1. Proof Method

Hi all,

I need the proof of the following:
If $\displaystyle A, B ,$ and $\displaystyle C$ are positive integers such that $\displaystyle C^2= A^2 + B^2$ then either $\displaystyle A$ or $\displaystyle B$ is an even integer.

I could be grateful if you help me.

Raed

2. If A and B are both odd, then C^2 is a multiple of 4. (It can't be even but not divisible by 4, right?) Then if you represent A and B as 2 times an even number plus 1, then the right-hand side is not divisible by 4.

3. Originally Posted by raed
Hi all,

I need the proof of the following:
If $\displaystyle A, B ,$ and $\displaystyle C$ are positive integers such that $\displaystyle C^2= A^2 + B^2$ then either $\displaystyle A$ or $\displaystyle B$ is an even integer.

I could be grateful if you help me.

Raed
You could also try to prove that the squares of a pair of odd integers do not sum to an integer square...

$\displaystyle A=2f+1,\;\;\;B=2j+1$

$\displaystyle C$ will have to be even, since $\displaystyle Odd^2+Odd^2=Even$

Odd squares are Odd and Even squares are Even, hence C is Even.

$\displaystyle C=2k$

$\displaystyle A^2+B^2=(2f+1)^2+(2g+1)^2=4f^2+4f+1+4g^2+4g+1$

This needs to equal $\displaystyle 4k^2$

$\displaystyle \Rightarrow\ 2\left(2f^2+2g^2+2f+2g+1\right)=2\left(2k^2\right)$

$\displaystyle 2\left(f^2+g^2+f+g\right)+1=2k^2$

However, this is contradictory as the LHS is Odd.