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 April 11th, 2010, 11:35 PM #1 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 solution Hello everyone! I'm new to this site and have a doubt. 1. x squared minus 92y squared =1 Please reply as I don't know latex
 April 12th, 2010, 02:27 AM #2 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: solution Hello bentick, What do you mean? $x^2-92y^2=1$ or x^2-(92y)^2=1 Hoempa
April 12th, 2010, 03:08 AM   #3
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Quote:
 Originally Posted by bentick I'm new to this site and have a doubt.

Quote:
 Originally Posted by bentick 1. x squared minus 92y squared =1
What are you supposed to do with this equation? What were the instructions? What have you tried, and how far have you gotten?

 April 12th, 2010, 01:05 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,191 Thanks: 1649 There are solutions such as x = 1, y = 0.
 April 12th, 2010, 10:05 PM #5 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 Re: solution Please solve this equation. IE. Solve for the value of x and y. According to me, as I have solved it x^2 -92y^2=1 x^2 =1+92y^2 then substitute the value of x^2. If it is solved by substituting then the answer comes as 1=1 Is this correct?
April 13th, 2010, 10:22 AM   #6
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Re: solution

Quote:
 Originally Posted by bentick Please solve this equation. IE. Solve for the value of x and y.
Unless I miss my guess, this is the equation of an hyperbola. Every point on the graph is a "solution" to the equation.

Since you have only one equation but two unknowns, there is no way to find "the" solution, only many solutions or the graph.

 April 13th, 2010, 01:27 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,191 Thanks: 1649 Are x and y required to be integers? If so, are they required to be positive integers?
 April 14th, 2010, 01:25 AM #8 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 Re: solution @skipjack I don't know about that. My question is this solvable than the other solution I posted, if so please reply
 April 14th, 2010, 04:33 AM #9 Global Moderator   Joined: Dec 2006 Posts: 19,191 Thanks: 1649 You hadn't previously asked whether the equation can be solved. Instead, you said that the task was to solve for both x and y, which disallows your suggested substitution (since it would need the value of x² to be known in advance). As is stands, the equation has infinitely many real solutions, and infinitely many complex solutions that are not real. However, this type of equation is often set in a context that implies that only integer solutions are sought, which is why I asked about that. There are infinitely many solutions in positive integers, but I suspect you haven't been taught how to find them. Do any of your textbooks even mention Diophantine equations?
 April 14th, 2010, 10:06 PM #10 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 Re: solution @skipjack I'm going to 9th grade this year. @skipjack no my text books doesn't mention anything like Diophantine equations. I just came across this question. But now I got to know the answer : IE. there are infinite solutions. I found the values for x,y too. x=1, y=0 or x=1151, y=120

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