Question 13

The number of ordered pairs of positive integers $$(x,y)$$ satisfying the equation $$x^2+4y=3x+16$$ is

Solution

Rearranging gives $$x^2-3x+4y-16=0$$. Treating this as a quadratic in $$x$$, the discriminant is $$73-16y$$ and must be a nonnegative perfect square. The possible positive values are $$y=3$$ and $$y=4$$, producing the positive solutions $$(4,3)$$ and $$(3,4)$$. Therefore there are $$2$$ ordered pairs.

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