A Complex Logic Puzzle!

We first note that the equation $D^{2015} - 1 = 0$ has $2015$ roots, so $D$ can be one of $2015$ different numbers, each of the form $D_{k} = \Large e^{ik*\frac{2\pi}{2015}}$ for integers $0 \le k \le 2014.$ Since we must have that $P = a^{6} \lt 2015$ for some integer $a \gt 1$ implies that $P$ can either be $2^{6} = 64$ or $3^{6} = 729,$ as $a^{6} \gt 2015$ for $|a| \ge 4.$

Now since this is enough information for Dhyan to know the value of $D^{P}$ means that

$D^{64} = D^{729} \Longrightarrow D^{729} - D^{64} = D^{64}(D^{655} - 1) = 0.$

Now $D^{64} = \Large e^{\frac{128\pi}{2015}}$ $\ne 0,$ so we must have that

$D^{655} = 1 = e^{i m*2\pi} \Longrightarrow 655*\dfrac{2k\pi}{2015} = m*2\pi \Longrightarrow 655k = 2015m \Longrightarrow 133k = 403m$

for some positive integers $k,m.$ Now as $133$ is prime and does not divide $403$ it must divide $m,$ and so we require that $k = 403n$ for some positive integer $n.$ Thus Dhyan's favorite number is of the form

$D = \Large e^{i 403n*\frac{2\pi}{2015}} = e^{i * \frac{n*2\pi}{5}},$ for which $D^{5} = e^{i n*2\pi} = 1.$

As $3, 13$ and $31$ are primes distinct from $5,$ we can conclude that $\boxed{D^{5}}$ is the only one of the given options that equals $1.$

From Pratik's first statement, we come to know that $P = a^{6}$ for some integer $a > 1$ . The first few possible values of $P$ are $64, 729, 4096 \ldots$

From Dhyan's reply to Pratik's first statement, we infer that $D$ is not a real number, and it is one of the 2015th roots of unity. That is, $D$ is of the form $\exp( \frac{2\pi k}{2015} i )$ , $k \in \{ 1, 2, 3, \ldots, 2014 \}$ . There are a total of $2014$ different values possible for $D$ .

In Pratik's second statement, Pratik says that there are more than $P$ possible values of $D$ . We know that there are $2014$ possible values for $D$ , so we get the inequality $P < 2014$ . We see that the only possible values of $P$ now are $64$ and $729$ .

From this deduction, Dhyan does not know whether $P$ is $64$ or $729$ , yet he knows the value of $D^{ P }$ . This means that $D^{64 }$ and $D^{729 }$ have the same value.

$\phantom{\implies} D^{64} = D^{729}$

$\implies \exp( \frac{2\pi k}{2015} \times 64 i ) = \exp ( \frac{2\pi k}{2015}\times 729 i )$

Since both the complex numbers are equal, their principle arguments are equal. That is, their arguments have a difference of a integral multiple of $2\pi$ .

$\implies \frac{2\pi k}{2015} \times 64 + 2n \pi = \frac{2\pi k}{2015} \times 729, ~~~n \in \mathbb{Z}$

$\implies 64 k + 2015n = 729k$

$\implies 665k = 2015 n$

$\implies 133 k = 403 n$

$\implies k =\dfrac{403 } {133 } n$

Since $k$ is an integer and $133$ is coprime with $403$ , $133$ must divide $n$ . This means $\frac{n}{133 }$ is an integer. Let's call it $m$ . We see that $m \in \{ 1, 2, 3, 4 \}$ .

$\implies k = 403 m$

$\implies D = \exp( \frac{2\pi \times 403 m }{2015} i )$

$\implies D = \exp( \frac{2\pi \times m }{5} i )$

We see that $D$ is a fifth root of unity not equal to $1$ , therefore $D^{5 } = 1$ and none of the other options are equal to $1$ .