Assume there are finitely many $n$ such that there are exactly two balanced numbers among $n, n+1, n+2, n+3$. Let $N$ be the largest integer $n$ satisfying this condition. It's easy to see that then, either for every $m>N$ there are more than two balanced numbers among $m, m+1, m+2, m+3$, or else for every $m>N$ there are less than two balanced numbers among $m, m+1, m+2, m+3$. Assume the first case. Let $l$ be an integer such that $2^l>N$. $2^l$ is unbalanced, so $2^l+1$ must be balanced. But $2^{l+1}$ and $2^{l+1}+2$ are unbalanced so we have a contradiction. In the second case we have to do the same trick with $2^l p$, where $p$ is an odd prime.

another solution. by Middle-Value theorem,it suffices to prove that the Balanced number in arbitrary $4$ consecutive numbers can be $\le or \ge 2$,both infinite times. this suffices to consider the following two equations: $x^2-6y^2=1$(Pell equation); $3x^2-2y^2=1$(OC's equation). easy to prove that they both have infitite solutions.

But not every $x^2$ is balanced. You missed word "DISTINCT". But your solution is correct solution to p4 from IV Romanian Master in Mathematics .

Swistak wrote: But not every $x^2$ is balanced. You missed word "DISTINCT". But your solution is correct solution to p4 from IV Romanian Master in Mathematics . oh yes.i discovered it later with Horizon's help. the modified solution is similar to your solution above.