Iranian Mathematical Olympiad

Define
$$A_n := \overline{a_n \dots a_1 a_0}, \quad B_n := \overline{b_n \dots b_1 b_0}, \quad K_n := \frac{B_n^2 + 999}{A_n^2 + 999}.$$
Claim 1. $\{K_n\}$ is eventually constant or $10^n \mid A_{n-1}^2 + 999$ for each positive integer $n$.

---

Proof. For each $n \ge 1$
$$B_n^2 + 999 = K_n (A_n^2 + 999), \quad B_{n-1}^2 + 999 = K_{n-1} (A_{n-1}^2 + 999) \quad (1)$$
From the Definition of $A_n, B_n$ we know that
$$A_n \equiv A_{n-1} \pmod{10^n}, \quad B_n \equiv B_{n-1} \pmod{10^n} \quad (2)$$
Then (1),(2) imply that
$$K_n (A_n^2 + 999) \equiv K_{n-1} (A_{n-1}^2 + 999) \pmod{10^n}. \quad (3)$$
If there exist $N \ge 1$ such that $10^N \nmid A_{N-1}^2 + 999$, then there is a prime number $p \in \{2, 5\}$ such that $p^N \nmid A_{N-1}^2 + 999$. (2) infers that $A_n \equiv A_{N-1} \pmod{p^N}$ for each $n \ge N-1$. So $p^N \nmid A_n^2 + 999$, meaning
$$v_p (A_n^2 + 999) \le N,$$
where $v_p(x)$ is the power of $p$ in prime factorization of $x$.
Define $d_n := \gcd(10^n, A_{n-1}^2 + 999)$. If $n \ge N+1$ then $d_n \nmid \frac{10^n}{p^{n-N}}$ because $v_p(A_n^2 + 999) \le N$. According to (3) there is $K_n \equiv K_{n-1} \pmod{\frac{10^n}{d_n}}$ so
$$K_n \equiv K_{n-1} \pmod{p^{n-N}}, \quad (4)$$
if $n \ge M$ is large enough such that $p^{n-N} > 100$ then for each $i \ge n-1$
$$A_i \ge 10^i \implies A_i^2 + 999 \ge 10^{2i} + 999 \\ B_i < 10^{i+1} \implies B_i^2 + 999 < 10^{2i+2} + 999.$$
Then
$$K_i = \frac{B_i^2 + 999}{A_i^2 + 999} < 100 < p^{n-N}$$
Hence (4) implies that $K_{n-1} = K_n$ for large enough values of $n$ so the claim is done. $\square$

As discussed above there are two cases:
1. $10^n \mid A_{n-1}^2 + 999$ for each $n \ge 1$.
2. $\{K_n\}$ is eventually constant.

Claim 2. *By assumption of the first case, $a_n = b_n$ for all non-negative integers $n$.*

---

Proof. Let $n \ge 1$ be any positive integer.
$$A_n^2 + 999 \mid B_n^2 + 999 \implies 10^n \mid B_{n-1}^2 + 999$$
With above assertions and some calculation it's deduced that $a_0 = b_0, a_1 = b_1$. Note that $A_n = 10^n a_n + A_{n-1}$ so
$$A_n^2 + 999 \equiv 2 \times 10^n a_n A_{n-1} + A_{n-1}^2 + 999 \pmod{10^{n+1}}.$$
Clearly, $2 \times 10^n \mid 10^{n+1} \mid A_n^2 + 999$ thus
$$2 \times 10^n \mid A_{n-1}^2 + 999 \mid B_{n-1}^2 + 999 \quad (5)$$
Which is stronger than assumed relation at beginning of the claim. The remaining part of proof is by using induction. Assume that $a_m = b_m$ for all $0 \le m \le n - 1$. Which means $A_{n-1} = B_{n-1}$. Using (5) implies for all $n \ge 1$
$$\begin{align*} A_n^2 + 999 &\equiv 0 \pmod{2 \times 10^{n+1}} \\ \implies (10^n a_n + A_{n-1})^2 + 999 &\equiv 0 \pmod{2 \times 10^{n+1}}. \end{align*}$$
Which implies
$$a_n A_{n-1} + \frac{A_{n-1}^2 + 999}{2 \times 10^n} \equiv 0 \pmod{10}. \quad (6)$$
Repeating above discussion with $B_n^2 + 999$ infers that
$$b_n B_{n-1} + \frac{B_{n-1}^2 + 999}{2 \times 10^n} \equiv 0 \pmod{10}. \quad (7)$$
According to (6), (7) and the assertion of $A_{n-1} = B_{n-1}$ (induction) it's known that $a_n A_{n-1} \equiv b_n A_{n-1} \pmod{10}$. Moreover $\gcd(A_n, 10) = 1$ since $10^{n+1} \mid A_n^2 + 999$. This implies $a_n \equiv b_n \pmod{10}$. Hence $a_n = b_n$ since $0 \le a_n, b_n \le 9$. Claim is proved. $\square$

The only remaining part of solution is the case that $\{K_n\}$ is eventually constant. Suppose that there exist positive integers $K, T$ such that $K_n = K$ for all $n \ge T$. Suppose that $n \ge T$ then
$$\left. \begin{array}{l} B_n^2 + 999 = K(A_n^2 + 999) \\ A_n = 10^n a_n + A_{n-1} \\ B_n = 10^n b_n + B_{n-1} \end{array} \right\} \\ \implies 10^n b_n^2 + 2 \times b_n B_{n-1} = K (10^n a_n^2 + 2 \times a_n A_{n-1}) \\ \implies 10^n (b_n^2 - K a_n^2) = 2 (K a_n A_{n-1} - b_n B_{n-1}).$$
So
$$\frac{10^n}{A_{n-1}} (b_n^2 - K a_n^2) = 2 \left( K a_n - b_n \frac{B_{n-1}}{A_{n-1}} \right). \quad (8)$$
It's known that $A_n, B_n \ge 10^n$ for large enough integers $n$
$$\implies A_n^2 \simeq A_n^2 + 999, \quad B_n^2 \simeq B_n^2 + 999.$$
Which implies
$$B_n^2 \simeq K A_n^2 \implies \lim_{n \to \infty} \frac{B_n}{A_n} = \sqrt{K}.$$
Define
$$C_n := \frac{B_n}{A_n} - \sqrt{K} \implies \lim_{n \to \infty} C_n = 0,$$
according to (8)
$$\begin{aligned} \frac{10^n}{A_{n-1}} (b_n^2 - K a_n^2) &= 2 (K a_n - b_n \sqrt{K}) - 2 b_n C_n \\ \implies \frac{10^n}{A_{n-1}} (b_n - \sqrt{K} a_n) (b_n + \sqrt{K} a_n) &= 2 \sqrt{K} (\sqrt{K} a_n - b_n) - 2 b_n C_n \\ \implies (b_n - \sqrt{K} a_n) \left( \frac{10^n}{A_{n-1}} (b_n + \sqrt{K} a_n) + 2 \sqrt{K} \right) &= -2 b_n C_n \end{aligned} \quad (9)$$
Right Hand Side of (9) converges to 0 since $\lim_{n \to \infty} C_n = 0$ and $0 \le b_n \le 9$. Also the second parenthesis in (9) is always greater than $2\sqrt{K}$. So
$$\lim_{n \to \infty} b_n - \sqrt{K} a_n = 0$$
But $b_n - \sqrt{K}a_n$ has finite values which infers that there exist positive integer $L \ge M$ such that $b_n = \sqrt{K}a_n$ for all $n \ge L$. Therefore according to (9) there must be $-2b_nC_n = 0$ for all $n \ge L$. Which implies $C_n = 0$ since $b_n \ne 0$ for $n \ge M$. Therefore
$$C_n = 0 \implies B_n = \sqrt{K} A_n$$
In addition with $B_n^2 + 999 = K(A_n^2 + 999)$ it's easy to see that $999K = 999$ which means $K = 1$. So for large enough positive numbers $n$,
$$B_n = A_n \implies \overline{b_n \cdots b_1 b_0} = \overline{a_n \cdots a_1 a_0}$$
Therefore $a_k = b_k$ for all $k \ge 0$ and we're done! ■