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

Murph Jones
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    Hint: An odd square has the form $(2n+1)^2=4n^2+4n+1$. If you add two such expressions, can the result be a multiple of $4$? – quasi Sep 27 '18 at 17:40

2 Answers2

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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 $2=4(\text{some integer})$.

Anurag A
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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.

fleablood
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