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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
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
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
04-02-2010, 03:24 PM
Post: #2
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
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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