BMO 2022 shortlist

Question

For every natural number x, let P(x) be the product of the digits of the number x. Is there a natural number n such that the numbers P(n) and P(n^2) are non-zero squares of natural numbers, where the number of digits of the number n is equal to a. 2021 b. 2022

Accepted Answer

The answers are affirmative in both cases.

a. Take

n = 33 \dots 3 2019 68

. Then

P (n) = (4 \cdot 3^{1010})^{2}

. Also,

n^{2} = (\frac{1 0 ^{2021} - 1}{3} + 35)^{2} = \frac{( 1 0 ^{2021} + 104 ) ^{2}}{9} = \frac{1 0 ^{4042} + 208 \cdot 1 0 ^{2021} + 10816}{9} = \frac{1 0 ^{4042} - 1 0 ^{2021}}{9} + 209 \cdot \frac{1 0 ^{2021} - 1}{9} + 1225 = 2021 1 \dots 1 2021 0 \dots 0 + 2021 2 \dots 2 202100 + 2021 9 \dots 9 + 1225 = 2019 1 \dots 1 2019 332 \dots 200 + 1 0^{2021} + 1224 = 2019 1 \dots 1 2017 342 \dots 23424 .

Thus P (n^{2}) = (3 \cdot 2^{2012})^{2} .

b. Take

n = 1133 \dots 3 2020

. Then

P (n) = (3^{1010})^{2}

. Also,

n^{2} = \frac{( 34 \cdot 1 0 ^{2020} - 1 ) ^{2}}{9} = \frac{1156 \cdot 1 0 ^{4040} - 68 \cdot 1 0 ^{2020} + 1}{9} = 128 \cdot 1 0^{4040} + \frac{4 ( 1 0 ^{4040} - 1 0 ^{2020} )}{9} - 7 \cdot 1 0^{2020} - \frac{1 0 ^{2020} - 1}{9} = 1280 \dots 0 4040 + 4 \dots 40 \dots 0 2020 - 7 \cdot 1 0^{2020} - 2020 1 \dots 1 = 1284 \dots 4368 \dots 89 20182019 .

Thus P (n^{2}) = (9 \cdot 2^{5049})^{2} .