# Thread: Nonsense proof I am afraid

1. ## Nonsense proof I am afraid

Prove that the sum of the squares of two odd integers cannot be a perfect square.

The author wrote the follow proof:

Assume, to the contrary, that there exist odd integers $\displaystyle x$ and $\displaystyle y$such that $\displaystyle x^2 + y^2 = z^2$, where $\displaystyle z \in \mathbb{Z}$. Then $\displaystyle x = 2a+1$ and $\displaystyle y = 2b+1$, where $\displaystyle a,b \in \mathbb{Z}$. Thus

$\displaystyle x^2+y^2 = (2a+1)^2+(2b+1)^2$
$\displaystyle =4a^2+4a+1+4b^2+4b+1$
$\displaystyle =4(a^2+1+b^2+b)+2=2[2(a^2+a+b^2+b)+1]=2s$

where $\displaystyle s = 2(a^2+a+b^2+b)+1$ is odd integer. If $\displaystyle z$ is even, then $\displaystyle z=2c$ for some integer $\displaystyle c$ and so $\displaystyle z^2 = 2(2c)$, where $\displaystyle 2c^2$ is an even integers; while $\displaystyle z$ is odd, than $\displaystyle z^2$ is odd. Produce a contradiction in each case.

Remark: I just don't see how z can be anything else be even. To me it's obvious $\displaystyle z^2=2s$ explicitly implied that $\displaystyle z^2$ is even, which by some theorem, $\displaystyle z$ is also even.

Question: Which part of the proof gave the author the reason to believe that it's odd? Further how in the case of it being even be a contradiction to $\displaystyle z^2$ being a perfect square?

I just can't see it. Could someone please show me the author's intent?

2. The proof is convoluted because the author did not use modular arithmetic. Here's my translation of the proof using modular arithmetic.

Suppose the sum of two odd squares $\displaystyle x^2,y^2$ is a square $\displaystyle z^2$; then it has to be a square of an even integer, so $\displaystyle z^2=(2s)^2=4s\equiv 0 \mod 4$. Now note that any odd square is $\displaystyle \equiv 1 \mod 4$. (Expand $\displaystyle (2a+1)^2$). Therefore $\displaystyle x^2+y^2 \equiv 1+1 \equiv 2 \mod 4$ which contradicts $\displaystyle z^2\equiv 0 \mod 4$.

3. Hello, novice!

What a convoluted proof . . .
This is a slight variation of Bruno J's solution.

Prove that the sum of the squares of two odd integers cannot be a perfect square.
Here's the proof I was shown many years ago.

We note that: .$\displaystyle (2n)^2 \:=\:4n^2$
. . The square of an even number is a multiple of 4.

And that: .$\displaystyle (2n+1)^2 \:=\:4n^2 + 4n +1 \:=\:4(n^2+n)+1$
. . The square of an odd number is one more than a mutliple of 4.

Hence, all squares are either a multiple of 4 or one more than a multiple of 4.

Consider two odd numbers: .$\displaystyle 2a+1\,\text{ and }\,2b+1$

The sum of their squares is: .$\displaystyle S \;=\;(2a-1)^2 + (2b-1)^2$

. . $\displaystyle S \;=\;4a^2 - 4a + 1 + 4b^2 - 4b + 1 \;=\;4(a^2-a+b^2-b) + 2$

The sum is two more than a multiple of 4; it cannot be a square.

4. Oh, gentlemen,

Thank you for coming to my rescue. I am so happy to learn that it's not a failure on my part to understand the proof.

Thank you both for showing me the good stuff.