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HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

Compute the number of ordered pairs of integers (a,b)(a, b), with 2≤a,b≤20212 \leq a, b \leq 2021, that satisfy the equation alog⁡b(a−4)=blog⁡a(ba−3).a^{\log _{b}\left(a^{-4}\right)}=b^{\log _{a}\left(b a^{-3}\right)}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Taking log⁡a\log _{a} of both sides and simplifying gives −4log⁡ba=(log⁡ab)2−3log⁡ab-4 \log _{b} a=\left(\log _{a} b\right)^{2}-3 \log _{a} b Plugging in x=log⁡abx=\log _{a} b and using log⁡ba=1log⁡ab\log _{b} a=\frac{1}{\log _{a} b} gives x3−3x2+4=0x^{3}-3 x^{2}+4=0 We can factor the polynomial as (x−2)(x−2)(x+1)(x-2)(x-2)(x+1), meaning b=a2b=a^{2} or b=a−1b=a^{-1}. The second case is impossible since both aa and bb are positive integers. So, we need only count the number of 1<a,b≤20211<a, b \leq 2021 for which b=a2b=a^{2}, which is ⌊2021⌋−1=43\lfloor\sqrt{2021}\rfloor-1=43.

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