Talk:Abc conjecture

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I've read that Fermat's Last Theorem is an easy consequence of this conjecture. If I get the chance I'll see if I can find out why and add it, unless someone else wants to do so in the meantime. --ScottAlanHill 16:54, 9 March 2006 (UTC)

Since the conjecture says nothing on the constant Cε, I don't see how it should rule out the existence of finitely many solutions for FLT. However, the abc-conjecture does easily imply that for n>3 there are at most finitely many solutions: If X,Y,Z are positive integers without common divisor with XN + YN = ZN , then rad(XNYNZN) = rad(XYZ) and rad(XYZ) \le Z^3 and therefore the abc-conjecture (if true) yields

Z^N < C_{\varepsilon} \operatorname{rad}(X^NY^NZ^N)^{1+\epsilon} \le C_{\varepsilon} Z^{3+3\epsilon} ,

which implies an upper bound for Z.

Joerg Winkelmann 20:22, 12 May 2006 (UTC)

[edit] Math-mode formulas

Is there a reason not to use math formulas here? These are the math-style formulas:

  • a + b = c
    a + b = c
  • rad(abc),
    \operatorname{rad}\,(abc)
  • rad(abc)/c
    \frac{\operatorname{rad}\,(abc)}{c}
  • rad(abc)1+ε/c
    \frac{{\operatorname{rad}\,(abc)}^{1+\epsilon}}{c}

Vegard 14:01, 29 September 2006 (UTC)

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