Can someone help me with proof by contradiction?

Question

How does one prove by contradiction that the sum of the squares of two odd integers cannot be the square of an even integer?

Answer 1

Hint:

Let $a,b$ be the two odd integers such that $a^2+b^2=c^2$. Assume that $c$ is even (because you are using proof by contradiction).

So $\exists \, m,n,k \in \mathbb{Z}$ such that $a=2m+1$ and $b=2n+1$ and $c=2k$. Now

\begin{align*} a^2+b^2 & =c^2\\ (2m+1)^2 + (2n+1)^2 & = (2k)^2. \end{align*}

From here see if you can arrive at something like $2=4(\text{some integer})$.

Answer 2

How does one prove by contradiction that the sum of the squares of two odd integers cannot be the square of an even integer?

By assuming $a = 2m + 1$ is odd and $b = 2n + 1$ is odd and $c = 2k$ and assuming that $a^2 + b^2 = c^2$ or in other words $(2m+1)^2 + (2n+1)^2 = (2k)^2$ and getting a contradiction.

If that's impossible then the sum of two odd squares can't be a square of an even number.

So do it:

$(2m+1)^2 + (2n+1)^2 = (2k)^2$

$4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4k^2$

$4m^2 + 4m + 4n^2 + 4n + 2 = 4k^2$ etc.