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Math Help - Number of solutions of x^2 + xy + y^2 = 27 in Q

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    Number of solutions of x^2 + xy + y^2 = 27 in Q

    How many solution does the x^2+xy+y^2=27 equation have in the set of \mathbb{Q}?
    Last edited by mr fantastic; September 25th 2010 at 02:19 AM. Reason: Edited title.
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    Quote Originally Posted by james_bond View Post
    How many solution does the x^2+xy+y^2=27 equation have in the set of \mathbb{Q}?
    There are infinitely many rational solutions to this equation. One family of solutions comes from taking y to have numerator 3, and using the continued fraction expansion of \sqrt3. That leads to the solutions (x,y) = (6,-3),\; \bigl(\frac{69}{13},-\frac3{13}\bigr),\; \bigl(\frac{942}{181},-\frac3{181}\bigr),\; \bigl(\frac{13101}{2521},-\frac3{2521}\bigr),\ldots\,.
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    How can we show that there exists infinitely many b integer so that 3(4b^2-1) is a square?
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    Quote Originally Posted by james_bond View Post
    How can we show that there exists infinitely many b integer so that 3(4b^2-1) is a square?
    If 3(4b^2-1) = a^2, suppose for convenience that a is a multiple of 3, a = 3c. Then 4b^2-1 = 3c^2, or \bigl(\frac{2b}c\bigr)^2 - \frac1{c^2} = 3. If c is large, then the left side will be close to \bigl(\frac {2b}c\bigr)^2 and so \frac {2b}c must be close to \sqrt3. So go to the continued fraction calculator and plug in "3" in the square root box. You will then see a list of the convergents for the continued fraction expansion of \sqrt3, and you will notice that every fourth item in the list has an even numerator: \frac21,\;\frac{26}{15},\;\frac{362}{209},\,\ldots  \,.

    Let 2p_n/q_n be the n'th term in that sequence. Then p_1=q_1=1 and the sequence grows by the inductive rules p_{n+1} = 7p_n + 6q_n, q_{n+1} = 8p_n + 7q_n. You should then be able to show by induction that 4p_n^2-1 = 3q_n^2. That gives an infinite family of integer solutions to the equation 4x^2-1 = 3y^2. Going back to the original problem, 3(4x^2-1) = (3y)^2, so that gives infinitely many integers x such that 3(4x^2-1) is a square.
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