Proving if something is even

Let A,B,C be integers such that A^2 + B^2 = C^2. Prove at least one of A and B is even.

Please help guide me through this problem so I can understand it better thanks.

I have another question also

let a and b be positive such that a^2 = b^3

given 4 divides b, prove that 8 divides a

Im hoping if someone can guide me through this problem also thanks.

Re: Proving if something is even

For the first question, have you taken a look at Euclid's method for generating Pythagorean triples?

For the second question, I would let b = 4n...what do you find?

Re: Proving if something is even

for the first one i want to try to use a proof by contradiction but i am kind of stuck on how to do it

and for the 2nd one when b = 4n

b^3 = (4n)^3

so a^2 = (4n)^3

so a = sqrt((4n)^3)

Re: Proving if something is even

What is the square root of 4 cubed?

Re: Proving if something is even

Edit: Nevermind I figred everything out thanks a lot for your help!

Re: Proving if something is even

Hello, gfbrd!

Quote:

$\displaystyle \text{Let }A,B,C\text{ be integers such that }A^2 + B^2 \,=\, C^2.$

$\displaystyle \text{Prove that at least one of }A\text{ and }B\text{ is even.}$

First, we will establish a theorem.

Consider the square of an integer, $\displaystyle n^2.$

If $\displaystyle n$ is even, we have: .$\displaystyle n = 2k$

. . Hence: .$\displaystyle n^2 \:=\:(2k)^2 \:=\:4k^2$

If $\displaystyle n$ is odd, we have: .$\displaystyle n = 2k+1$

. . Hence: .$\displaystyle n^2 \:=\:(2k+1)^2 \:=\:4k^2 + 4k + 1 \:=\:4(k^2+k)+1$

Therefore, the square of an integer is either:

. . (1) a multiple of 4, or

. . (2) *one more* than a multiple of 4.

We are given: .$\displaystyle A^2 + B^2\,=\,C^2$

Suppose $\displaystyle A$ and $\displaystyle B$ are *both** *__odd__.

We have: .$\displaystyle A \,=\,2h+1,\;B \,=\,2k+1$

The we have:

. . $\displaystyle A^2 + B^2\;=\:(2h+1)^2 + (2k+1)^2$

. . . . . . . . .$\displaystyle =\:4h^2+4h+1 + 4k^2+4k+1$

. . . . . . . . .$\displaystyle =\;4(h^2+h + k^2+k) + 2$

Hence: $\displaystyle A^2+B^2$ is *two* more than a multiple of 4.

. . . . . .It can*not* equal a square, $\displaystyle C^2.$

Therefore, at least one of $\displaystyle A$ and $\displaystyle B$ must be even.