# My guess ... easy proof ?

• Jul 22nd 2010, 05:05 AM
simplependulum
My guess ... easy proof ?
This is my guess , i hope that's true .

Let's say a triangle is a Heron triangle if the lengths of its three sides as well as its area
are integers.

I guess : If $p ,q$ are primes such that $p^2 + 1 = 2q$ , then all the non-isosceles Heron triangles with $p$ as the length of one of the sides are right-angled . Therefore , we can only find out one or two possible Heron triangle(s) for it .

For example , let $p = 5$ we have $5^2 + 1 = 2(13)$

Then the only Heron triangles are $(3,4,5) ~ ,~ (5,12,13)$ ?
• Jul 22nd 2010, 05:36 AM
TheCoffeeMachine
How about $\left\{13, 14, 15\}$ or $\left\{12, 35, 37\}$?

EDIT (Oops!):

$13^2+1 = 170 = 2(75)$ and $75$ is not prime.
$37^2+1 = 1370 = 2(685)$ and [LaTeX ERROR: Convert failed] is not prime.
• Jul 22nd 2010, 07:55 PM
simplependulum
Oh , i missed a thing so what i guessed is not necessarily correct !

For example , $29^2 + 1 = 2(421)$ but $(29,52,69)$ is not right-angled .
• Jul 23rd 2010, 03:28 AM
TheCoffeeMachine
Had it been true, it would have been a curiously nice result; and we would have called it Simplependulum's
postulate (because it oddly reminded me of Bertrand's postulate). Keep guessing, my friend; keep guessing! (Yes)