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Prove that \(25 x^2 -6x -95 = y^2-y\quad \) has no positive integer solutions \((x,y)\).
03-26-2010, 11:51 AM
Post: #1
Prove that \(25 x^2 -6x -95 = y^2-y\quad \) has no positive integer solutions \((x,y)\).
Please try in number theory's way
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04-02-2010, 03:24 PM
Post: #2
RE: Prove that \(25 x^2 -6x -95 = y^2-y\quad \) has no positive integer solutions \((x,y)\).
Solve \(y \quad (y>0)\) to get \(y=(1/2)(1+\sqrt{100x^2-24x-379})\)
So it's necessary to have \(x,n \in \mathbf{N}^+\) such that \(100x^2-24x-379 = n^2\)
thus \(x = (6+\sqrt{25n^2+9511})/50\) and so \(25n^2+9511\) is a perfect square
hence \(25n^2+9511=m^2\) for some \(m \in \mathbf{N}^+\).
Now \(9511\) is prime and \((m+5n)(m-5n)=9511\), we see that
\(5n+m=9511,\quad m=5n+1\) and so \(n=951\). But then \(x\) would not be an integer.
We thus get a contradiction.
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