By the Prime Number Theorem, which states $\pi(x)\approx\frac{x}{\ln x}$, we can take the approximation $\pi(2x)-\pi(x)\approx \frac{2x}{\ln 2x}-\frac{x}{\ln x}=\frac{2x-x\left(1+\frac{\ln2}{\ln x}\right)}{\ln 2x}$, which can be made as close to $\frac{x}{\ln 2x}$ as we like, which can be made far greater than 2015 (If you want to be rigorous, you can use several well known bounds on $\pi(x)$.) Pick an arbitrary large composite $x$ (to avoid boundary cases) that satisfies $\pi(2x)-\pi(x)>2015$. It is clear that for any $n<2x$, $n$ has at most $1+\log_2 x$ prime factors (the maximal case is when $n$ is a power of 2, of course). Let $q=\lceil 1+\log_2 x\rceil$. Now, let $S$ be the set of primes less than $x$. Consider the numbers $m=\prod_{p\in S} p^q(p-1)$, and $k=\prod_{p\in S} p^{q+1}$. It is clear $\varphi(k)=m$, and $n<x\implies n|k$. Now, for any prime $p\in (x,2x)$, consider the number $\frac{pk}{p-1}$. It is obvious that $p-1|k$, as otherwise there must exist a prime divisor of $p-1$ above $x$, but $p-1<2x$, and $2|p-1$, so this is absurd. Therefore $\frac{pk}{p-1}$ is an integer, and it is easy to verify that $\varphi\left(\frac{pk}{p-1}\right)=(p-1)\prod_{r\in S} r^{q-\nu_r(p-1)}(r-1)=(p-1)\frac{m}{p-1}=m$. This works for any $p\in(x,2x)$, which there are over 2015 of, thus we have found over 2015 solutions.

hmm here's the solution I foudn during the test Lemma: For a prime $p$, we have $\phi (p(p-1)) = \phi ((p-1)^2)$ Proof: Since phi is multiplicative, this is trivial. Lemma 2: Let $S = [a_1,a_2,..a_n]$ be a set and let $S' = [b_1,b_2,..b_m]$ be a set. Suppose $\phi (a_1) = \phi (a_2) = .. = \phi (a_n) $ and $\phi (b_1) = \phi (b_2) = ... = \phi (b_m)$. Furthermore, suppose that for any $i, j$, if $d | \gcd (a_i, b_j)$, then $d$ divides all the $a$s and all the $b$s. Then $\phi (a_ib_j)$ is fixed for all $i,j$. This basically says we take the "product" of two sets, the phi's are still equal. Proof: Let $P$ be the set of all distinct prime divisors of $\gcd(a_i, b_j)$ and use the fact that $\phi$ is multiplicative over and over again. Basically we want to show $\phi (a_xb_y) = \phi(a_wb_z)$ but then write $\phi(a_x b_y) = \phi ( \prod_{p \in P} p^{v_p (a_xb_y)} ) \phi(quotient)$ and then use algebra. Now, we induct. Let $p_1 = 7$ and for any $i$, take $A_i = [(p_i -1)^2, p_i (p_i -1)]$, and define $A_1 = S_1$. Define $S_i$ to consist of the productts of all numbers in $S_{i-1}$ with all numbers in $A_i$. Define the $p_i$'s so that $\gcd ( p_i-1 , (p_1-1)(p_2-1)..(p_{i-1}-1) ) = 1$ for all $i>1$ (this is easy with Dirichlet + CRT). Then by lemma 2, we can basically let $i>11$ to get $|S_i| > 2015$ and we have over 2015 numbers with the same phi value.

By the Generalized Prime Number theorem, the number of integers $n < x$ such that $n$ has at most $N$ distinct prime divisors is $\frac{x(\log\log x)^{N - 1}}{(N-1)!\log x}(1 + o_N(1))$, which for a fixed $N$ is $o(x)$. Pick $N = 11$; then there exists an $x$ such that less than $0.001x$ integers less than $x$ have at most 11 prime factors. At least $0.999x$ integers $n < x$ have at least 12 distinct prime factors, and in this case, $2^{11}\vert \phi(n)$ and $\phi(n) < x$, so $\phi(n)$ can take on at most $\frac{x}{2048}$ different values. Thus some value $m$ is taken on at least $\frac{0.999x}{x/2048} > 2015$ times.

Does there exist $ m $ that there is exactly 2015 ?

I consider the following ELEVEN PRIME NUMBERS: $S = \left\{ 11, 13, 17, 19, 29, 31, 37, 41, 43, 61, 71 \right\}$. It has the property that for any $p \in S$, all prime factors of $p-1$ are one digit. Let $N = (210)^{\text{billion}}$, and consider $M = \phi\left( N \right)$. For any subset $T \subset S$, we have \[ M = \phi\left( \frac{N}{\prod_{p\in T} (p-1)} \prod_{p \in T} p \right). \] Since $2^{\left\lvert T \right\rvert} > 2015$ we're done. $\blacksquare$ This solution was motivated by the DEEP FACT that \[ (2^2-2)(3^2-3) = (7-1)(2-1)(3-1) \implies \phi(2^2\cdot3^2) = \phi(2\cdot3\cdot7). \]

LOL this is actually a theorem: https://en.wikipedia.org/wiki/Euler's_totient_function Conclusion/Trend: #5's are theorems (same with IMO #5 and USAMO #5).

Ford's theorem is in fact much stronger.

Excuse me, but can you tell me what is Ford's theorem??

toto1234567890 wrote: Excuse me, but can you tell me what is Ford's theorem?? It's in the linked Wikipedia article...

I don't know if this is of interest to anyone, but I did a little research of the topic and here is what I found. Erdos proved that if we let $N(x)$ denote the number of integers $\le x$ that can be represented (not counting multiplicity) as the Euler Totient function of some other integer, then $$N(x)=o\left(x(\log x)^{\varepsilon-1}\right),$$ for any $\varepsilon>0.$ Now, if we let $N'(x)$ denote the same function, but counting multiplicity then $$N'(x)=\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}x+O\left(\dfrac{x}{(\log x)^{\varepsilon}}\right),$$ for any $\varepsilon>0.$ (Proving $N'(x)>>x$ would be more than sufficient for our needs though.) In particular, if this problem where not true, then we would get $$N'(x)<2015N(x)$$ which would lead to contradiction. I just wanted to point this out, since this is an analytic approach but an approach that I think is much easier than Ford's Theorem mentioned in this thread.

My solution: We define the following sets: $C_1=\{15,16,20,24,30\}$ if $x\in C_1$ $\Longrightarrow $ $\varphi(x)=8$ $C_2=\{11.37,13.31\}$ if $x\in C_2$ $\Longrightarrow $ $\varphi(x)=360$ $C_3=\{19.97,17.109\}$ if $x\in C_3$ $\Longrightarrow $ $\varphi(x)=18.108$ $C_4=\{23.127,43.67\}$ if $x\in C_4$ $\Longrightarrow $ $\varphi(x)=66.42$ $C_5=\{29.101,71.41\}$ if $x\in C_5$ $\Longrightarrow $ $\varphi(x)=28.100$ $C_6=\{7.461,61.47\}$ if $x\in C_6$ $\Longrightarrow $ $\varphi(x)=46.60$ $C_7=\{307.53,157.103\}$ if $x\in C_7$ $\Longrightarrow $ $\varphi(x)=102.156$ $C_8=\{139.661,277.331\}$ if $x\in C_8$ $\Longrightarrow $ $\varphi(x)=276.330$ $C_9=\{199.541,397.271\}$ if $x\in C_9$ $\Longrightarrow $ $\varphi(x)=270.396$ $C_{10}=\{211.457,421.229\}$ if $x\in C_{10}$ $\Longrightarrow $ $\varphi(x)=210.456$ Let $x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}$ intergers positive such that $x_i\in C_i$ for all $1\leq i\leq 10$ the we get: $\varphi(x_1.x_2.x_3.x_4.x_5.x_6.x_7.x_8.x_9.x_{10})=\varphi(x_1).\varphi(x_2).\varphi(x_3).\varphi(x_4).\varphi(x_5).\varphi(x_6).\varphi(x_7). \varphi(x_8).\varphi(x_9).\varphi(x_{10})$ $\varphi(x_1).\varphi(x_2).\varphi(x_3).\varphi(x_4).\varphi(x_5).\varphi(x_6).\varphi(x_7). \varphi(x_8).\varphi(x_9).\varphi(x_{10})=8.360.1944.2772.2800.2760.15912.91080.106920.95760$ $\Longrightarrow $ $\varphi(x_1.x_2.x_3.x_4.x_5.x_6.x_7.x_8.x_9.x_{10})$ is constant $\Longrightarrow $ the number of ways of choosing $x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}$ is: $5.2.2.2.2.2.2.2.2.2=2560\geq 2015$ $\Longrightarrow$ $\varphi(x)=8.360.1944.2772.2800.2760.15912.91080.106920.95760$ has at least $2015$ solutions in $x$...

I think there's an easier solution for this problem. Let $m=2^{N_1}\cdot3^{N_2}\cdot5^{N_3}$ where $N_1,N_2,N_3$ are integers large enough. Let $S=\{7,11,13,17,19,31,37,41,61,101,109,163,251,257,65537\}$ be a set of primes. Noted that for every $p\in S$ ,$p-1$ only have prime factors in $\{2,3,5\}$ and $\phi(3)=2,\phi(5)=2^2$. Therefore $m$ can be written as $\phi (2^k\cdot3^l\cdot5^m\cdot \prod_{p_i \in S} {p_i}^{x_i})$ , herein $x_i\in \{0,1\}$ . Then there are at least $2^{|S|}>2015$ solutions in $n$ .

Actually I've found a solution which is basically the same as Evan Chen's idea. That is, to find a set $P$ comprising of prime numbers such that $\forall p,q\in P, q\nmid \varphi(p)=p-1$ We can then get $2^{|P|}$ solutions. We can use Dirichlet's theorem to construct such $P$, so $|P|$ can be arbitrarily large

Consider $>10$ sets consisting of two elements constructed in the following manner: If we have $(c,d)$ under $(a,b)$ the set is $\{ad,bc \}$ and do this for every two $2k,2k+1-$th pairs from underneath. Since the every pair is $(p,2p-1)$ where both of them are prime it is clear that we have $\ge 2^{11}>2015$ solutions for $n,$ $m$ being the product of Euler functions of all sets (since we have $\varphi(ad)=\varphi(bc)$). 7219 14437 7297 14593 7369 14737 7411 14821 7507 15013 7537 15073 7561 15121 7621 15241 7639 15277 7681 15361 7687 15373 7867 15733 7951 15901 8017 16033 8167 16333 8191 16381 8209 16417 8287 16573 8317 16633 8329 16657 8461 16921 8521 17041 8527 17053 8539 17077 8629 17257 8647 17293 8689 17377 8941 17881 9007 18013 9049 18097 9067 18133 9091 18181 9109 18217 9127 18253 9151 18301 9157 18313 9199 18397 9241 18481 9277 18553 9319 18637 9397 18793 9619 19237 9721 19441 9739 19477 9859 19717 9901 19801 9907 19813 9931 19861 P.S Big numbers are taken since if we did this for smaller ones, there could've been easily repeating of some numbers, violating the condition that all numbers must be coprime, which we need for the proof.

v_Enhance wrote: I consider the following ELEVEN PRIME NUMBERS: $S = \left\{ 11, 13, 17, 19, 29, 31, 37, 41, 43, 61, 71 \right\}$. It has the property that for any $p \in S$, all prime factors of $p-1$ are one digit. Let $N = (210)^{\text{billion}}$, and consider $M = \phi\left( N \right)$. For any subset $T \subset S$, we have \[ M = \phi\left( \frac{N}{\prod_{p\in T} (p-1)} \prod_{p \in T} p \right). \]Since $2^{\left\lvert T \right\rvert} > 2015$ we're done. $\blacksquare$ This solution was motivated by the DEEP FACT that \[ (2^2-2)(3^2-3) = (7-1)(2-1)(3-1) \implies \phi(2^2\cdot3^2) = \phi(2\cdot3\cdot7). \] Yeah lol when solving this problem for practice, I let $m = (30)^{1000}$, and we consider the prime numbers $\{7, 11, 13, 17, 37, 41, 97, 101, 109, 257, 641, 65537\}$. This is more than enough prime numbers, and each element has the property that it's $1$ more than a number that has only $2, 3, 5$ in its prime factorization. Then, I finish the same way as v_Enhance, except we use the fact that $\phi(3)$ and $\phi(5)$ are both powers of $2$(after picking the subset of primes, we choose the corresponding power of $5$ that makes $v_5$ the right amount, then the corresponding power of $3$ that makes $v_3$ the right amount, then finally picking the corresponding power of $2$ that makes $v_2$ the right amount; GCD with the product of primes in $S$ is unique to the choice of subset, so we have different numbers each time that produce the same totion). Honestly, I don't know why this was a TSTST 5, if it could be solved with a simple bash. It seems easier than J1 difficulty.

^ Easy to say in retrospect... By the Prime Number theorem, we can find a large $N$ such that there are at least $2015$ primes $N < p_1 < p_2 < \dots < p_n < 2N$. Notice that since $\frac{p_i - 1}{2} < N$, its prime factors are less than $N$. Let $q_1 < q_2 < q_3 < \dots < q_k < N$ be the primes less than $N$, and let $P = \prod_{i=1}^k q_i$. Then if we choose a very large $K$ and $m = \phi(P^K)$, then the multiplicativity of $\phi$ ensures that $$\phi(p_j \cdot \prod_{i=1}^k q_i^{K - v_{q_i} (p_j - 1)}) = m.$$This leads to at least $2015$ solutions.

raxu wrote: Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$. Proposed by Iurie Boreico Hmm, existential proof By Dirichlet's Theorem, pick a bunch of primes $q_1, \dots, q_s>10^9$ with $q_{i+1} \equiv 2 \pmod{q_1 \dots q_i}$ for all $1 \le i<s$. Consider all primes $p_1, \dots, p_k$ that divide $(q_j-1)$ for some $1 \le j \le s$. Let $M$ sufficiently large and consider $$m=(p_1 \dots p_k)^M (p_1-1) \dots (p_k-1).$$Then any number $n=\left(\frac{q_{\ell}}{q_{\ell}-1}\right) \cdot (p_1 \dots p_k)^{M+1}$ is a solution of the equation $\phi(n)=m$, for all $1 \le \ell \le s$. Consequently, we can let $s=2017$ to obtain the desired conclusion. Remark: Since the structure $\phi(a)=\phi(b)$ is quite complex in general; it is best suited if we "perturb" a solution only slightly. Thus, we seek to append a single prime but also hope it'd cancel somehow. Our quest yields this solution

Since I don't see such a solution above, here is a completely elementary solution (I did the problem this way live when I took the contest back in 2015). We show the following lemma by induction, which directly finishes the problem. Lemma: Let $p_1, p_2, \dots, p_k$ be the first $k$ primes. Then there exist at least $k$ distinct positive integers $n$ such that $\phi(n) = \phi\left(\prod_i p_i \right)$, and each of these $n$ only has prime divisors among $p_1, p_2, \dots, p_k.$ Proof: We use induction. Let $n_1, n_2, \dots, n_k$ satisfy $\phi(n_j) = \phi\left(\prod_{i=1}^k p_i \right)$ for all $j$. Say that $n_1 = \prod_i p_i$. Then note that obviously, $\phi\left(n_j \cdot p_{k+1}\right) = \phi\left(\prod_{i = 1}^{k+1} p_i \right).$ To finish the induction, we need one more number. To get this number, write $\phi(p_{k+1}) = p_{k+1} - 1 = \prod_{i=1}^k p_i^{e_i}$, as $p_{k+1}-1$ only has prime divisors among $p_1, \dots, p_k.$ Then it is clear that \[ \phi\left(\prod_{i=1}^k p_i^{e_i} \cdot \prod_{i=1}^k p_i \right) = \phi\left(\prod_{i=1}^{k+1} p_i \right), \]which completes the induction.

Let $p_1<p_2<\cdots $ be the set of primes.Replace $2015$ by $k$,we will prove the result that there exists $m$ such that $\phi(n)=m$ has atleast $k$ solutions.Indeed define numbers $x_i$ such that $$N=p_1p_2\cdots p_k; x_i=N(1-\tfrac{1}{p_i})$$I claim that $\phi(N)=\phi(x_i)$.Indeed note that all prime factors of $p_i-1$ belong to the primes ${p_1,\cdots p_{i-1}}$.Thus $$\dfrac{\phi(x)}{x_i}=\prod_{p\mid x} \left(1-\dfrac{1}{p}\right)=\dfrac{\prod_{j=1 }^{j=k} (1-\dfrac{1}{p_j})}{\left(1-\dfrac{1}{p_i}\right)}=\dfrac{\dfrac{\phi(N)}{N}}{\left(1-\dfrac{1}{p_i}\right)}=\dfrac{\phi(N)}{x_i}\implies \phi(x_i)=\phi(N)$$

I claim we can construct an infinite sequence of primes $p_i$ for which no $p_i$ divides another $p_j - 1$. We do this inductively. For the base case, say we simply pick $p_1 = 5$ ad $p_2 = 7$. Thereafter, for every $k \geq 3$, we pick $p_k \equiv 2 \pmod {p_1p_2\ldots p_{k-1}}$ which clearly exists by Dirichlet. Construct this sequence up to a certain $n$. I claim that the $2^n$ possible numbers $X = x_1x_2\ldots x_n$ generated by $x_i \in \{p_i(p_i - 1), (p_i - 1)^2\}$ all satisfy $\phi(X)$ being equal to the same value. Indeed, consider $X_1$ with primes in $S_1$ contributing $x_i$ of the former form and $X_2$ with primes in $S_2 \neq S_1$ contributing $x_i$ of the latter form. Then, since each $p_i$ is relatively prime to each other $p_j - 1$, we have\[\phi(X_1) = \phi(x_1\ldots x_n) = \left(\prod_{p_i \in S_1} \phi(p_i)\right) \phi\left(\prod (p_i - 1)^e\right)\]where $e = 1$ if $p_i \in S_1$ and $2$ otherwise. Similarly, \[\phi(X_2) = \phi(x_1\ldots x_n) = \left(\prod_{p_i \in S_2} \phi(p_i)\right) \phi\left(\prod (p_i - 1)^e\right)\]where $e = 1$ if $p_i \in S_2$ and $2$ otherwise. Divide them and quickly see that the quotient is $1$. Indeed, so taking sufficietly large $n$, we have $2^n > 2015$ numbers for which $\phi$ generates the same value.

Interesting! Clearly, $\varphi(p(p-1))= \varphi((p-1)^2)$. Clearly, by induction, Dirichlet, and CRT we can construct a set of primes $P = \{p_1,p_2,...,p_{\text{big }} \}$ where $p_i-1$ is not divisible by any element in $P$. For any $i$, we can either choose $p_i(p_i-1)$ or $(p_i-1)^2$ and each is independent since they are relatively prime. Clearly, $2^{\text{big}} \geq 2015$. So we are done.

Let $p_1,p_2,p_3,......$ be an increasing sequence of primes and fix a positive integer $k$ Clearly $N=p_1 p_2 p_3.....p_k$ and $x_i=N \cdot (1 - \frac{1}{p_i})$ satisfies $\phi(N)=\phi(x_i)$ since $\frac{\phi(x_i)}{x_i}$$=\prod(1-\frac{q}{p_j}) \cdot \frac{p_i}{p_i-1}$$=$${\phi(N) \over N }\cdot$$ \frac{p_i}{p_i-1}$$=$$\frac{\phi(n)}{x_i}$,as required.

$m=2^{20}\cdot 3^9\cdot 5^3\cdot 7^2\cdot 11^2$ works. We can let $\nu_{p}{(n)}$ equal $0$ or $1$ for every prime in $\{13, 17, 19, 23, 29, 31, 37, 41, 43, 61, 67\}$ and adjust the number of factors of $2,3,5,7,11$ in $n$ as needed. The number $m$ is equivalent to $$15296168853504000$$

Solved with GoodMorning. We claim $m=\boxed{2^{41}\cdot 3^{19}}$ works. Let $S$ be the set of primes $\{5,13,17,37,73,97,109,163,193,577,769\}$. We chose these primes because they are $1$ more than a power of $2$ times a power of $3$. Note that $m=\varphi\left({\prod_{x\in S}x}\right)$. Claim: There exists a positive integer $x$ with $\varphi(x)=2^a\cdot 3^b$, for all positive integers $a$, nonnegative integers $b$, and $a=b=0$. Furthermore, $x$ has no prime factors except $2$ and $3$. Proof: Set $x=2^a\cdot 3^{b+1}$. $\blacksquare$ Take any subset $T\subset S$. We have \[\frac{\varphi\left(\prod_{x\in S}x\right)}{\varphi\left(\prod_{x\in T}x\right)}=\varphi\left(\prod_{x\in S, x\not\in T}x\right)\]Note that this value can be written in the form $2^a3^b$, where $a>0$. Setting $n=2^a\cdot 3^{b+1}\cdot \prod_{x\in T}x$ gives $\varphi(n)=m$. Also setting $n=\prod_{x\in S}x$ works. There are $2^{11}-1=2047$ ways to choose $T$, and one additional way, which gives at least $2048>2015$ different values of $n$.

Oops. We claim that \[m=312 \cdot 408 \cdot 576 \cdot 660 \cdot 936 \cdot 1152 \cdot 1800 \cdot 1320 \cdot 1008 \cdot 1200=149967695669069120274432000000\]works. Choose $n=a_1 \cdot a_2 \cdots a_{11}$, where \begin{align*} a_1 &\in \{1,2\}\\ a_2 &\in \{313,3 \cdot 157\}\\ a_3 &\in \{409,5 \cdot 103\}\\ a_4 &\in \{577,7 \cdot 97\}\\ a_5 &\in \{661,11 \cdot 67\}\\ a_6 &\in \{937,13 \cdot 79\}\\ a_7 &\in \{1153,17 \cdot 73\}\\ a_8 &\in \{1801,19 \cdot 101\}\\ a_9 &\in \{1321,23 \cdot 61\}\\ a_{10} &\in \{1009,29 \cdot 37\}\\ a_{11} &\in \{1201,31 \cdot 41\} .\end{align*}Since there are $2^{11}>2015$ choices for $n$, we are done.

Am I getting trolled? We claim that $m=2^{10000}3^{10000}$ works. Consider the eleven primes $$5,17,257,65537$$$$7,19,163$$$$13,37,109$$$$73.$$Each of these primes has the property that the number 1 less than it contains no prime factors other than possibly 2 or 3. We will generate 2048 different $n$ for which $\phi(n)=2^{10000}3^{10000}$ in the following manner: Take any subset of the aformentioned 11 primes (which there are 2048 of), multiply all of those primes, and then multiply that result by 6. Call this intermediate result $I$. Clearly, $\phi(I)=2^a3^b$ for $a,b<10000.$ Then, since $I$ is already a multiple of 6, we then have $$\phi(I\cdot 2^{10000-a}3^{10000-b})=2^{10000}3^{10000},$$which gives us a solution for each of the 2048 values of $I$, so we are done

Wait a minute.. Just take primes $p_1,p_2,...,p_2015$ such that $p_i$ does not divide $p_j - 1$ for all $i,j$, this is possible by Dirichilet. Now, take all the prime divisors $q_i$ of $p_i-1$ for all $i$ and let $C$ be a big integer. Take numbers $p_i.q_1^{a_1}.q_2^{a_2}...q_j^{a_j}$, choosing $a_i$ in the following way: noticing $q_i\neq p_j$ for all $i,j$ by our choice, we choose $a_i$ such that $v_{q_i}(p_i-1)+a_i=C$, which solves the problem as all the v_q's of the numbers are equal, and all the $q-1$ all cancel out.

For a positive integer $n=p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$, we have that $\phi(n)=\left(p_1^{\alpha_1}-p_1^{\alpha_1-1}\right)\cdot\ldots\cdot \left(p_k^{\alpha_k}-p_k^{\alpha_k-1}\right)$. Note that if we have $\phi(n)=x$ has at least $a$ solutions and $\phi(n)=y$ has at least $b$ solutions, then $\phi(n)=xy$ has at least $ab$ solutions as long as each of the $a$ solutions are coprime to each of the $b$ solutions. Letting the $a$ solutions for the first equation be $\{a_1,\ldots,a_a\}$ and the $b$ solutions for the second equation be $\{b_1,\ldots,b_b\}$, we see that this is true because if $a_i$ and $b_j$ are coprime, then $\phi(a_i\cdot b_j)=\phi(a_i)\cdot\phi(b_j)$. Therefore, finding enough numbers $m$ for which $\phi(n)=m$ has more than one solution will do the trick. For this,, just pick $\left\lceil\log_2(2015)\right\rceil=11$ pairs of paiwise distinct primes $p,q$ such that $r=(p-1)(q-1)+1$ is prime, and which are each also pairwise distinct from those constructed numbers $r$ (just do small cases, sorry ). As then $\phi(r)=\phi(pq)=(p-1)(q-1)$, we get what we want.

I feel like I have some kind of obsession with Dirichlet now :/ Pick $2015$ primes $p \equiv 1 \pmod {10}$ by Dirichlet, and let $n$ be the LCM of $(p-1)^2$ for these primes multiplied by $10^{1000}$. Notice that we have $$\phi(n) = \phi\left((p-1) \cdot \frac n{p-1}\right) = \phi\left( p \cdot \frac n{p-1}\right)$$for each of these primes $p$ because $p-1$ divides $\frac n{p-1}$. It follows that by varying across all $p$ we can construct $2015$ solutions to $\phi(k) =\phi(n)$, as needed.

Let $n=30^{1434143414341434}$. Let $p$ be the product of any subset of $\{ 7, 11, 13, 17, 31, 37, 41, 61, 97, 151, 181\}$ besides $1$. Note that there are $2047$ values of $p$. Note that $\phi(\frac{pn}{\phi(p)})=\phi(p)*\phi(\frac{n}{\phi(p)})=\phi(n)$, since for all $p$, $\phi(p)$ is the product of all the prime factors of $p$ minus one, which only has prime factors $2,3,$ or $5$, so it divides $n$ and dividing by $\phi(p)$ does not impact the prime factors and only divides the final value by $\phi(p)$, which is then multiplied back. All values of $p$ work, which is more than $2015$ values.

Consider the $11$ pairs of primes $(3,5), (7,13), (31,61) ,(37,73) ,( 57,113), (79,147) , (97,193) , (139,277) , (157, 313) , ( 199,397), (211, 421)$. The $2048$ numbers generated from picking one prime from each pair and taking the product have Totient functions differing by powers of $2$, so it suffices to multiply each of those numbers by a power of $2$, and we are done.

Let $p_1, p_2, \dots, p_N$ be the first $N$ primes for some sufficiently large $n$, and consider \[m=\phi(p_1p_2 \cdots p_N) = (p_1-1)(p_2-1)\cdots(p_N-1).\]The claim is that for all $N/2 < k < N$, we can construct an $n$ such that $p_k \nmid n$, $p_i \mid n$ for all $i \neq k$, and $\phi(n) = m$. Clearly these $n$ are distinct, so they exhibit our desired examples. To do this, let $q_1, q_2, \dots, q_\ell$ be the maximal prime powers that divide $p_k - 1$. By construction, for each $i$ there exists an index $j < k$ such that $q_i$ is equal to the $a_i$th power of $p_j$. Then taking \[n = (p_k - 1)\prod_{i \neq k} p_i\]works for each $k$. Remark: The construction for $n$ is oddly simplistic by number theory standards.