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Thread: solutions of equations, using rings

  1. #1
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    solutions of equations, using rings

    My problem is to show that $\displaystyle m^3+14n^3-12=0$ has no solution in the integers.
    I've done other problems like this with one variable, where you show that there are no roots in, say $\displaystyle Z_3$ or $\displaystyle Z_5$. However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

    Thanks a lot!
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  2. #2
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    Quote Originally Posted by kimberu View Post
    My problem is to show that $\displaystyle m^3+14n^3-12=0$ has no solution in the integers.
    I've done other problems like this with one variable, where you show that there are no roots in, say $\displaystyle Z_3$ or $\displaystyle Z_5$. However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

    Thanks a lot!
    This is probably not the intended method, but it works fine...

    Assume towards a contradiction that there are integers $\displaystyle m,n$ with $\displaystyle m^3+14n^3-12$. Then

    $\displaystyle m^3=2(6-7n^3)$ $\displaystyle \implies$ $\displaystyle 2\big|m^3$ $\displaystyle \implies$ $\displaystyle m=2a$ for some integer $\displaystyle a$.

    So $\displaystyle 8a^3=12-14n^3$, that is, $\displaystyle 7n^3=2(3-2a^3)$, which implies $\displaystyle 2\big|7n^3$

    $\displaystyle \implies$ $\displaystyle n=2b$ for some integer $\displaystyle b$.

    So $\displaystyle 7(8b^3)=6-4a^3$, that is, $\displaystyle 2(14b^3+a^3)=3$. But this implies $\displaystyle 2\big|3$, which is a contradiction.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by kimberu View Post
    My problem is to show that $\displaystyle m^3+14n^3-12=0$ has no solution in the integers.
    I've done other problems like this with one variable, where you show that there are no roots in, say $\displaystyle Z_3$ or $\displaystyle Z_5$. However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

    Thanks a lot!
    Mod 14 works. However, so does mod 7.

    Working mod 7, we want to find m such that $\displaystyle m^3 \equiv 5 \text{ mod }7$.

    $\displaystyle 1^3 = 1$
    $\displaystyle 2^3 = 8=1$
    $\displaystyle 3^3 = 3*9=3*2=6=-1$
    $\displaystyle 4^3 = (-3)^3 = (-1)^33^3 = 1$
    $\displaystyle 5^3 = (-2)^3=-1$
    $\displaystyle 6^3 = (-1)^3=-1$

    And so none are congruent to 5 mod 7, and so your equation has no roots...
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