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Thread: Nonsense proof I am afraid

  1. #1
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    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?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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 $\displaystyle z^2\equiv 0 \mod 4$.
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  3. #3
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    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.

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  4. #4
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    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.
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