Quadratic Diophantine Equations

Since this problem’s reduction is basically the same as last week’s, let’s skip ahead to it now.

The problem: Quadratic Diophantine Equations.  This is problem AN8 in the appendix.

The description: Given positive integers a,b, and c, can we find positive integers x and y such that ax2 + by = c?

Example: Let a=1, b = 2, and c=5.  Then we’re asking if positive integers x and y exists to solve x2 + 2y = 5.  This is true when x=1 and y = 2.  If we change b to 3, then there are no positive integers x and y where x2 + 3y = 5.  (y has to be positive, so the only y that has a chance of working is y=1, which would require x to be √2)

Reduction: This is in the same paper by Manders and Adleman that had the reduction for Quadratic Congruences.  In fact, the reduction is almost exactly the same.  We go through the same process of creating the Subset Sum problem, and build the same H and K at the end.  The only difference is the instance: We build the quadratic (K+1)3*2*8m+1*(H2-x2) + K(x2-r2)-2K*8m+1y = 0.  We can multiply out the vales to get a,b, and c.

The rest of the reduction uses similar crazy algebra to the last problem.

Difficulty: 9, for the same reasons last week’s problem was.

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