# Find the smallest positive number p that satisfies the condition

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?

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

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.