Always Square

Checking with the first three $N$ , we found two acceptable triplets of $(A, B, C)$ as follows:

$\begin{cases} N = 0 & \implies \begin{cases} 16 = 4^2 \\ 49 = 7^2 \end{cases} \\ N = 1 & \implies \begin{cases} 1156 = 34^2 \\ 4489 = 67^2 \end{cases} \\ N = 2 & \implies \begin{cases} 111556 = 334^2 \\ 444889 = 667^2 \end{cases} \end{cases}$

The two acceptable triplets of $(A, B, C)$ appear to be $(1, 5, 6)$ and $(4, 8, 9)$ . Let us prove by induction that they are true for all $N \ge 0$ by proving the following two claims.

Claim 1 : $\ \underbrace{333\cdots 3}_{N \ 3's}4^2 = \underbrace{111 \cdots 1}_{(N+1) \ 1's}\underbrace{555 \cdots 5}_{N \ 5's}6$

Assume that the claim is true for $N$ , then we have:

$\begin{aligned} \underbrace{333\cdots 3}_{{\color{#3D99F6}(N+1)} \ 3's}4^2 & = (3 \cdot 10^{N+1} + \underbrace{333\cdots 3}_{N \ 3's}4)^2 \\ & = 9 \cdot 10^{2N+2} + 6 \cdot 10^{N+1} \cdot \underbrace{333\cdots 3}_{N \ 3's}4 + \underbrace{333\cdots 3}_{N \ 3's}4^2 \\ & = 9 \underbrace{000\cdots 0}_{(2N+2) \ 0's} + 2 \underbrace{000 \cdots 0}_{N \ 0's} 4 \underbrace{000 \cdots 0}_{(N+1) \ 0's} + \underbrace{111 \cdots 1}_{(N+1) \ 1's}\underbrace{555 \cdots 5}_{N \ 5's}6 \\ & = \underbrace{111 \cdots 1}_{({\color{#3D99F6}(N+1)}+1) \ 1's}\underbrace{555 \cdots 5}_{{\color{#3D99F6}(N+1)} \ 5's}6 \end{aligned}$

The claim is true for $\color{#3D99F6}N+1$ , hence it is true for all $N \ge 0$ .

Claim 2 : $\ \underbrace{666 \cdots 6}_{N \ 6's}7^2 = \underbrace{444 \cdots 4}_{(N+1) \ 4's}\underbrace{888 \cdots 8}_{N \ 8's}9$

Assume that the claim is true for $N$ , then we have:

$\begin{aligned} \underbrace{666 \cdots 6}_{{\color{#3D99F6}(N+1)} \ 6's}7^2 & = (6 \cdot 10^{N+1} + \underbrace{666 \cdots 6}_{N \ 6's}7)^2 \\ & = 36 \cdot 10^{2N+2} + 12 \cdot 10^{N+1} \cdot \underbrace{666 \cdots 6}_{N \ 6's}7 + \underbrace{666 \cdots 6}_{N \ 6's}7^2 \\ & = 36 \underbrace{000\cdots 0}_{(2N+2) \ 0's} + 2 \underbrace{000 \cdots 0}_{(N+1) \ 0's} 4 \underbrace{000 \cdots 0}_{(N+1) \ 0's} + \underbrace{444 \cdots 4}_{(N+1) \ 4's}\underbrace{888 \cdots 8}_{N \ 8's}9 \\ & = \underbrace{444 \cdots 4}_{({\color{#3D99F6}(N+1)}+1) \ 4's}\underbrace{888 \cdots 8}_{{\color{#3D99F6}(N+1)} \ 8's}9 \end{aligned}$

The claim is true for $\color{#3D99F6}N+1$ , hence it is true for all $N \ge 0$ .

Therefore, the answer is $\boxed{2}$ .