If $n = p_1^{e_1}\cdot p_2^{e_2}\cdots$ with $p_i$ prime, then the number of divisors $\tau(n) = (e_1 + 1)(e_2 + 1)\cdots.$

Therefore, if the number of divisors $\tau(n)$ is a prime number $p$ , then $n = q^{p-1}$ , with $q$ prime. This allows us to write $a^b = q^4,\ b^a = r^6\ \ \ \ \ \ \text{with}\ q,r\ \text{prime};$ we see that $b | 4$ and $a | 6$ , and both $a$ and $b$ must be powers of a prime number. This allows for only two solutions of each number: $a = 2$ or $3$ , and $b = 2$ or $4$ .

However, since $b^a = r^6$ is a sixth power, we need $a = 3$ and $b = 4$ , so that $b^a = 4^3 = 2^6$ .

It follows that $\tau(a \times b) = \tau(2^2 \cdot 3^1) = (2 + 1)(1 + 1) = \boxed{6}.$

The only time you can have primes for the amount of divisors is if the number can be expressed as $n = x^{p-1}$ where x and p are prime

$a^{b} = x^{4}$

$b^{a} = y^{6}$

After fiddling around with it you get $a = 3$ and $b = 4$

See Arjen Vreugdenhil's answer for a more comprehensive explanation XD

By the fundamental theorem of arithmetic we can say that $a = \prod p_i^{a_i}, b = \prod q_i ^{b_i}$ . Then $a^b = \prod p_i^{ba_i}, b^a = \prod q_i ^{ab_i}$ . And if we apply the product representation of the divisor function we get the equations $\prod (ba_i + 1) = 5, \prod (a b_i + 1) = 7$ . Notice that in either of these products, each term of the product has to be larger than or equal to $2$ , but the right hand side of the equalities is a prime number, so there can only be one term in each of these products so we only have: $ba_1 + 1 = 5, a b_1 + 1 = 7$ implying that both $a$ and $b$ have only one prime factor. Simplifying these equations we get $b a_1 = 4, a b_1 = 6$ .

To solve this system let's substitute in $a = p_1^{a_1}, b = q_1^{b_1}$ and first let's try to solve the first equation: $q_1^{b_1} a_1 = 4$ . First, because $q_1$ is prime and the right hand side is a $4=2^2$ , necessarily this prime number has to be $q_1 = 2$ . Now, there are two solutions: $q_1 = 2, b_1 = 2, a_1 = 1$ and $q_1 = 2, b_1 = 1, a_1 = 2$ . Now look at the second equation, $p_1^{a_1} b_1 = 6$ . Because the right hand side is square-free, $a_1 = 1$ which then implies $q_1 = 2, b_1 = 2$ so
$p_1 = 3$ .

Finally this implies $a = 3^{1} = 3, b = 2^{2} = 4$ . So the solution is the number of positive divisors of $3 \times 4 = 12$ which is 6.

Any number with an odd number of divisors must be square. Therefore, we know that both $a^b$ and $b^a$ are square. Notice that a number can only have $5$ positive divisors if it is in the form of $\text{prime number}^4$ and can only have $7$ positive divisors if it is in the form $\text{prime number}^6$ . We first try to see if both $a$ and $b$ are prime, but this is impossible because 4 and 6 aren't prime. We then see if one is a prime and the other is a square of a prime. We test a as prime and b as a square of a prime. We get a = 3, b = 4. 3*4=12, and 12 has 6 divisors.

$a$ must be greater than 1 since then it would have only 1 divisor. So trying $a = 2$ , $b$ must be 4 to get 5 divisors, but that didn't produce 7 divisors for $b ^ a$ . So I tried $a = 3$ and $b = 4$ and that did work so I only needed to work out the divisors of 12.

If $a^b$ has $5$ divisors, then it has three divisors different to $1$ and $a^b$ . Since $a$ is factor of $a^b$ (because $b\neq 1$ since $b^a$ has $7$ divisors) then $a^{b-1}$ is also a factor. The only left factor then is $a^{\frac{b}{2}}$ . So we can conclude that

1. $a$ is prime
2. $b=4$

Then we have to find the number $a$ such $4^a=2^{2a}$ has $7$ divisors. Since $2$ is prime, $a=3$ so that $2a=6$ .

Finally, $ab=12$ which has $6$ factors

By inspection: tentatively assume a is prime. Then a^4 has 5 positive divisors (1, a, a^2, a^3, and a^4). Then we need to find a s.t. 4^a has 7 divisors. 4^a = 2^2a, which has 7 divisors if a = 3 (1, 2, 4, 8, 16, 32, 64). This meets the assumption that a is prime. 3^4 has 5 divisors, and 4^3 has 7 divisors. 4*3 = 12, which has 6 divisors (1, 2, 3, 4, 6, and 12).

Easy Python 3 code:

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def divisors(n):
    divs = []
    for i in range(1, n+1):
        if n % i == 0:
            divs.extend([i])
    return divs


flg = 0
for a in range(1, 20):
    for b in range(1, 20):
        if len(divisors(a**b))==5 and len(divisors(b**a)) == 7:
            flg=1
            break
    if flg == 1:  
        break

print('a = {} and b = {} found.'.format(a, b))
print('Answer: {}×{}={} has {} divisors.'.format(a, b, a*b, len(divisors(a*b))))

Output:

1
2

a = 3 and b = 4 found.
Answer: 3×4=12 has 6 divisors.

If X=(p1)^a(p2)^b......

has no. of divisors (a+1)(b+1)....... as a can be from {0,1,2....,a} that's ( a+1) choices and similarly for b

now for all at a time (intersection=>multiplication)

but 5 can only be 5x1 hence a is a prime number and hence b=4

from b^a=7 we get a=3

so 3x4=12 has 6 divisors

Only perfect squares have an odd number of divisors so 5 and 7 are discarded (because a cannot be equal to b because the result in both cases will have the same amount of divisors), also tells us that both numbers must be even. The resulting number is not prime so it will have at least 4 divisors (1, a , b and a*b). We can check that the numbers are not 2 and four because 2^4 and 4^2 are both 16 with has 5 divisors, so, from the fact that they're even we conclude they have more than 4 divisors, wich leaves only one option, 6.