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Problem 1423

AIME late

Algebra Difficulty 5.8 Prove it Belarusian Mathematical Olympiad · Belarus

Prove that there exist infinitely many positive integers $n$ so that $n$ and the sum of its digits are perfect squares and

a) the decimal representation of $n$ contains at most one $0$ ;

b) the decimal representation of $n$ does not contain $0$ . (T. Lasy)

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

Note that the sum of the digits of this number is equal to
$S(X_n) = 1 \cdot n + 5 \cdot (n - 1) + 6 = 6n + 1.$
Further,
$\begin{aligned} X_n &= \frac{10^{2n} - 1}{9} + \frac{4 \cdot (10^n - 1)}{9} + 1 \\ &= \frac{10^{2n} - 1 + 4 \cdot (10^n - 1) + 9}{9} \\ &= \frac{10^{2n} + 4 \cdot 10^n + 4}{9} \\ &= \left( \frac{10^n + 2}{3} \right)^2 \end{aligned}$
is a perfect square. (Since $S(10^n + 2) = 3$ , we have $10^n + 2 \vdots 3$ .) It suffices to note that if $n = 6m^2 + 2m$ ( $m \in \mathbb{N}$ ), then $S(X_n) = 6n + 1 = 6(6m^2 + 2m) + 1 = (6m + 1)^2$ . Therefore, the numbers $X_{6m^2+2m}$ satisfy the problem condition for all $m \in \mathbb{N}$ , and we have infinitely many such numbers.

Remark. Also we can consider
$(2 \cdot 10^n + 1)^2$

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