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Math Help - Integer Solutions

  1. #1
    Member Haven's Avatar
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    Integer Solutions

    Help!

    Show that 2*y^2 = 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 y = {p_1}^{a_1} *....* {p_t}^{a_t}

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

    I get if any of the exponents are even, then (\frac{{p_i}^{a_i}}{17}) = 1
    If any of the exponents are odd, then (\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; July 30th 2009 at 08:59 PM. Reason: LaTeX
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  2. #2
    Super Member PaulRS's Avatar
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    Let p be a prime dividing y. Then <br />
x^4  - 17 \equiv 0\left( {\bmod .p} \right)<br />

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

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

    But clearly, reading the equation module 17 we have: <br />
2y^2  \equiv x^4 \left( {\bmod .17} \right)<br />
since <br />
y \equiv z^2 \left( {\bmod .17} \right)<br />
for some integer z, coprime to 17, it follows that: <br />
2z^4  \equiv x^4 \left( {\bmod .17} \right) \Rightarrow 2 \equiv \left( {z^{ - 1} x} \right)^4 \left( {\bmod .17} \right)<br />
which is impossible since 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 ( a being coprime to p) x^n\equiv _{p} a has solutions if and only if <br />
a^{\tfrac{{p - 1}}<br />
{{\left( {p - 1,n} \right)}}}  \equiv 1\left( {\bmod .p} \right)<br />
, in our case n=4 and p = 17 and we see that <br />
2^{\tfrac{{16}}<br />
{{\left( {16,4} \right)}}}  = 2^4  \equiv  - 1\left( {\bmod .17} \right)<br />
-

    (*) The prime p cannot be 17, because y cannot be divisible by 17, otherwise the equation would imply that 17 also divides x. Let y=17\cdot y' and x=17 \cdot x' then <br />
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<br />
and note that the LHS is dividible by 17 while the RHS is not.
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