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I am trying to find relation of pairs $(x,y)$ and $(x_1,y_1)$ such that $f(x,y)\equiv f(x_1,y_1)(\mbox{mod }n)$ where $f$ is a quadratic expression like $ax^2+bxy+cy^2+dx+ey+f$ Thanks in advance

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Sorry guys. I found the answer myself. This needs factorization of number n. Solution goes as follows. We know that any quadratic equation $ax^2+bxy+cy^2+dx+ey+f=0$ can be written as $X^2-dY^2=N$ and hence it is sufficient to characterize roots of x^2-dy^2-n. Let $f(x,y)\equiv f(x_0y_0)$, then $x^2-dy^2\equiv x_0^2-dy_0^2$. clearly, every pair satisfying $x^2\equiv x_0^2(\mbox{mod }n)$ and $y^2\equiv y_0^2(\mbox{mod }n)$ will be solution to the question if $(x_0,y_0)$ is a solution. But every solution of $x^2\equiv x_0^2(\mbox{mod n})$ have either gcd$(x+x_0,n)$ or gcd$(x-x_0,n)$ not same as 1 and hence will be factor of n. This will give characterization of some of the roots.

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Any suggestions are invited at karun3kumar@gmail.com. Sending mail please give this page as reference. –  Wantanswer Oct 13 '12 at 14:50
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