Proof Writing For Pythagorean Triple “Reduction” Problem

Question

You are given a primitive Pythagorean triple $(a,b,c)$ such that $a,b,c$ are integers and $a<b<c$. “Reduce” the triple via the transformation $(a,b,c)\to (a,\sqrt{b^2-a^2},b)$, rearranging if necessary such that we have our new terms in increasing order. Show that, under sufficiently many “reductions”, each primitive Pythagorean triple “reduces” to $(1,1,\sqrt{2})$.

I’ve experimentally verified this for the $(3,4,5)$ and $(5,12,13)$ triangles, and the friend of mine who is best at math claims that the answer is a “simple” proof by infinite descent. I want to know how to produce a rigorous proof.

Answer 1

The statement can be slightly generalized to the following statement: Let the ordered triple $a<b<c$ of positive real numbers be such that $$a^2<b^2<c^2~{\rm are~integers,~}a^2+b^2=c^2,{\rm ~and~}\gcd(a^2,b^2)=1. \cdots\cdots (1)$$ Then after a finite number of reduction steps described in the question, the original triple can be brought to $(1,1,\sqrt{2})$.

The proof goes as follows:

Case 1: If $a=\sqrt{b^2-a^2},$ then $$2a^2=b^2$$ $$\Rightarrow b^2=2,a^2=1~({\rm since~}\gcd(a^2,b^2)=1)$$ $$\Rightarrow a=1,b=\sqrt{2},$$ i.e. one gets the triple $(1,1,\sqrt{2})$.

Case 2: If $a<\sqrt{b^2-a^2}<b$ or $\sqrt{b^2-a^2}<a<b$, then the new triple satisfies the same conditions as in (1), so one can continue the descent.

But since $c^2$ is in a decreasing sequence of integers bounded below by $2$, Case 2 cannot continue infinitely. So after a finite number of steps, Case 1 is reached and the proof is complete.

Answer 2

I suppose it would be cleaner to start from the squares of a primitive Pythagorean triple (PPT) and then describe reducing in the form of $(a^2,b^2,c^2)\mapsto(a^2,b^2-a^2,b^2)$ (i.e. $(x,y,z)\mapsto(x,y-x,y)$), so that it would be clearer that every reduction didn't risk creating a series of nested radicals.

Beyond that, your proof just needs to demonstrate that starting from a PPT that every reduction along that path from a triple other than $(1,1,2)$ leads to a triple with a smaller sum where the two smallest terms are not the same. That last point is essential because a triple like $(15,15,17)$ would go to $(15,0,15)$ and would be a counterexample, so your proof would need to show that something like that wouldn't happen when you start from a PPT. I suspect your friend didn't think about that point when claiming that the proof would be "simple", although the details might not be too horrible.

Answer 3

The key property of primitive Pythagorean triples that underlies the result is the relative primality of the two legs of the triangle. (If the two legs are not coprime, the triple is not primitive.) So, we'll prove a more general result about pairs of coprime numbers, from which the desired result follows as a corollary.

Suppose $a_0$ and $b_0$ are nonnegative integers such that $0 < a_0 < b_0$ and $a_0$ and $b_0$ are coprime (which we'll write $a_0 \bot b_0$). We define the following recurrence for $i \geq 0$: \begin{align*} a_{i+1} &= \min(a_i, b_i - a_i) \\ b_{i+1} &= \max(a_i, b_i - a_i) \enspace. \end{align*} We claim that there exists a $j$ such that $a_j=1$ and $b_j=2$.

Let's prove our claim. It's not difficult to see that for all $i \geq 0$, $a_i \bot b_i$ and $a_i, b_i \geq 0$. Moreover, unless $a_i = 0$, \begin{equation*} a_{i+1} < b_{i+1} < b_i \enspace. \end{equation*} These observations imply that eventually $a_k = 0$ for some $k$. Now we recall that the only nonnegative coprime integers whose difference is $0$ are $1$ and $1$. Hence, if $a_k = 0$, it means that $a_{k-1} = b_{k-1} = 1$. But $a_{k-1} = b_{k-1} = 1$ means that \begin{align*} \min(a_{k-2}, b_{k-2} - a_{k-2}) &= 1 \\ \max(a_{k-2}, b_{k-2} - a_{k-2}) &= 1 \enspace. \end{align*} This implies \begin{equation*} a_{k-2} = b_{k-2} - a_{k-2} = 1 \end{equation*} and so $b_{k-2} = 2$. In conclusion, $j=k-2$ and the claim is proved.

On to the corollary. Note that if $a \bot b$, which is the case of the legs of a primitive Pythagorean triple, then $a^2 \bot b^2$. If we set $a_0 = a^2$ and $b_0 = b^2$, we'll reach a point where $a_j=1$ and $b_j=2$. In the recurrence with the square root, whose relation to our recurrence is explained by @MatthewDaly, we get to $(1, \sqrt{2}, x)$ for some $x$ and, at the next iteration we get $(1, 1, \sqrt{2})$.