Finding Pythagorean triples, with a conjecture.

Question

A right triangle with integer side lengths ($a , b , c$ where $a < b < c$) can be found (to a certain degree) with the inner radius ($r$), the formulas for the area of a triangle and Pythagoras theorem.

We use these to find a relationship between $r$ and the sides of $a$ and $b$.

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

$a^2 = c^2 - b^2$

$a^2 = (c + b)(c -b)$

We now can sub 2 expressions that are equivalent to $c + b$ and $c- b$ into the theorem:

Inner radius formula:

$r = ( a + b - c)/2$

$2r = a + b - c$

$2r - a = b - c$

$a - 2r = c - b$

Inner radius multiplied by semi perimeter formula (A formula for area in right triangles):

$ ab/2 = r (a + b + c)/2$

$ ab = r(a + b + c)$

$ab/r = a + b + c$

$ab/r - a = b + c $

Now we can substitute for $c - b$ and $b + c$.

$a^2 = (ab/r - a)(a - 2r)$

which when rearranged is:

$b=\frac{2r\left(a-r\right)}{a-2r}$

If we substitute in any value of $a$ we can create a function in terms of $b$ and $r$ where $b = b(r)$

For example, lets write this for the Pythagorean triple, where $a = 3.$

$b(r) = 2r(3 - r)/ 3 - 2r$,

We can conjecture that there is only one positive integer value of $b$ where $r$ is also an integer within the domain of $0 < r < a/2$, which means we can find a direct solution for both $b$ and $r$ by graphing.

https://www.desmos.com/calculator/zzxbryrahc

Note: If you look at when $r$ is $1$, you'll see that $b$ is $4$, which is true in the case of a Pythagorean triple with side lengths of ($3,4,5$).

My question is if there's a way to decrease the number of possibilities more than just $ 0 < r < a/2$ this method would become nearly impossible without the use of a graphing calculator for larger Pythagorean triples, Any other way to improve on this would be much appreciated.

Answer 1

You have $$b=\frac{2r\left(a-r\right)}{a-2r}$$

You can rewrite this as:

$$b = r - \frac{a}2 + \frac{a^2}{2(a-2r)}$$

For this to be an integer, you want $a^2$ to be divisible by $a-2r$. So if $a$ is given, look for the divisors $d|a^2$ for which $d<a$, and then let $r=\frac{a-d}2$. Of course you want this to be an integer too, so you also want $d$ to have the same parity are $a$.

Note that not all of these divisors will necessarily work, not only because these don't always give integer solution due to those halves in the expression, but also because there is no guarantee that even with integer $a$, $b$ and $r$ that they actually give a Pythagorean triangle.

This is all easy to do even for very large $a$ provided you have its prime factorization.

This also suggests that there will be several values of $r$ which could work with larger $a$, so that your conjecture is probably false even when restricted to only Pythagorean triangles.

Edit:
Note that Pythagorean triangles always have integer inradius, so your conjecture essentially says that no two Pythagorean triangles have the same shortest side. Here are two (primitive) Pythagorean triangles with the same shortest side: $(20, 21, 29)$ and $(20, 99, 101)$. Here is another pair sharing a shortest side with an odd length: $(105, 208, 233)$ and $(105,608,617)$.

Answer 2

Whilst not directly related to your method, some points I'd like to give are that this is the case of Fermat's last Theorem with $n=2$ and so you could use part of the proof from that to find values, which stems from the fact that we can say: $$a^2+b^2=c^2\Rightarrow (a/c)^2+(b/c)^2=1$$ so it is equivalent to trying to find integer coordinates $(a/c,b/c)$ on a unit circle.

There is also Euclid's formula for which two positive integers $m,n$ are coprime and give: $$\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}m^2-n^2\\2mn\\m^2+n^2\end{pmatrix}$$ Although this will not find all of the pythagorean triples that exist.