$x,y\in \mathbb Z^+$ Show that $P(x)=P(y) \implies x=y$ 
Let $x,y\in \mathbb Z^+$, we define $P(a)$ as the product of all the divisors of $a$ including $1$ And $a$, Show that $$P(x)=P(y)\implies x=y.$$

Im gonna show my Attempt, which is based on cases.
Case 1: if  $x,y$ Are primes.
$$x,y\in \mathbb P\implies \cases{P(x)=x\\P(y)=y}\implies (P(x)=P(y) \iff x=y)$$
Where $\mathbb P$ Is the set of prime numbers.
Case 2: if $x,y$ are some powers of prime.
$$\cases{x=p^n\\ y=q^m } \implies \cases{P(x)=1\cdot p\cdot p^2...p^n=p^{\frac{n(n+1)}{2}}\\P(y)= 1\cdot q\cdot q^2...q^m= q^{\frac{m(m+1)}{2}}} $$ So to prove the statement We set $$p^{\frac{n(n+1)}{2}} =q^{\frac{m(m+1)}{2}}$$
The next step is to show that this statement implies $x=y$, I have no idea how to do that. (By the way the next case is to consider $x,y$ As a product of primes.)
 A: Deep breath.
So every positive integer is going to have a unique prime factorization.  Each factor is going to be a product of the these primes to some power and so $P(x)$ is going to be a product of these factors is going to be a product of these primes to powers.
So if $x = \prod p_i^{k_i}$ is the unique prime factorization of $x$ then $P(x) = \prod p_i^{m_i}$ where $\{m_i\}$ is a distinct set of natural numbers based on some manipulation of the values of $k_i$ and possibly, but probably not, the values of $p_i$.  Just what this manipulation is ... well, we'll figure that out later.
So if $P(x) = P(y) = M = \prod p_i^{m_i}$ then it must be that $x$ and $y$ have the exact same prime factors, namely $\{p_i\}$ and $x = \prod p_i^{k_i}$ and $y = \prod p_i^{w_i}$ and that somehow the set of $\{k_i\}$ and the set of $\{w_i\}$ both get manipulated to the same set $\{m_i\}$.
We have to show nothing more or less than if two set's of $\{k_i\}$ and $\{w_i\}$ both get manipulated to the same set of $\{n_i\}$ then that would imply that each $k_i$ is equal $w_i$ respectively.
Okay... so just what is the manipulation[1]?
......
Well.  Consider $x = \prod p_i^{k_i}$ and lets consider one particular prime designated $Q$ (with a capital) so that $x = Q^K \prod_{p_i\ne Q} p_i^{k_i}$ and let $N = \frac x{Q^K}$ and so that $p\not \mid N$.
Okay.... Let $F = \{$ factors of $N\}$ and so each factor of $x$ will be expressible as $Q^jf$ for some $f\in F$ and some power $j$ where $w = 0,...., K$.
So $P(x) = \prod_{j= 0}^K [\prod_{f\in F} Q^j f]=$
$(\prod_{j=0}^K Q^j)^{|F|}\prod_{f\in F} f$
Okay... $\prod_{f\in F} f = P(N) = P(\frac x{Q^K})$ and $\prod_{j=0}^K Q^j= Q^{\sum_{j=0}^K j} =Q^{\frac {K(K+1)}2}$ and $|F|$....
well.... if a number has $\prod p_i^{k_i}$ as it's prime factorization then a factor $f$ is of the form $\prod p_i^{a_i}$ where $0 \le a_i \le k_i$.  There are $k_i +1$ options for $a_i$ so there are $\prod (k_i + 1)$ factors.
So $|F| =\prod_{p_i \ne Q} (k_i + 1)$.
So we have $P(x) = Q^{\frac K2\cdot (K+1)\prod_{p_i \ne Q} (k_i + 1)}\cdot P(\frac x{Q^K})$ but we can put that $(K+1)$ inside the product so $P(x) = Q^{\frac K2\prod(k_i + 1)}\cdot P(\frac x{Q^K})$
And so recursively we have

$P(x) = \prod p_j^{\frac {k_j}2 \prod(k_i + 1)}$.

........
So if $P(x) = P(y) = \prod p_j^{m_j}$ then $x = \prod p_j^{k_j}$ and $y = \prod p_j^{w_j}$ and we have for each $j$ than
$m_j = \frac {k_j}2(\prod (k_i +1)) =\frac {w_j}2(\prod (w_i +1))$ and our task is simply to show that implies each $k_j = w_j$.
....Phew.....
Okay so we know that
each $k_j = w_j \cdot \frac {\prod(w_i+1)}{\prod(k_i+ 1)}$
but $\frac {\prod(w_i+1)}{\prod(k_i+ 1)}=\prod \frac {w_i + 1}{k_i + 1}$ is a positive constant.
If we call it $C$ we have:
$C = \prod \frac {w_i + 1}{k_i + 1} = =\prod \frac {w_i + 1}{Cw_i + 1}$
Now we just have to show that is only possible if $C = 1$.
And if $C < 1$ then $\frac {w_i + 1}{Cw_i + 1} > 1$ and $C = \prod \frac {w_i + 1}{Cw_i + 1} > 1$.  A contradiction.
And if $C > 1$ then $\frac {w_i + 1}{Cw_i + 1} < 1$ and $C = \prod \frac {w_i + 1}{Cw_i + 1} < 1$.  A contradiction.
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[1] Credit where credit is due.  Calvin Lin had a for more elegant calculation and expression of $P(N)$ that escaped me.
Let $x =\prod p_i^{k_i}$ and let $\tau(x) = \prod(k_i + 1)=$ the number of factors of $x$.  Now if $d|x\iff\frac xd|x$ obviously. So....
$P^2(x) = (\prod_{d|x}) d\cdot (\prod_{d|x})\frac xd = \prod_{d|x} N= N^{\tau(x)} = N^{\prod(k_i + 1)}$
So $P(x) = N^{\frac {\prod(k_i + 1}2}$....
Which as $\sqrt{N} = \prod p_i^{\frac {k_i}2}$ is the same result as mine.
A: Recall that if $ N = \prod p_i ^ {a_i}$ is the prime factorization of $N$, then $N$ has $\tau(N) = \prod (a_i + 1)$ factors, and $P(N) = N^{\tau(N)/2 } $ (by pairing up the factor $d$ with $N/d$).
Let $M = \prod p_i ^{b_i}$ (with the same ordering of primes, and finitely many exponents are non-zero).
Suppose that $P(N) = P(M)$.
Then $N^{\prod(a_i + 1) } = M^{\prod (b_i + 1) }$.
Considering prime $p_k$, this gives us $ a_k \prod(a_i + 1 ) = b_k \prod(b_i + 1)$ for all $k$.
Hence $a_k / b_k$ is the constant $\prod (b_i+1)/(a_i+1)$.
If $a_k / b_k < 1$, then clearly $a_k \prod (a_i + 1) < b_k \prod (b_i + 1)$, which is a contradiction. Likewise for $a_k / b_k  >1$.
Hence, $a_k / b_k = 1$ and so $N = M$.
A: Let $\mathcal E(x)$ be the exponents of the prime factorization of $x$, for example $\mathcal E(2^7 \times 3^0 \times 5^3) = [7, 0, 3]$.
Let $X = \mathcal E(n)$ and $Y = \mathcal E(P(n))$.  As noted in Calvin Lin's Answer, $P(n) = n^{\tau(n)/2}$, so $Y_k = X_k \times \tau(n)/2$.  So given $Y$, to determine $\tau$:
$$\tau = (X_0 + 1)(X_1 + 1)\dots(X_{z - 1} + 1) = (2Y_0/\tau + 1)(2Y_1/\tau + 1)\dots(2Y_{z-1}/\tau + 1)$$
$$\tau^{z+1} = (\tau + 2Y_0)(\tau + 2Y_1)(\tau + 2Y_2) \dots (\tau + 2Y_z)$$
By Descartes' Sign Rule, the above only has 1 positive solution for $\tau$ if it has any, so $P$ is injective.
