2019 AMC 10 A

Prime factorizing , we get . A perfect square must have even powers of its prime factors,so our possible choices for our exponents of a perfect square are , , , , for both and . This yields perfect squares.

Perfect cubes must have multiples of for each of their prime factors' exponents, so we have either , , , or for both and , which yields perfect cubes, for a total of . Subtracting the overcounted powers of (, , , and ), we get .

Observe that . Now divide into cases:

Case : The factor is . Then we can have , , , , , or .

Case : The factor is . This is the same as Case .

Case : The factor is some combination of and .

This would be easy if we could just have any combination, as that would simply give . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for .

is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.

is a "cube" because it would give a factor of this number that is a perfect cube. More generally,it is a multiple of .

is a "square"

is interesting, since it's both a "square" and a "cube". Don't count this as either because this would doublecount, so we will count this in another case.

is a "square"

is a "cube".

Now let's consider subcases:

Subcase : The squares are with each other.

Since we have square terms, and they would pair with other square terms, we get possibilities.

Subcase : The cubes are with each other.

Since we have cube terms, and they would pair with other cube terms, we get possibilities.

Subcase : A number pairs with .

Since any number can pair with (as it gives both a square and a cube), there would be possibilities. Remember however that there can be two different bases ( and ), and they would produce different results. Thus, there are in fact  possibilities.

Finally, summing the cases gives .