# Proof for interesting pattern in Pythagorean triples?

In my research on right triangles I have found an interesting observation. It would be better if I explained it to you with an example. Please keep in mind that I am talking about integer right-triangles. Consider the triple 3,4,5. The sum of the (successor of 3) and 4, is 4 + 4 = 8. Here, 8 is a number which is the leg of more than 2 triangles: 6,8,10 and 8,15,17 etc. Again add 8 to 8. 8+8=16. 16 is the leg of another two right triangles (or probably many): 12,16,20 and 16,30,34. Again add 8 to 16. 8+16=24. 24 is the leg of two (probably many) triangles: 7,24,25 and 18, 24, 30.

This keeps going on and on for any triple you take and follow this procedure:

Add the successor of the smallest leg to the greater leg. Take it as x.

...(((X+x)+x)+x).... Are numbers which are definitely the legs of different right triangles. Is there any generalised proof for this or is there one triple which doesn't satisfy the procedure? My research tells me that there aren't any. I thank you for your help in advance.

Thanks,

S Sandeep

## 1 Answer

Note that any number $b$ that is even and such that $b/2$ is not prime (nor the square of a prime) is the leg of more than one right triangles: We can write $b=2uv$ with $u>v>1$ in such a case and have two distinct Pythagorean triples with $b$ as leg: \begin{align} (u^2-v^2)^2+b^2&=(u^2+v^2)^2\\(u^2v^2-1)^2+b^2&=(u^2v^2+1)^2.\end{align} (For the sake of completeness: If $b=2p^2$ with $p$ an odd prime, we get two Pythagorean triples from $(p^4-1)^2+b^2=(p^4+1)^2$ and $(p^3-p)^2+b^2=(p^3+p)^2$)

On the other hand, if $b=2p$ then there is only one Pythagorean triple with $b$ as leg because the factorization of $4p^2=c^2-a^2=(c+a)(c-a)$ into two distinct factors of equal (hence even) parity is only possible as $c+a=2p^2$, $c-a=2$.

Now if you start from any primitive Pythagorean triangle $a^2+b^2=c^2$, then one of $a,b$ is odd and one is even, hence $x:=(a+1)+b$ is even. So the question is: Can it happen that $x=2p$? Yes, it can: $45^2+28^2=53^2$ gives you $45+1+28=2\cdot 37$, which is a leg only of $1368^2+74^2=1370^2$.