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Thread: Solve a Integer equation system

  1. #1
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    Solve a Integer equation system

    $\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2 = 3b^3+2 = 5c^5+3$
    Find the smallest $\displaystyle d$ above
    What if $\displaystyle d = 2a^2+1 = 3b^3+2 = 5c^5+3$ ?
    Last edited by elim; May 4th 2010 at 08:02 PM.
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  2. #2
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    Quote Originally Posted by elim View Post
    $\displaystyle a,b,c \in \mathbb{N}^+, \quad d = 2a^2+1 = 3b^3+2 = 5c^5+3$
    Find the smallest $\displaystyle d$ above
    If $\displaystyle 2a^2+1 = 3b^2+2$ then $\displaystyle 2a^2\equiv 1\!\!\!\pmod3$, which is impossible. So no solutions there. But the remaining equations $\displaystyle d = 2a^2+1 = 5c^5+3$ have at least one solution d = 163 (with a = 9 and c = 2).
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    Thanks Opalg! Let's forget that half of the problem
    What about the smallest $\displaystyle d=2a^2=3b^3=5c^5+3$?
    Last edited by elim; May 4th 2010 at 08:04 PM.
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    By Chinese Remainder Theorem,
    $\displaystyle d = 38+30k$ for some $\displaystyle k$
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    Integer solution for $\displaystyle d=2a^2=3b^3+2=5c^5+3$
    become very tough. I heard that $\displaystyle d < 10^{50}$ has no solutions
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    So $\displaystyle \frac{d}{2}$ is a square, $\displaystyle \frac{d-2}{3}$ is a cube, and $\displaystyle \frac{d-3}{5}$ is a fifth power.

    If you look at these conditions modulo some small numbers, you might find something.

    - Hollywood
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