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Problem 1749

National Olympiad, first round

Number theory Difficulty 6.5 Prove it Belarusian Mathematical Olympiad · Belarus

For two positive integers $a$ and $b$ the number $\overline{a.b}$ is equal to the decimal fraction which we have if after the number $a$ we put the decimal point and then write the number $b$ . For example, for $a = 20, b = 13$ we get $\overline{a.b} = 20.13$ , and $\overline{b.a} = 13.2$ .
Prove that there are infinite number of natural $n$ , such that the equation $\overline{a, b} \cdot \overline{b, a} = n$ has no positive integer roots $a$ and $b$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Show that if $n = 9k \pm 3$ , $k \in \mathbb{N}$ , then the given equation has no natural solutions.
Let the decimal representations of $a$ and $b$ consist of $m$ and $l$ digits respectively. Then the initial equation is equivalent to the equation
$\left(a + \frac{b}{10^l}\right) \left(b + \frac{a}{10^m}\right) = 9k \pm 3 \quad \text{or} \quad ab + \frac{ab}{10^{m+l}} + \frac{a^2}{10^m} + \frac{b^2}{10^l} = 9k \pm 3,$
i. e.
$10^{m+l}ab + ab + a^2 10^l + b^2 10^m = 10^{m+l} \cdot (9k \pm 3). \quad (*)$
Since $10^t - 1 = \underbrace{9\dots9}_{t \text{ times}}$ for any positive integer $t$ , i. e. $10^t = 9A + 1$ for some positive integer $A$ , we can replace all powers of 10 in (*) by their presentations, then we obtain
$(9A_1 + 1)ab + ab + (9A_2 + 1)a^2 + (9A_3 + 1)b^2 = (9A_4 + 1) \cdot (9k \pm 3)$
or $(a+b)^2 = 9B \pm 3$ for some positive integer $B$ . But this equality is impossible because the left hand side of it is the square number and the left hand side is a number which is divisible by 3 but is not divisible by 9.

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