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

Printable View

• September 24th 2010, 10:38 PM
james_bond
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 $\mathbb{Q}$?
• September 25th 2010, 03:46 AM
Opalg
Quote:

Originally Posted by james_bond
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 $(x,y) = (6,-3),\; \bigl(\frac{69}{13},-\frac3{13}\bigr),\; \bigl(\frac{942}{181},-\frac3{181}\bigr),\; \bigl(\frac{13101}{2521},-\frac3{2521}\bigr),\ldots\,.$
• September 26th 2010, 10:48 AM
james_bond
How can we show that there exists infinitely many $b$ integer so that $3(4b^2-1)$ is a square?
• September 27th 2010, 12:37 AM
Opalg
Quote:

Originally Posted by james_bond
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.