b^2 = c^2 + d^2$$ Now what separates a prime number from any other (composite) number?

Solution to finding distinct integer solutions: $$a^2 + b^2 = c^2 + d^2 = \{n \iff d(n) \geq 4 : d(n) = \text{number of divisors of $n$}, \ \forall n\in \mathbb{R}\}$$

nguyen quang do · Answer 5 · 2017-09-16T13:26:25.833

First note that in the equation $a^2+b^2=c^2+d^2 (*)$ you can adjust the signs of the variables, which can thus be taken in $\mathbf Z$ in order to work in the Gaussian ring $\mathbf Z[i]$. Denoting by $N$ the norm map of $\mathbf Q(i)/\mathbf Q$, one can write $(*)$ as $N(a+ib)=N(c+id)$, or $(a+ib)=u(c+id)$, with $N(u)=1$. Every $u \in \mathbf Q(i)$ having norm $1$ is of the form $(e+if)/(e-if)$, where $e,f $ can be taken in $\mathbf Z$ for reasons of homogeneity. By developping the products in the equation $(a+ib)(e-if)=(c+id)(e+if)$, one gets immediately the following parametrization of the solutions of $(*)$ : $(a, b, c, d)=(AE+BF, BE-AF, CE-DF, DE+CF))$, with $A, B, C, D, E, F \in \mathbf Z$.

Note that this kind of problem can be generalized to $N_{1}(z_1)=\lambda N_2(z_2)$, where the $N_h$'s are the norm maps of two quadratic fields, see e.g. For which values of $a\in\mathbb{Q}$ does integer solutions to $x^2+x+1=a(y^2+1)$ exist?

Distinct integer solutions to $a^2+b^2=c^2+d^2$

5 Answers5