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freemind
Riemann Hypothesis
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Location: MIT or Moldova
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Post Posted: Mar 29, 2008, 7:41 am • # 1 


Find all solutions (x,y)\in \mathbb{R}\times\mathbb R of the following system: \begin{cases}x^3 + 3xy^2 = 49, \\ x^2 + 8xy + y^2 = 8y + 17x.\end{cases}

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Last edited by freemind on Mar 29, 2008, 10:33 am, edited 1 time in total.
 
 
silouan
Birch & Swinnerton Dyer
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Post Posted: Mar 29, 2008, 10:31 am • # 2 


Here is my solution to this problem .
Clearly x\neq 0 so from the first equation we get y^2 = \frac {49}{3x} - \frac {x^2}{3}

Subtituting in the second equation we get x^2 + 8xy+\frac {49}{3x} - \frac {x^2}{3} - 17x - 8y = 0 or equivalently

(x - 1)(2x^2 + 24xy - 49x - 49) = 0 . Now if x = 1 we get y = 4 or y = - 4

If 2x^2 + 24xy - 49x - 49 = 0 we get 2x^2 + 24xy = 49 + x^3 + 3xy^2 or

2x + 24y = 49 + x^2 + 3y^2 or (x - 1)^2 + 3(y - 4)^2 = 0 so we get again the same solution x = 1 and y = 4

Finally the solutions are (x,y) = \left\{(1,4),(1, - 4)\right\}

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