Since \(f(n) \mid g(n)\) for infinitely many \(n\), \(\deg f < \deg g\). Divide \(g\) by \(f\) in \(\mathbf{Q}[X]\) such that \(g = qf + r\). Let the numbers such that \(f(n) \mid g(n)\) be \(n_1\), \(n_2\), \(\dots\). Mutliplying \(g = qf + r\) by some big LCM of denominators, we get \(Lg = q'f + r'\), where \(q'\) and \(r'\) are integer polynomials. For \(n_i\), this becomes \[Lkf(n_i) = Lg(n_i) = q'(n_i)f(n_i) + r'(n_i) \implies f(n_i) \mid r'(n_i) \;\; \forall i\] But \(|f(n_i)| > |r'(n_i)|\) for sufficiently large \(i\) due to \(r'\) having lesser degree than \(f\). Thus \(r'(n_i)\) is \(0\) for all large \(i\), implying it (and thus \(r\)) is the zero polynomial.

PFTB deals with this problem in basically the same way but phrases it as: \(\{Lr(n_i)/f(n_i)\} = \{r'(n_i)/f(n_i)\}\) is an integer sequence that converges to \(0\), which means it stabilizes at \(0\) eventually. Note that \(\{r(n_i)/f(n_i)\}\) also converges to \(0\), but since it isn't an integer sequence, that really doesn't mean anything.

Plugging \(n = 0\) implies \(c = y^2\) for some \(y\). \[\left(\sqrt{a} n + \frac{b}{2\sqrt{a}}\right)^2 = an^2 + bn + \frac{b^2}{4a} = x_n^2 - \left(c - \frac{b^2}{4a}\right)\] where \(\{x_i\}\) is the integer sequence \(\sqrt{an^2 + bn + c}\). The sequence \(\{x_n - \sqrt{a} n\}\) is convergent, but not an integer sequence. To obtain a convergent integer sequence from this, we can just notice that \(\{x_{n+1} - x_n\}\) converges to \(\sqrt{a}\), implying \(\sqrt{a}\) is an integer - \(a = x^2\). Now we can say that \(\{x_n - x n\}\) is a convergent integer sequence, so \(x_n = xn + r\) for large \(n\). This means \(ax^2 + bx + c = (xn + r)^2\) for infinitely many \(n\), thus meaning the polynomials are the same, and we are done.

Let \(a_i = \{a_i\}\), so \(0 \le a_i < 1\). Then assuming the contrary we get that \[\frac{\lfloor a_1 p \rfloor + \lfloor a_2 p \rfloor + \dots + \lfloor a_k p\rfloor}{p}\] is an integer sequence for sufficiently large prime \(p\). But how does it behave? Well, it converges to \(a_1 + a_2 + \dots + a_k\), so that must be an integer. Moreover, we get \[\lfloor a_1 p \rfloor + \lfloor a_2 p \rfloor + \dots + \lfloor a_k p\rfloor = a_1p + a_2p + \dots + a_kp \implies a_i p \in \mathbf{Z} \;\; \forall i\] But this is also true for all sufficiently large primes, so this means \(a_i = 0\) for all \(i\), a contradiction to the problem statement.

Let the degree of this polynomial be \(d > 0\) (constant polynomials \(= 11\dots 1\) work). We have the condition \[f\left(\frac{10^i - 1}{9}\right) = \frac{10^{g(i)} - 1}{9}\] \[\frac{10^{g(i)-di}}{c/9^d} \sim \frac{f\left(\frac{10^i - 1}{9}\right)}{{c \cdot 10^{di}}/{9^d}} \to 1 \implies 10^{g(i) - di} \sim \frac{c}{9^{d-1}}\] Therefore \(\{g(i) - di\}\), an integer sequence, converges to a value \(x\) and thus becomes constant at it eventually. Thus \(g(i) = x + di\) for large \(i\), and \[f\left(\frac{10^i - 1}{9}\right) = \frac{10^{di + x} - 1}{9} \implies f(n) = \frac{10^x(9n+1)^d-1}{9}\]

Let \(d\) be the degree of \(f\), and \(x_n\) be the positive integer such that \(f(x_n) = 2^n\). Then \(x_n \sim 2^{n/d}\). We know that \(\lim_{n \to \infty} x_{n+d}/2x_n = 1\). Consider the sequence \(\{x_{n+d} - 2x_n\}\) - using \(f(x_{n+d}) = 2^d f(x_n)\), we get \[\sum_{i=0}^d a_i(2^d x_n^i - x_{n+d}^i) = 0 \implies \sum_{i=0}^{d-1} a_i(2^d x_n^i - x_{n+d}^i) = x_{n+d}^d - (2x_n)^d \implies x_{n+d} - 2x_n = \frac{\sum_{i=0}^{d-1} a_i(2^d x_n^i - x_{n+d}^i)}{\sum_{i=0}^{d-1} (2x_n)^i x_{n+d}^{d-i-1}}\] Using \(x_{n+d}/x_n \to 2\), we obtain that \(\{x_{n+d} - 2x_n\}\) is indeed convergent, and being an integer sequence, after some point we have \(x_{n+d} = 2x_n + e\) for some \(e\). Thus we have \(f(2x + e) = 2^d f(x)\) for all sufficiently large \(x\), and since \(2^d\) is a constant, taking the polynomial \(g(x) = f(2x + e) - 2^df(x)\), we get that it is identically zero - thus \(f(2x + e) = 2^d f(x) \) for all \(x\).

Now consider any complex root \(z\) of \(f\), in which case \(2z + e\), \(4z + 3e\), and so on, all become roots of \(f\). In order for the number of roots to be finite, \(2^r z + (2^r - 1) e = 2^s z + (2^s - 1) e\) must hold, which means the only root is \(z = -e\), giving the polynomial \(f(x) = (x + e)^d\). However, this polynomial does not yield positive integer solutions to \(f(x) = 2^n\) for all \(n\) unless \(d = 1\), so we are done.

Suppose this is not true, in which case after a certain number \(N\), no \(n\) satisfies \(\pi(n) \mid n\). Choose the first prime \(p_1 > N\). Let \(\pi(p_1) = \psi\). Then, the next prime \(p_2\) must satisfy \(p_2 < p_1 + \psi - 1\), \(p_3\) must satisfy \(p_3 < p_1 + (\psi - 1) + (\psi)\). Continuing a weak bound like this gives nothing.

Having taken a hint from the book, now I know that all integer values are to be covered in the sequence \(n/\pi(n)\). Suppose \(n/\pi(n)\) does not take the value \(k\). The only values of \(n\) for which it could have taken the value \(k\) are obviously of the form \(km\). Let \(M\) be the biggest value such that \(kM/\pi(kM) < k\), which exists since \(n/\pi(n) \to \infty\). This means \(\pi(kM) \ge M + 1\). Then \[k < \frac{k(M+1)}{\pi(k(M + 1))} \le \frac{k(M+1)}{\pi(kM)} \le k\] which is a clear contradiction to our assumption. Thus, \(n/\pi(n)\) indeed covers all sufficiently large positive integers, which means \(\pi(n) \mid n\) for infinitely many \(n\), and we are done.

Surely this is the reciprocal trick. Note that if there are only finitely many non-power terms, then from a certain point on, everything is powers. The sums of reciprocals of all the powers is something that we know converges because \[\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} \frac{1}{n^i} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)} = \sum_{n=2}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n}\right) = 1 - \frac{1}{n_{\to \infty}} = 1\] Since there are only finitely many non-power terms, the sum of reciprocals of all the terms in this arithmetic progression must converge. However, supposing \(a < kd\) for some \(k\), \[\sum_{i=0}^{\infty} \frac{1}{a+id} > \sum_{i=0}^{\infty} \frac{1}{(k+i)d} = \frac{1}{d} \sum_{i=0}^{\infty} \frac{1}{k+i} \to \infty \] so we're done.

Well, of course, another 'trivial' way to prove this is to cite Dirichlet.