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It was a problem that took place on college entrance exam.

Find the smallest positive number p for which the product of $$\boldsymbol p*(2\sqrt3-3\sqrt2)$$ is a whole number (integer).

Is there a general formula for solving this type of problem?

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    $\begingroup$ Yes: $a^2 - b^2 = (a-b)(a+b)$. $\endgroup$ Commented Jul 8, 2022 at 12:33
  • $\begingroup$ @eyeballfrog that was my goto when I tried solving it, but it doesn't produce the smallest possible number. $\endgroup$
    – kesetovic
    Commented Jul 8, 2022 at 12:36
  • $\begingroup$ Can you do better than the reciprocal? $\endgroup$
    – lulu
    Commented Jul 8, 2022 at 12:38
  • $\begingroup$ @lulu The number is negative, so its reciprocal doesn't work. $\endgroup$ Commented Jul 8, 2022 at 12:39
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    $\begingroup$ @eyeballfrog So take the negative of the reciprocal. Any other positive number, yielding an integer smaller than $-1$ has to be larger. $\endgroup$
    – lulu
    Commented Jul 8, 2022 at 12:42

1 Answer 1

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Since $p \ne 0$, the smallest choice of $p$ will have $|p(2\sqrt{3}-3\sqrt{2})| = 1$. Since $2\sqrt{3}-3\sqrt{2} < 0$ and $p >0$, that product must be $-1$. So we have $$ p = \frac{1}{3\sqrt{2}-2\sqrt{3}}. $$ Alternatively, we can use difference of squares to find $$ (2\sqrt{3}+3\sqrt{2})(2\sqrt{3}-3\sqrt{2}) = 12-18 = -6. $$ Divide both sides by $6$ to get $$ \left(\frac{2\sqrt{3}+3\sqrt{2}}{6}\right)(2\sqrt{3}-3\sqrt{2}) = -1\Longrightarrow p = \frac{2\sqrt{3}+3\sqrt{2}}{6}. $$ Despite the difference in form, these two values for $p$ are equal.

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  • $\begingroup$ Thank you, that was the exact correct solution given by the exam organization. $\endgroup$
    – kesetovic
    Commented Jul 8, 2022 at 13:52

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