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Thread: Number Theory , an easy proof .

  1. #1
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    Number Theory , an easy proof .

    Let's first recall some Pythagorean Triples :


    $\displaystyle (3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61) $ etc .


    Can you see a pattern ? In every triple , we must find one integer from $\displaystyle (x,y,z) $ , such that it is the mutiple of $\displaystyle 5 $ .

    $\displaystyle 5,5,15,40,60$ etc .


    I would like you to prove this fact , the proof shouldn't be so long , i believe ...
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Let (a,b,c) be a pythagorean triple. Observe that $\displaystyle x^2 \equiv 0,1,4$ (mod) 5.

    Suppose neither $\displaystyle a,b,c\equiv 0$ (mod) 5. Then $\displaystyle a^2,b^2,c^2\equiv 1,4$ (mod) 5

    But then we can never have $\displaystyle a^2+b^2\equiv c^2$ (mod) 5

    Contradiction.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by simplependulum View Post
    Let's first recall some Pythagorean Triples :


    $\displaystyle (3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61) $ etc .


    Can you see a pattern ? In every triple , we must find one integer from $\displaystyle (x,y,z) $ , such that it is the mutiple of $\displaystyle 5 $ .

    $\displaystyle 5,5,15,40,60$ etc .


    I would like you to prove this fact , the proof shouldn't be so long , i believe ...
    The integers $\displaystyle m^2-n^2,\ 2mn,\ m^2+n^2$ form a Pythagorean triple.

    Consider modulo 5.

    If $\displaystyle m\equiv 0\ (mod\ 5)$ or $\displaystyle n\equiv 0\ (mod\ 5)$, then $\displaystyle 2mn\equiv 0\ (mod\ 5)$.

    If $\displaystyle m\equiv\pm n\ (mod\ 5)$, then $\displaystyle m^2-n^2\equiv 0\ (mod\ 5)$.

    If $\displaystyle m\equiv\pm2\ (mod\ 5)$ and $\displaystyle n\equiv\pm1\ (mod\ 5)$ or $\displaystyle m\equiv\pm1\ (mod\ 5)$ and $\displaystyle n\equiv\pm2\ (mod\ 5)$, then $\displaystyle m^2+n^2\equiv 0\ (mod\ 5)$.

    This concludes all the cases.
    Last edited by alexmahone; Aug 2nd 2010 at 06:18 AM.
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  4. #4
    Super Member Bacterius's Avatar
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    Hello,
    The proof is trivial. Let $\displaystyle x \equiv a \pmod{5}$, $\displaystyle y \equiv b \pmod{5}$, $\displaystyle z \equiv c \pmod{5}$. Now consider a Pythagorean triple :

    $\displaystyle x^2 + y^2 = z^2$

    Since all terms $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ are modulo $\displaystyle 5$, we may equate and say that :

    $\displaystyle a^2 + b^2 \equiv c^2 \pmod{5}$

    According to the conjecture, this congruence may only be satisfied if and only if either $\displaystyle a, b, c \equiv 0 \pmod{5}$ (if one of these integers is divisible by $\displaystyle 5$). Recall the quadratic residues modulo $\displaystyle 5$ :

    $\displaystyle 0^2 \equiv 0 \pmod{5}$
    $\displaystyle 1^2 \equiv 1 \pmod{5}$
    $\displaystyle 2^2 \equiv 4 \pmod{5}$
    $\displaystyle 3^2 \equiv 4 \pmod{5}$
    $\displaystyle 4^2 \equiv 1 \pmod{5}$

    Now consider all the possibilities (I omitted those that didn't satisfy the conjecture because there's too much of them, this has been double-checked by computer) :

    $\displaystyle 0^2 + 0^2 = 0^2 \pmod{5}$
    $\displaystyle 0^2 + 1^2 = 1^2 \pmod{5}$
    $\displaystyle 0^2 + 1^2 = 4^2 \pmod{5}$
    $\displaystyle 0^2 + 2^2 = 2^2 \pmod{5}$
    $\displaystyle 0^2 + 2^2 = 3^2 \pmod{5}$
    $\displaystyle 0^2 + 3^2 = 2^2 \pmod{5}$
    $\displaystyle 0^2 + 3^2 = 3^2 \pmod{5}$
    $\displaystyle 0^2 + 4^2 = 1^2 \pmod{5}$
    $\displaystyle 0^2 + 4^2 = 4^2 \pmod{5}$
    $\displaystyle 1^2 + 0^2 = 1^2 \pmod{5}$
    $\displaystyle 1^2 + 0^2 = 4^2 \pmod{5}$
    $\displaystyle 1^2 + 2^2 = 0^2 \pmod{5}$
    $\displaystyle 1^2 + 3^2 = 0^2 \pmod{5}$
    $\displaystyle 2^2 + 0^2 = 2^2 \pmod{5}$
    $\displaystyle 2^2 + 0^2 = 3^2 \pmod{5}$
    $\displaystyle 2^2 + 4^2 = 0^2 \pmod{5}$
    $\displaystyle 3^2 + 0^2 = 2^2 \pmod{5}$
    $\displaystyle 3^2 + 0^2 = 3^2 \pmod{5}$
    $\displaystyle 3^2 + 1^2 = 0^2 \pmod{5}$
    $\displaystyle 3^2 + 4^2 = 0^2 \pmod{5}$
    $\displaystyle 4^2 + 0^2 = 1^2 \pmod{5}$
    $\displaystyle 4^2 + 0^2 = 4^2 \pmod{5}$
    $\displaystyle 4^2 + 2^2 = 0^2 \pmod{5}$
    $\displaystyle 4^2 + 3^2 = 0^2 \pmod{5}$
    $\displaystyle 2^2 + 1^2 = 0^2 \pmod{5}$

    By analyzing the triples that do satisfy the conjecture, we find that all those triples have one, and only one, integer that is a multiple of $\displaystyle 5$ ... except the first one. However, apart from $\displaystyle 0^2 + 0^2 = 0^2$ which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by $\displaystyle 25$ cannot possibly be divisible by $\displaystyle 25$ (the proof is left as an exercise to the reader ).

    Therefore the conjecture is verified and valid.

    $\displaystyle \mathfrak{QED}$

    (okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
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  5. #5
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    Any integer is congruent to $\displaystyle 0,\pm 1,\pm 2$ modulo 5 (complete residue system). So any square is congruent modulo 5 to $\displaystyle 0,\pm 1$.
    Assume that both $\displaystyle a^2$ and $\displaystyle b^2$ are not congruent modulo 5 to $\displaystyle 0$, then $\displaystyle a^2+b^2\equiv 0,2,$ or $\displaystyle -2(mod\ 5)$.
    Since $\displaystyle c^2=a^2+b^2$ and $\displaystyle c^2 $ is a square, we must have $\displaystyle c^2\equiv 0(mod\ 5)$.
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  6. #6
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    Thanks , indeed , all the solutions here are not so long . I want to include my proof in this post :


    Suppose all the primitive solutions of the equation $\displaystyle x^2 + y^2 = z^2 $ can be represented by $\displaystyle (m^2 - n^2 , 2mn , m^2 + n^2 ) $ . $\displaystyle x $ is odd while $\displaystyle y $ is even , $\displaystyle m , n $ are coprime and one odd one even . (It is true but the proof not included here . )

    Then take modulo $\displaystyle 5 $ , we have

    $\displaystyle m^2 - n^2 = (m-n)(m+n) $

    $\displaystyle m^2 + n^2 \equiv m^2 - 4n^2 = (m-2n)(m+2n) $

    consider the product $\displaystyle xyz $ , assume $\displaystyle n $ is prime to $\displaystyle 5 $ , otherwise , $\displaystyle y $ is divisible by $\displaystyle 5 $ .

    We have $\displaystyle xyz \equiv 2n(m-2n)(m-n)(m)(m+n)(m+2n) \equiv 2n^5 (s-2)(s-1)(s)(s+1)(s+2) \bmod{5} $ where $\displaystyle s \equiv m(n^{-1}) \bmod{5}$ . We can see that the product $\displaystyle (s-2)(s-1)(s)(s+1)(s+2) $ is divisible by $\displaystyle 5$ so is $\displaystyle xyz $ , this finishes the proof .
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