Problem 1323

1. Lemma 1: $a_n = 0$ if and only if $n$ is a perfect square. Otherwise, $a_n = 1$.

Proof: Write $n$ as its prime factorization: $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ where $p_i$ are distinct primes and $a_i \in \mathbb{Z}$. The number of divisors of $n$ is given by $(a_1 + 1)(a_2 + 1) \cdots (a_r + 1)$. This product is odd if and only if each $a_i$ is even, which means $n$ is a perfect square.

2. Lemma 2: A number $m = 0.b_1b_2b_3\ldots$ is rational if and only if the sequence $b_1, b_2, b_3, \ldots$ is periodic.

Proof: Every rational number $x = 0.c_1c_2c_3\ldots$ has a periodic decimal representation. Let $x = \frac{p}{q}$ where $\gcd(p, q) = 1$ and $q > p > 0$. There exist integers $r$ and $s$ such that $q \mid 10^s(10^r - 1)$. Let $s = \max \{ e_2(q), e_5(q) \}$, where $e_p(n)$ is the maximal exponent of $p$ dividing $n$. Let $g = \frac{q}{2^{e_2(q)} 5^{e_5(q)}}$, so $q \mid 10^s(10^r - 1) \iff g \mid 10^r - 1$. Choose $r = \phi(g)$, where $\phi$ is Euler's totient function. Thus, $q \mid 10^s(10^r - 1)$. Let $k = \frac{10^s(10^r - 1)}{q}$. Then $x = \frac{p \cdot k}{q \cdot k}$. Define the digits $d_1, d_2, \ldots, d_t$ of $p \cdot k$ so that $p \cdot k = d_1d_2\ldots d_t$. Then $x = \frac{d_1d_2\ldots d_t}{99\ldots 99}$ with $t$ nines. Hence, $x = 0.d_1d_2\ldots d_t d_1d_2\ldots d_t \ldots$, showing a periodic decimal representation.

3. We want to prove that the number $x = 0.a_1a_2a_3\ldots$ is irrational. Assume, for contradiction, that $x$ is rational. By Lemma 2, the sequence $a_1, a_2, \ldots$ is periodic. Let $n$ be its period, meaning $a_k = a_{n+k}$ for all $k \in \mathbb{N}$.

4. Since $a_1 = 1$, it follows that $a_{kn+1} = 1$ for all $k \in \mathbb{N}$. By Lemma 1, $1, n+1, 2n+1, 3n+1, \ldots$ must all be perfect squares.

5. Consider $kn+1 = m^2$ for some $k$. Then $(m+1)^2 = m^2 + 2m + 1 > m^2 + n = (k+1)n + 1$. Thus, $(k+1)n + 1$ lies between two squares, $m^2$ and $(m+1)^2$, and cannot be a square. Therefore, $a_{(k+1)n+1} = 0$, contradicting the periodicity of $a_1, a_2, \ldots$.

6. This contradiction implies that our assumption that $x$ is rational is false. Hence, $x$ must be irrational.

$\blacksquare$