The positive integers $a,b,c$ are pairwise relatively prime, $a$ and $c$ are odd and the numbers satisfy the equation $a^2+b^2=c^2$. Prove that $b+c$ is the square of an integer.
Problem
Source: Baltic Way 1995
Tags: number theory, relatively prime, number theory proposed
applepi2000
08.10.2011 18:36
It's well known that for primitive pythagorean triples we can paramaterize $a=m^2-n^2, b=2mn, c=m^2+n^2$ so $b+c=(m+n)^2$...
We have $(c-b)(c+b)=a^2$, and since $(b+c, c-b)=(b+c, 2c)=(b+c, c)=(b, c)=1$, we have $b+c$ is a perfect square.
aahmeetface
14.06.2012 20:46
From Pythagorean triple, solution is very easy.
mathbuzz
16.06.2012 09:17
using general characterization of pythagorean triples a=t(r^2-s^2) , b=2rst , c=t(r^2+s^2) is enough for solution.because the problem condition (b,c)=1 forces that t=1 and we get that b+c=(r+s)^2
too easy
WakeUp
16.06.2012 15:00
Both have you have just regurgitated the first part of applepi2000's post...