Given $1 \leq a \leq b \leq n^{2}$, how many perfect squares divisible by $m$, where $1 \leq m \leq n$ exist in the interval $[a,b]$. This number be computed by doing a linear search. Can this be done any faster (say in time $log_{2}(b - a)$?
Count the number of perfect squares divisible by $m$ in the closed interval $[a,b]$.
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number-theory
elementary-number-theory
1 Answers
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Write $m=k^2n$ with $n$ square free.
Then, if $x$ is a perfect square divisible by $m$, $x$ must have the form $$x=mny^2$$
Therefore the problem becomes: How many perfect squares $y^2$ exists between $\frac{a}{mn}$ and $\frac{b}{mn}$. This is equivalent to figuring how many integers $y$ exists in the interval $[\sqrt{\frac{a}{mn}}, \sqrt{\frac{b}{mn}}]$.
So, as long as you can find the prime factorisation of $m$, the answer can be calculated without any search.