## A Hilbert number is any positive integer of the form $4k+1$ for integer $k\geq 0$. We shall define a squarefree Hilbert number as a Hilbert number which is not divisible by the square of any Hilbert number other than one. For example, $117$ is a squarefree Hilbert number, equaling $9\times13$. However $6237$ is a Hilbert number that is not squarefree in this sense, as it is divisible by $9^2$. The number $3969$ is also not squarefree, as it is divisible by both $9^2$ and $21^2$.

There are $2327192$ squarefree Hilbert numbers below $10^7$.

How many squarefree Hilbert numbers are there below $10^{16}$?

### This mathematics problem is very advanced and requires understanding and application of number theory. The task is to calculate the number of squarefree Hilbert numbers less than 10^16. The problem specifies that a Hilbert number is a positive integer of the form 4k+1, meaning that the number should be one more than a multiple of 4.

It’s important to note that these numbers are the same as the numbers of the form 4n – 3 (since if n is an integer, then 4n – 3 = 4(n – 1) + 1, which is of the form 4k + 1)

One thing to note is that a number of the form 4n – 3 can only have prime factors also of the same form. This is because if a prime factor, p, of a number of the form 4n – 3 didn’t fall into the form 4n – 3 then it must fall into the form 4n – 1 or 4n + 1, the product of a number of form 4n – 3 and a number falling into one of the other forms would result in a remainder of 1 when divided by 4, meaning it would no longer be a Hilbert number.

However, calculating the total count of squarefree Hilbert numbers would be computationally expensive to directly calculate. Thus, this problem is not only a test of number theory understanding, but also a problem-solving and algorithm design challenge as well.

The squarefree criterion adds another layer of complexity. Instead of finding all Hilbert numbers, we now want numbers that cannot be written as the product of a square Hilbert number and some other number. This prevents us from simply squaring each prime of the form 4n – 3 and counting the numbers that are larger than the square and less than the maximum number of 10^16.

To solve this problem, it requires sophisticated computational techniques or deep mathematical insights to derive a closed form solution. These approaches go beyond the scope of high school mathematics and are typically found in advanced academic research in number theory.

Nevertheless, one potential approach involves enumerating all primes of the form 4n – 3 up to the square root of the maximum number (10^16), calculating the product of selections of these primes and ensuring each product is less than the max number and count the total number of these products. These would constitute our squarefree Hilbert numbers.

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