Some numbers p have the property that multiplying two consecutive integers and then adding p generates only prime number outputs, so long as both consecutive integers are less than p in size. (By symmetry of multiplication under negation, it suffices to consider just consecutive natural numbers less than p)

In particular, by considering the case of zero times one, we need p itself to be prime (unless p is one or smaller, a trivial loophole in the wording above).

When \(p = 2\), we satisfy the condition, but this can be seen as another kind of too small triviality. Beyond this, any prime \(p\) must be odd, and as we will see below, this brings us into a realm of nicer theory.

The relevant range of this monotonic polynomial \(G(n) = n(n + 1) + p = n^2 + n + p\) then stretches from \(G(0) = p\) up through \(G(p - 2)\), after which we hit \(G(p - 1) = p^2\). In particular, within the relevant range, all the outputs have size less than \(p^2\); if any value of such size had any nontrivial factor, it would have a nontrivial factor less than \(\sqrt{p^2} = p\).

Thus, to establish that \(p\) has the relevant property, it suffices to establish that \(G(n)\) is never divisible by any prime less than \(p\) [if it were divisible by such a value at any point, it would have to be divisible by such a value at some point in the relevant range, by the fact that G(n) mod x = G(n mod x) mod x].

So long as \(p\) is odd, we will always have that \(G(n) = n(n + 1) + p\) is odd as well [the product of two consecutive values is always even], ensuring that \(G(n)\) is never divisible by \(2\). So what remains is to ensure that \(G(n)\) is never divisible by any odd prime \(q < p\).

For this consideration, we can apply the quadratic formula, to see that \(n^2 + n + p = 0\) has solutions in mod \(q\) land just in case there is a square root of the discriminant \(1 - 4p\) in mod \(q\) land.

TODO: Finish writing this out.

It turns out this all also happens just in case the ring of integers over \(Q(\sqrt{1 - 4p})\) has unique factorization (i.e., class number one). And it turns out the largest \(p\) with this property is \(41\) (though this was only proven in the mid-20th century, and thus is probably a difficult result).

TODO: Speak of j-invariant as well.