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Thread: Integer Solutions

  1. #1
    Member Haven's Avatar
    Jul 2009

    Integer Solutions


    Show that $\displaystyle 2*y^2$ = $\displaystyle x^4$- 17 has no integer solutions.
    (Hint: Show using quadratic reciprocity that y is a square mod 17 by considering each prime
    divisor of y. Use this to show 2 is a fourth power mod 17, a contradication.)

    Using the fundamental theorem of arithmetic, i assume $\displaystyle y = {p_1}^{a_1} *....* {p_t}^{a_t}$

    By Euler's Criterion (y/17) $\displaystyle \equiv$$\displaystyle y^8$ (mod 17)

    I get if any of the exponents are even, then $\displaystyle (\frac{{p_i}^{a_i}}{17}) = 1$
    If any of the exponents are odd, then $\displaystyle (\frac{{p_i}^{a_i}}{17}) = (\frac{p_i}{17})$

    This is where i get stuck. (Sorry about the bad notation, i haven't quite figured out LaTeX)
    Last edited by Haven; Jul 30th 2009 at 08:59 PM. Reason: LaTeX
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  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    Let $\displaystyle p$ be a prime dividing $\displaystyle y$. Then $\displaystyle
    x^4 - 17 \equiv 0\left( {\bmod .p} \right)

    That is: $\displaystyle
    x^4 \equiv 17\left( {\bmod .p} \right)
    $ so 17 is a quadratic residue module p. So it follows that $\displaystyle
    \left( {\tfrac{{17}}
    {p}} \right)_L = 1
    $ $\displaystyle (*)$

    By the Quadratic Reciprocity Theorem: $\displaystyle
    \left( {\tfrac{p}
    {{17}}} \right)_L \left( {\tfrac{{17}}
    {p}} \right)_L = \left( { - 1} \right){}^{\tfrac{{p - 1}}
    {2} \cdot \tfrac{{17 - 1}}
    {2}} = 1
    $ hence it follows that $\displaystyle
    \left( {\tfrac{p}
    {{17}}} \right)_L = 1
    $ , that is, all primes dividing $\displaystyle y$ are quadratic residues module 17, hence so is $\displaystyle y$.

    But clearly, reading the equation module 17 we have: $\displaystyle
    2y^2 \equiv x^4 \left( {\bmod .17} \right)
    $ since $\displaystyle
    y \equiv z^2 \left( {\bmod .17} \right)
    $ for some integer $\displaystyle z$, coprime to $\displaystyle 17$, it follows that: $\displaystyle
    2z^4 \equiv x^4 \left( {\bmod .17} \right) \Rightarrow 2 \equiv \left( {z^{ - 1} x} \right)^4 \left( {\bmod .17} \right)
    $ which is impossible since $\displaystyle 2$ is not the 4th power of an integer module 17. Thus we conclude that the equation can have no solutions.

    - an easy way to check this it to remember that ( $\displaystyle a$ being coprime to p) $\displaystyle x^n\equiv _{p} a$ has solutions if and only if $\displaystyle
    a^{\tfrac{{p - 1}}
    {{\left( {p - 1,n} \right)}}} \equiv 1\left( {\bmod .p} \right)
    $ , in our case n=4 and p = 17 and we see that $\displaystyle
    {{\left( {16,4} \right)}}} = 2^4 \equiv - 1\left( {\bmod .17} \right)

    $\displaystyle (*)$ The prime p cannot be 17, because $\displaystyle y$ cannot be divisible by 17, otherwise the equation would imply that 17 also divides $\displaystyle x$. Let $\displaystyle y=17\cdot y'$ and $\displaystyle x=17 \cdot x'$ then $\displaystyle
    2y^2 = x^4 - 17 \Rightarrow 2\left( {y'} \right)^2 \cdot \left( {17^2 } \right) - \left( {x'} \right)^4 \cdot \left( {17^4 } \right) = - 17
    $ and note that the LHS is dividible by 17 while the RHS is not.
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