There are about 4 billion 64 bit integers for each 32 bit integer.
The chance of a random 64 bit integer being a 32 bit integer is 0.0000000233 %
The chance of a random 64 bit integer being a product of two 32 bit integers is 17%
Nice
The chance of a random 64 bit integer being a 32 bit integer is 0.0000000233 %
The chance of a random 64 bit integer being a product of two 32 bit integers is 17%
Nice
Therefore the fact that relatively few 64-bit numbers are products of 32-bit integers means that a lot of pairs of 32-bit integers give by multiplication the same product.
X = ab and aY < 2^32 and bY < 2^32:
X × Y = X/a × aY = X/b × bY = Y × X = aY × X/a = bY × X/b
Which is 6 pairs resulting in the same product. This will be reduced if e.g. aY = X, but still...
This is more than just the prime numbers. For example, a 41-bit prime can be multiplied by 16 and it will still fit into 64 bits.
If you say that you want to be doing modular arithmetic instead of arithmetic, it doesn't look like 2 and 3 are enough. You're looking for a solution to
2ᵃ * 3ᵇ ≡ n (mod 2⁶⁴)
3ᵇ ≡ n (mod 2⁶⁴)
If I'm reading wikipedia correctly, we know that this will fail for some n:
https://en.wikipedia.org/wiki/Primitive_root_modulo_n
> In symbols, g is a primitive root modulo n if for every integer a coprime to n, there is some integer k for which gᵏ ≡ a (mod n).
This is what we want, with g = 3, k = b, a = n₁, and n₂ = 2⁶⁴. Our restriction that n₁ (our n) is odd satisfies the requirement that a be coprime to n₂ (wikipedia's n, the modulus).
The article continues:
> a primitive root exists modulo n if and only if n is 4, pᵏ or 2pᵏ for some odd prime number p and some k ≥ 0.
2⁶⁴ does not satisfy this requirement and therefore there is no primitive root modulo 2⁶⁴. As such, 3 is not a primitive root modulo 2⁶⁴.
I did say things would get complicated.