# Finding two non-congruent right-angle triangles

The map $g: B \to A, \ (x,y) \mapsto \left(\dfrac {x^2 - 25} y, \dfrac {10x} y, \dfrac {x^2 + 25} y \right)$ is a bijection where $A = \{ (a,b,c) \in \Bbb Q ^3 : a^2 + b^2 = c^2, \ ab = 10 \}$ and $B = \{ (x,y) \in \Bbb Q ^2 : y^2 = x^3 - 25x, \ y \ne 0 \}$.

Given the Pythagorean triple,

\begin{aligned} a&=\frac {x^2 - 25} y\\ b&=\frac {10x} y\\ c&=\frac {x^2 + 25} y \end{aligned} so $a^2+b^2 = c^2$ and $y\neq 0$. We wish to set its area $G = 5$. Thus,

$$G = \frac{1}{2}ab = \frac{1}2 \frac {10x} y \frac {(x^2 - 25)} y = 5$$

Or simply,

$$x^3-25x = y^2\tag1$$

Can you help find two non-congruent right-angled triangles that have rational side lengths and area equal to $5$?

• How are the first part of the question and the second part related? – user228113 Dec 3 '15 at 17:27
• there are two points where P(-4,6) and Q(0,0) on the elliptic curve y^2=x^3-25x – Gabriel Dec 3 '15 at 17:32
• @G.Sassatelli: Don't you think the set $A$ describes right-angled triangles of the required type? Given the first part, the question about triangles is thus reduced to finding rational points on the elliptic curve (other than the 2-torsion points). – Jyrki Lahtonen Dec 3 '15 at 21:01
• Have you tried what the point doubling formula does to $P=(-4,6)$? Or just adding one of the points $(0,0)$ or $(\pm5,0)$ to $P$? – Jyrki Lahtonen Dec 3 '15 at 21:06
• Please reopen this question. I think that it makes a lot of sense now and people are forced to provide real answers in comments. – Justpassingby Dec 4 '15 at 7:13

## 1 Answer

Update: It turns out that for rational $a,b,c$, if,

$$a^2+b^2 = c^2$$

$$\tfrac{1}{2}ab = n$$

then $n$ is a congruent number. The first few are $n = 5, 6, 7, 13, 14, 15, 20, 21, 22,\dots$

After all the terminology, I guess all the OP wanted was to solve $(1)$. He asks, "... find two non-congruent right-angled triangles that have rational side lengths and area equal to 5".

We'll define congruence as "... two figures are congruent if they have the same shape and size." Since we want non-congruent, then two such right triangles are,

$$a,b,c = \frac{3}{2},\;\frac{20}{3},\;\frac{41}{6}$$

$$a,b,c = \frac{1519}{492},\;\frac{4920}{1519},\;\frac{3344161}{747348}$$

However, since the elliptic curve,

$$x^3-25x = y^2\tag1$$

has an infinite number of rational points, then there is no need to stop at just two. Two small solutions to $(1)$ are $x_1 = 5^2/2^2 =u^2/v^2$ and $x_2 = 45$. We can use the first one as an easy way to generate an infinite subset of solutions. Let, $x =u^2/v^2$, so,

$$\frac{u^2}{v^6}(u^4-25v^4) = y^2$$

or just,

$$u^4-25v^4 = w^2\tag2$$

a curve also discussed in the post cited by J. Lahtonen. Here is a theorem by Lagrange. Given an initial solution $u^4+bv^4 = w^2$, then further ones are,

$$X^4+bY^4 = Z^2$$

where,

$$X,Y,Z = u^4-bv^4,\; 2uvw,\; (u^4+bv^4)^2 + 4 b u^4 v^4$$

Thus, using Lagrange's theorem recursively, we have the infinite sequence,

$$u,v,w = 5,\;2,\;15$$

$$u,v,w = 41,\;12,\;1519$$

$$u,v,w = 3344161,\; 1494696,\; 535583225279$$

and so on.