Question 10

The number of pairs of relatively prime positive integers $$(a, b)$$ such that $$\frac{a}{b} + \frac{15b}{4a}$$ is an integer is

Solution

Writing the sum as $$\frac{4a^2+15b^2}{4ab}$$, divisibility forces $$a$$ to divide 15 and $$b$$ to divide 4 with $$b$$ even, giving candidates $$a\in\{1,3,5,15\}$$, $$b\in\{2,4\}$$. Checking each coprime pair against $$4ab \mid 4a^2+15b^2$$ shows only $$b=2$$ works for every value of $$a$$, giving the 4 pairs $$(1,2),(3,2),(5,2),(15,2)$$.

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The number of pairs of relatively prime positive integers $$(a, b)$$ such that $$\frac{a}{b} + \frac{15b}{4a}$$ is an integer is

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