1993 OIM Problems/Problem 6

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Problem

Two non-negative integers $a$ and $b$ are cuates if the decimal expression $a+b$ consists only of zeros and ones. Let $A$ and $B$ be two infinite sets of non-negative integers, such that $B$ is the set of all numbers that are cuates of all the elements of $B$.

Prove that in one of the sets $A$ or $B$ there are infinitely many pairs of numbers $x$, and $y$ such that $x-y = 1$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe8.htm