# Given the Pythagorean triple, <mtable columnalign="right left right left right left right left

Given the Pythagorean triple,
$\begin{array}{rl}a& =\frac{{x}^{2}-25}{y}\\ b& =\frac{10x}{y}\\ c& =\frac{{x}^{2}+25}{y}\end{array}$
so ${a}^{2}+{b}^{2}={c}^{2}$ and $y\ne 0$. We wish to set its area $G=5$. Thus,
$G=\frac{1}{2}ab=\frac{1}{2}\frac{10x}{y}\frac{\left({x}^{2}-25\right)}{y}=5$
Or simply,$\begin{array}{}\text{(1)}& {x}^{3}-25x={y}^{2}\end{array}$
Can you help find two non-congruent right-angled triangles that have rational side lengths and area equal to 5?
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Sawyer Day
It turns out that for rational $a,b,c$, if
${a}^{2}+{b}^{2}={c}^{2}$
$\frac{1}{2}ab=n$
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,
$a,b,c=\frac{3}{2},\phantom{\rule{thickmathspace}{0ex}}\frac{20}{3},\phantom{\rule{thickmathspace}{0ex}}\frac{41}{6}$
$a,b,c=\frac{1519}{492},\phantom{\rule{thickmathspace}{0ex}}\frac{4920}{1519},\phantom{\rule{thickmathspace}{0ex}}\frac{3344161}{747348}$
However, since the elliptic curve,
$\begin{array}{}\text{(1)}& {x}^{3}-25x={y}^{2}\end{array}$
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,
$\frac{{u}^{2}}{{v}^{6}}\left({u}^{4}-25{v}^{4}\right)={y}^{2}$
or just,
$\begin{array}{}\text{(2)}& {u}^{4}-25{v}^{4}={w}^{2}\end{array}$
Here is a theorem by Lagrange. Given an initial solution ${u}^{4}+b{v}^{4}={w}^{2}$, then further ones are,
${X}^{4}+b{Y}^{4}={Z}^{2}$
where,$X,Y,Z={u}^{4}-b{v}^{4},\phantom{\rule{thickmathspace}{0ex}}2uvw,\phantom{\rule{thickmathspace}{0ex}}\left({u}^{4}+b{v}^{4}{\right)}^{2}+4b{u}^{4}{v}^{4}$
Thus, using Lagrange's theorem recursively, we have the infinite sequence,
$u,v,w=5,\phantom{\rule{thickmathspace}{0ex}}2,\phantom{\rule{thickmathspace}{0ex}}15$
$u,v,w=41,\phantom{\rule{thickmathspace}{0ex}}12,\phantom{\rule{thickmathspace}{0ex}}1519$
$u,v,w=3344161,\phantom{\rule{thickmathspace}{0ex}}1494696,\phantom{\rule{thickmathspace}{0ex}}535583225279$
and so on.