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.
Best Answer
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.