A trigonometrical question inspired by Otto Bretscher

My approach is not as systematic as Arjen's.

We are asked to find the smallest number $n$ such that $1440=\phi(n)$ . Since $1440+1$ isn't prime, we are looking for a factorization $1440=a*b$ where $a$ and $b$ are as close as possible to each other (close to $\sqrt{1440}\approx 38$ ) and, ideally, such that $a+1$ and $b+1$ are prime. This approach works beautifully here: We have $1440=36*40=\phi(37)*\phi(41)= \phi(\boxed{1517})$ We have 11 other factorizations $1440=a*b$ into even factors, for example, $1440=6*240=\phi(7)\phi(241)=\phi(1687)$ and $1440=72*20=\phi(73)\phi(25)=\phi(1825)$ but the product is higher in these cases since the factors are further apart (there is no need to check them one by one).

Let the prime decomposition of $n$ be $n = p_1^{e_1}\cdot p_2^{e_2}\cdots.$

Lemma $P(n) = \phi(n) = \prod_i p_i^{e_i-1}\cdot (p_i-1).$

In this case, $2015 = 5^1\cdot 13^1\cdot 31^1\ \ \ \Longrightarrow\ \ \ P(2015) = (5-1)\cdot(13-1)\cdot(31-1) = 1440.$

Now to find other integers for which this is the case. This requires a factorization of 1440 into factors of the form $p^e\cdot(p-1)$ . In order to keep the number as low as possible, it is best to keep the number of factors as well as $e$ minimal. (E.g. the best case scenario would be $n = 1441$ if it were prime, but sadly it is a multiple of 11.)

The prime factorization of 1440 is $2^5\cdot 3^2\cdot 5$ , which means that it has $6\cdot 3\cdot 2 = 36$ divisors. We analyze them as follows:

$\begin{array}{rll} d & p^{e-1}(p-1) & p^e \\ \hline 1 & 2-1 & 2 \\ 2 & 3-1 & 3 \\ 2 & 2^1(2-1) & 4 \\ 4 & 5-1 & 5 \\ 4 & 2^2(2-1) & 8 \\ 6 & 7-1 & 7 \\ 6 & 3^1(3-1) & 9 \\ 8 & 2^3(2-1) & 16 \\ 10 & 11-1 & 11 \\ 12 & 13-1 & 13 \\ 16 & 17-1 & 17 \\ 16 & 2^5(2-1) & 32 \\ 18 & 19-1 & 19 \\ 18 & 3^2(3-1) & 27 \\ 20 & 5^1(5-1) & 25 \\ 30 & 31-1 & 31 \\ 36 & 37-1 & 37 \\ 40 & 41-1 & 41 \\ 240 & 241-1 & 241 \\ \hline \end{array}$

The remaining divisors of 1440, which are 3, 9, 15, 24, 32, 45, 48, 60, 72, 80, 90, 96, 120, 144, 160, 180, 288, 360, 480, 720, 1440, cannot be written in this form.

Using these factors, we write $1440 = 6\cdot 240 = \phi(7)\cdot \phi(241) = \phi(1687); \\ 1440 = 6\cdot 240 = \phi(9)\cdot \phi(241) = \phi(2169); \\ 1440 = 36\cdot 40 = \phi(37)\cdot \phi(41) = \phi(\boxed{1517}).$

Other factorizations are also possible, but the increased number of factors gives higher values. For instance, $1440 = 6\cdot 30\cdot 8 = \phi(7)\cdot \phi(31)\cdot \phi(16) = \phi(3472); \\ 1440 = 20\cdot 12\cdot 6 = \phi(25)\cdot\phi(13)\cdot \phi(7) = \phi(2275);$ and so on.