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LolWolf

  • 3 years ago

*Math challenge* Prove: \[ x^2+y^2=z^2\implies 60|xyz\\ x,y,z\in \mathbb{Z} \] Why IS this site so addicting?

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  1. Chipper10
    • 3 years ago
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    Can you explain the notation on the right?

  2. LolWolf
    • 3 years ago
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    Yep, \(a|b\) means \(a\) divides \(b\), or a more formal definition: \[ a, b\in \mathbb{Z}\\ a|b\implies\exists k\in\mathbb{Z}\text{ s.t. }ak=b \]

  3. Chipper10
    • 3 years ago
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    So 60 is a factor of xyz or a multiple?

  4. LolWolf
    • 3 years ago
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    60 is a factor of \(xyz\).

  5. Chipper10
    • 3 years ago
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    Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.

  6. LolWolf
    • 3 years ago
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    The correction is up.

  7. experimentX
    • 3 years ago
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    the pythagoras triplets are given by {2m, m^2-1, m^2+1} 2m * (m^2 - 1) * (m^2+1) = 60 * k where k is an integer.

  8. experimentX
    • 3 years ago
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    60 = 2^2*3*5 let m=2

  9. mukushla
    • 3 years ago
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    if \(a^2+b^2=c^2\) then general form of primitive pyth triples is in the form\[a=m^2-n^2\]\[b=2mn\]\[c=m^2+n^2\]where \(\gcd(m,n)=1\) and one of \(m,n\) is even

  10. experimentX
    • 3 years ago
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    it seems all multiples of n* 3,4,5 is solution of the given condition;

  11. experimentX
    • 3 years ago
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    for n=3, 2*3*(3^2 - 1) * (3^2 +1) = 2 * 3 * 2^3 * 2*5 <-- a multiple of 60 so, n * 6,8,10 is also multiple of 60

  12. experimentX
    • 3 years ago
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    n=4, 2* 4 * 15* 17 <-- multiple of 60

  13. mukushla
    • 3 years ago
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    exper u let n=1 and i think we must work on general form

  14. experimentX
    • 3 years ago
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    n=5, 10*24*26 n=6 .... probably yes for n>1

  15. experimentX
    • 3 years ago
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    let's try induction.

  16. mukushla
    • 3 years ago
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  17. experimentX
    • 3 years ago
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    2m * (m^2 - 1) * (m^2+1) = 60 k 2* m * (m+1)(m-1)(m^2 + 1) ^ ^ ^ ^ ^ 1 2 3 4 5 let's try that for m=7 2*7*8*6*50

  18. mukushla
    • 3 years ago
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    one of m,n is even so \[4|2mn\]so\[4|abc\]we need to prove \(3|abc\) and \(5|abc\)

  19. mukushla
    • 3 years ago
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    if \(3|m\) or \(3|n\) then \(3|abc\) . if not then\[m^2 \equiv1 \ \ \text{mod} \ 3\]\[n^2 \equiv1 \ \ \text{mod} \ 3\]and\[m^2-n^2 \equiv0 \ \ \text{mod} \ 3\]so \(3|abc\)

  20. experimentX
    • 3 years ago
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    m * (m+1)(m-1) for any value of m, this is divisible by 3

  21. experimentX
    • 3 years ago
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    m * (m+1)(m-1)(m^2 + 1) ^ ^ ^ can be of the form 5n+2 or 5n+3

  22. experimentX
    • 3 years ago
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    (5n+2)^2 +1 = .... 2^2 + 1 = 5 (5n + 3)^2 + 1 = .... 3^2 + 1 = 10 both cases divisible by 5

  23. experimentX
    • 3 years ago
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    for all n>1, the Pythagorean triplet is divisible by 60 (probably)

  24. mukushla
    • 3 years ago
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    if \(5|m\) or \(5|n\) then \(5|abc\) . if not then\[m^2 \equiv 1,4 \ \ \text{mod} \ 5\]\[n^2 \equiv 1,4 \ \ \text{mod} \ 5\]now if \(m^2 \equiv n^2 \ \ \text{mod} \ 5\) then \(5|m^2-n^2\) and \(5|abc\) otherwise \(m^2 \not\equiv n^2 \ \ \text{mod} \ 5\) \(c=m^2+n^2 \equiv1+4\equiv0 \ \ \text{mod}\ 5 \) so \(5|abc\)

  25. mukushla
    • 3 years ago
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    so \(3,4,5|abc\) it gives \(60|abc\).

  26. Herp_Derp
    • 3 years ago
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    FERMAT'S LAST THEOREM: Let \(n, a, b, c \in \mathbb{Z}\) with \(n > 2\). If \(a^n +b^n = c^n\) then \(abc= 0\). Proof: The proof follows a program formulated around 1985 by Frey and Serre. By classical results of Fermat, Euler, Dirichlet, Legendre, and Lamé, we may assume \(n = p\), an odd prime \(>11\). Suppose \(a, b, c \in \mathbb{Z}\), \(abc \ne 0\), and \(a^p + b^p = c^p\). Without loss of generality we may assume \(2|a\) and \(b \equiv 1 \mod 4\). Frey observed that the elliptic curve \(E : y^2 = x(x - ap)(x + bp)\) has the following "remarkable" properties: (1) \(E\) is semistable with conductor \(N_E =\Pi_{\ell|abc}\ell\); and (2) \(\bar{\rho}_{E,p}\) is unramified outside 2p and is flat at at p. By the modularity theorem of Wiles and Taylor-Wiles, there is an eigenform \(f \in \mathcal{S}_2(\Gamma_0(N_E))\) such that \(\rho_{f,p} =\rho_{E,p}\). A theorem of Mazur implies that \(\bar{\rho}_{E,p}\) is irreducible, so Ribet's theorem produces a Hecke eigenform \(g \in \mathcal{S}_2(\Gamma_0(2))\) such that \(\rho_{g,p} \equiv \rho_{f,p} \mod\wp\) for some \(\wp\). But \(X_0(2)\) has genus zero, so \(\mathcal{S}_2(\Gamma_0(2)) = 0\). This is a contradiction and Fermat's Last Theorem follows. \(\mathbf{Q.E.D.}\)

  27. mukushla
    • 3 years ago
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    o.O

  28. experimentX
    • 3 years ago
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    what the hell is that ....

  29. experimentX
    • 3 years ago
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    can't follow ... probably not my stuff.

  30. Herp_Derp
    • 3 years ago
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    @mukushla I copied it from your pdf

  31. mukushla
    • 3 years ago
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    lol

  32. LolWolf
    • 3 years ago
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    Fermat's? Lol, only like 100 people in the world understand Wiles' proof of the Taniyama–Shimura conjecture XD But, anyways, you guys are correct, but missing one essential step: Prove that if both of \(m,n\) are odd, then it is not a primitive solution. Otherwise, you've got it, nice job!

  33. zhess21
    • 3 years ago
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    4. extra points to whomever gets the reference

  34. mukushla
    • 3 years ago
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    if both of \(m,n\) are odd\[m^2-n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]

  35. LolWolf
    • 3 years ago
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    Yep. Nice job!

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