Smallest solutions to show that a number is congruent On wikipedia, under the 'smallest solutions' section, there is a table of congruent numbers, where for each congruent number $n$ they put a triple $(a, b, c)$ with $a^2 + b^2 = c^2$ and $n = ab/2$. It says that the numerator for $c$ is the smallest. Lets call this the 'simplest solution' to a congruent number problem, then my question is if I have any solution $(a', b', c')$, how can I go about finding the simplest solution?
Edit If the method is super complicated, a link to a book/paper describing it would answer my question too.
What I know is that there is a correspondence between the triples $(a, b,c)$ and rational points $(x,y)$ on the elliptic curve $y^2 = x^3 - n^2 x$. So if you have some triple, you can convert it to a point on the curve, and use the addition on the curve to generate infinitely many points. But it doesn't seem clear how this method would allow you to hit all rational points, and even if you could find all rational points on the curve, how do you find the rational point which can then be converted back to the 'simplest' triple.
Thank you.
 A: The bijection is shown on the wikipedia page you link. To repeat given $(a,b,c)$ we set $x = n(a+c)/b$ and $y = 2n^2 (a + c)/b^2 \neq 0$, then
$$y^2 = x^3 - n^2x$$
and conversely given $(x, y)$ solving this equation with $y$-coordinate not $0$, set $a = (x^2 - n^2)/y$, $b = 2nx/y$ and $c = (x^2 + n^2)/y$.
Let $E_n$ be the elliptic curve
$$E_n : y^2 = x^3 - n^2x$$
It is possible to show that the only torsion on $E_n$ is 2-torsion (i.e., where $y=0$). Thus if you can find a set of generators $P_1, ..., P_r$ for the free part of $E_n(\mathbb{Q})$ you can recover all of the triples solving the congruent number problem.
I'll give $2$ different sorts of examples showing that this may not be achieved by simply adding the point corresponding to a single triple.

The point is not a generator
Consider the $(3,4,5)$ where $n = 6$. This gives the point $P = (12, 36)$. Then $2P = (25/4 , -35/8)$ - which gives the triangle $(7/10, 120/7, 1201/70)$.
But the point $P$ has infinite order, so cannot be recovered from $2P$ by additions of the latter. Thus if you were to find the second triple, you would not be able to naively recover the first.

The rank is $>1$
Consider $n = 41$. In this case we can check that that $P_1 = (-9, -120)$ and $P_2 = (18081/400, 1023729/8000)$ lie on $E$, and are independent (in fact they generate $E(\mathbb{Q})$ but this is much harder to show - since they have 2-torsion, congurent number curves admit the method of "descent by 2-isogeny").
But if we were given $P_1$ or $P_2$, we would have no way of recovering the other.
Other such examples can be obtained where $n = 34, 65$.

Finally I suspect that the "simplest solution" you want corresponds to the point of least height on $E$, (which should be a generator) although I haven't really thought about this deeply.
