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Math Help - Prove: if (a,b,c) is a Pythagorean triple, then a and b cannot both be odd

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    Member courteous's Avatar
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    Question Prove: if (a,b,c) is a Pythagorean triple, then a and b cannot both be odd

    Intro exercise:
    Show that any integer square leaves remainder 0 or 1 on division by 4.
    a_{even}=2k: a^2=(2k)^2=4(k^2)=0\mod{4}
    a_{odd}=2k-1: a^2=(2k-1)^2=4(k^2-k)+1=1\mod{4}  \square
    Deduce from above exercise that if (a,b,c) is a Pythagorean triple, then  a and  b cannot both be odd.
    c^2=a^2+b^2 \xrightarrow{?} c^2\mod{4}=(a^2+b^2)\mod{4}=\left\{ \begin{array}{ll}<br />
0 & \mbox{; $a^2=0\mod{4} \text{ AND } b^2=0\mod{4}$};\\<br />
1 & \mbox{; $a^2=1\mod{4} \text{ XOR } b^2=1\mod{4}$}.\end{array} \right. \square

    Are my two tries above worth the halmos \square sign?
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by courteous View Post
    Intro exercise: a_{even}=2k: a^2=(2k)^2=4(k^2)=0\mod{4}
    a_{odd}=2k-1: a^2=(2k-1)^2=4(k^2-k)+1=1\mod{4}  \square
    c^2=a^2+b^2 \xrightarrow{?} c^2\mod{4}=(a^2+b^2)\mod{4}=\left\{ \begin{array}{ll}<br />
0 & \mbox{; $a^2=0\mod{4} \text{ AND } b^2=0\mod{4}$};\\<br />
1 & \mbox{; $a^2=1\mod{4} \text{ XOR } b^2=1\mod{4}$}.\end{array} \right. \square

    Are my two tries above worth the halmos \square sign?
    (Edited)

    Hmm for the intro normally I would do the easy thing and just check all equivalence classes mod 4 exhaustively.

    0^2 \equiv 0\bmod{4}
    1^2 \equiv 1\bmod{4}
    2^2 \equiv 0\bmod{4}
    3^2 \equiv 1\bmod{4}

    But it seems that to check squares mod n^2 it is sufficient to check the equivalence classes mod n, and what you did is good.

    For the second part, I think it's best to rewrite since the theorem we want to prove involves a,b odd, for which we know a^2 + b^2 \equiv 1+1 \equiv 2 \bmod{4}. (You had this but it's not really explicit the way you wrote it, since you omitted the case a,b odd.)

    by the way to get \,\equiv in LaTeX you can use the command \equiv.
    Last edited by undefined; September 25th 2010 at 12:02 PM.
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