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Algebra Difficulty 5.0 AIME Find the answer

3. Given that aa and bb are integers, and satisfy
(1a1a−1b−1b1a+1b)(1a−1b)11a2+1b2=23.  \left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }

Then a+b=a+b= \qquad .

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

Solution

3.3.
 Left side =aba−b=23,∴(3b−2)(3a−2)=4. \begin{array}{l} \text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\ \therefore(3 b-2)(3 a-2)=4 . \end{array}

Given that a≠ba \neq b and they are integers, hence 3b−2,3a−23 b-2, 3 a-2 can only take the values 1, 4 or -1, -4.
(1) Suppose 3b−2=1,3a−2=43 b-2=1, 3 a-2=4.
Solving gives b=1,a=2b=1, a=2. Therefore, a+b=3a+b=3.
(2) Suppose 3b−2=−1,3a−2=−43 b-2=-1, 3 a-2=-4.
This results in a,ba, b being fractions, which we discard.
Thus, a+b=3a+b=3.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.