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Thread: Another Diophantine Question

  1. #1
    qwath
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    Another Diophantine Question

    Hello community. Recently, I received a homework problem that has completely stumped me.

    Show that the Diophantine equation: x^4 - 2y^4 = 1 has no nontrivial solutions. My hint is to consider congruences modulo 5.

    Thank you for any assistance you may offer.
    Last edited by qwath; Dec 19th 2006 at 04:22 AM. Reason: Wrote down the wrong equation, its x^4-2y^4=1
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  2. #2
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    Quote Originally Posted by qwath View Post
    Hello community. Recently, I received a homework problem that has completely stumped me.

    Show that the Diophantine equation: x^4 - 2y^4 = 1 has no nontrivial solutions. My hint is to consider congruences modulo 5.

    Thank you for any assistance you may offer.
    To consider congruences modulo 5 is to consider all the remainders,
    $\displaystyle 5k$
    $\displaystyle 5k+1$
    $\displaystyle 5k+2$
    $\displaystyle 5k+3$
    $\displaystyle 5k+4$
    When we raise to the exponent of 4 the forms take respectively,
    $\displaystyle 5k$
    $\displaystyle 5k+1$
    $\displaystyle 5k+1$
    $\displaystyle 5k+1$
    You might have realized that is Fermat's Little Theorem.

    But anyways the two possible forms of $\displaystyle a^4$ (a fourth power) are: $\displaystyle 5k,5k+1$.

    Thus, the possibilities are in the Diophantine Equation:
    $\displaystyle (5k)+2(5j)=5\cdot 0+1$---> Impossible, RHS different form.
    $\displaystyle (5k)+2(5j+1)=5\cdot 0+1$---> Impossible
    $\displaystyle (5k+1)+2(5j)=5\cdot 0+1$---> Possible
    $\displaystyle (5k+1)+2(5j+1)=5\cdot 0+1$---> Impossible.

    Thus, what your homework was asking was not correct.
    The second case, is still possible.
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