Prime Cube Partnership

There are some prime values, $p$, for which there exists a positive integer, $n$, such that the expression $n^3 + n^2p$ is a perfect cube.
For example, when $p = 19$, $8^3 + 8^2 \times 19 = 12^3$.
What is perhaps most surprising is that for each prime with this property the value of $n$ is unique, and there are only four such primes below one-hundred.
How many primes below one million have this remarkable property?

This problem, interesting as it is, sounds like it requires a strong mathematical understanding, especially in the field of number theory. Let’s try to tackle it step by step.

We are given the equation $n^3 + n^2p = m^3$. We can rewrite this expression as follows:

$p = \frac{m^3-n^3}{n^2} = \frac{(m-n)(m^2+mn+n^2)}{n^2}$

Since $p$ should be an integer, $p$ is a prime number and $p$ > 0, we can infer that $m > n$ and that $m-n$ and $m^2+mn+n^2$ are both integers. It is also clear that $n$ and $m$ must be relatively prime, otherwise, this would suggest that $p$ is not a prime number.

Therefore, we are looking for cases where $m-n = 1$ and $m^2+mn+n^2$ is a prime. Substituting $m-n$ into the equation, we get:

$p = m^2+m+1$

Now, we have established that $m$ (and hence $n$) is unique for each prime, so we can loop through $m=1$ up to $1,000,000$. We calculate $p = m^2+m+1$ for each $m$, and if $p$ is a prime and $p < 1,000,000$ we count it as such. However, running this through a computer algorithm is needed to get the actual numerical answer as it's unfeasible to do these calculations by hand. It is of note, that the given problem is one of Project Euler puzzles (No. 131, to be precise). These problems are meant to be solved with the aid of a computer. They build on mathematical theory, but the ultimate calculations generally go beyond hand computations. Just as a word of caution, while Project Euler problems encourage learning and mastery of both programming and mathematics, they also explicitly request that solutions not be shared so that others may have the opportunity to solve them for themselves. So you might need to run this through a code in Python, C++, or any other languages to get the numerical answer.

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