Diophantine Equations with Factor Exponents I'm trying to prove that the following equations have no solutions to finish a problem. They're intuitively impossible but I'm looking for rigorous arguments (if they are actually possible, then prove that too).
Let $d(n)$ be the number of positive divisors of a positive integer $n$.
For positive integers $a$ and $b$ that aren't perfect squares, prove that there are no solutions to the equation $$a^{d(a)/2} = b^{d(b)/2}$$
For positive integers $a^2$, $b^2$, prove that there are no solutions to the equation $$a^{2d(a) + 1} = b^{2d(b) + 1}$$
For positive integers $a^2$, $b$ where $b$ isn't a perfect square, prove that there are no solutions to the equation $$a^{2d(a) + 1} = b^{d(b)/2}$$

Let us solve the first equation we can assume that $a>1,b>1$:
$$a^{d(a)}=b^{d(b)} \tag{1}$$
Given a prime $p$ let $v_p(x)$ the greatest integer $k$ such that $p^k$ divides $x$, and let $\omega(x)$ be the number of prime factors of $a$, in other words $\omega(x)$ is the number of primes $p$ such that $v_p(x)\neq 0$, Now the equation $(1)$ is equivalent to: for every prime $p$
$$v_p(a)d(a)=v_p(b)d(b) \tag2$$
whence $\omega(a)=\omega(b)$, we want to prove that $d(a)=d(b)$, WLOG we can assume that  $d(a)\leq d(b)$ so using $(2)$ we obtain for every prime $p$:
$$v_p(b)\leq v_p(a)\tag3$$
But using the formulas :
$$d(a)=\prod_{p\in P}(v_p(a)+1)\text{ and }d(b)=\prod_{p\in P}(v_p(b)+1)\tag4$$
we conclude using $(3)$ that 
$$d(b)\leq d(a) $$
finally $d(a)=d(b)$ and from $(1)$ we conclude that $a=b$.
you can apply the same method for the other equations
