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Thread: Odd squares

  1. #1
    Senior Member I-Think's Avatar
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    Odd squares

    Good night.
    I have a proof question here that I needs validating. I was also wondering if I could receive alternative solutions for the question.
    Thanks in advance for the help

    Question
    Prove that the square of an odd integer is always of the form $\displaystyle 8q+1$ where $\displaystyle q $ is an integer

    Answer
    We wish to prove that $\displaystyle 8q+1=k^2$, where $\displaystyle k $ is an odd integer
    Let $\displaystyle k $ be an odd integer of the form: $\displaystyle 2I+1$
    $\displaystyle (2I+1)^2=4I^2+4I+1$

    As $\displaystyle I $is an integer, it can either be odd or even
    Case 1: $\displaystyle I $is odd, i.e. it is of the form:$\displaystyle 2m+1$
    $\displaystyle (2I+1)^2= 4(2m+1)^2+4(2m+1)+1$
    $\displaystyle 16m^2+16m+4+8m+4+1=16m^2+24m+9$
    $\displaystyle (16m^2+24m+8)+1 $
    $\displaystyle 8(2m^2+3m+1)+1$ Let $\displaystyle 2m^2+3m+1=q$
    $\displaystyle 8q+1$

    Case 2:$\displaystyle I $is even, i.e. it is of the form $\displaystyle 2m$
    $\displaystyle (2I+1)^2= 4(2m)^2+4(2m)+1$
    $\displaystyle 16m^2+8m+1=(16m^2+8m)+1$
    $\displaystyle 8(2m^2+m)+1$ Let $\displaystyle 2m^2+m=q$
    $\displaystyle 8q+1$

    Proven in both cases, hence the statement is true.
    End of solution. Q.E.D

    I welcome your critiques and say thanks in advance,
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  2. #2
    Super Member

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    Hello, I-Think!

    Your proof is correct.


    Prove that the square of an odd integer is always of the form $\displaystyle 8q+1$ where $\displaystyle q $ is an integer.

    Let the odd integer be: .$\displaystyle n \:=\:2p+1\,\text{ for some integer }p.$

    . . Then: .$\displaystyle n^2 \:=\:(2p+1)^2 \:=\:4p^2 + 4p + 1$


    Here is a very sneaky step . . .

    $\displaystyle \text{We have: }\;n^2 \;=\;4\underbrace{p(p+1)} + 1\;\;{\color{blue}[1]}$
    . . . . . . . .
    two consecutive integers

    With two consecutive integers, one of them is even.
    . . Hence, their product is even: .$\displaystyle p(p+1) \:=\:2q\,\text{ for some integer }q.$

    Then $\displaystyle {\color{blue}[1]}$ becomes: .$\displaystyle n^2 \;=\;4(2q) + 1 \;=\;8q+1 \quad\hdots\;\text{ta-}DAA!$

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