Does there exist a function that only generates numbers that are the product of at least 3 prime numbers, and $n$ of the form $m^2−1$?

Question

I posted this question an hour ago, but was a bit unclear, so I'll explain better what I'm looking for:

I want to know if there exist a $P(n)$ such that every number it generates, is the product of at least 3 prime numbers, and of the form $m^2-1$ with $m ∈ \mathbb N$;

Numbers it can generate:

$24$, because $24= 5^2-1$ and $24= 2*2*2*3$

$48$, because $48= 7^2-1$ and $24= 2*2*2*2*3$

$63$, because $63= 8^2-1$ and $24= 3*3*7$

Numbers it can't generate:

$3$, because $3= 2^2-1$, but $3=3$

$35$, because $35= 6^2-1$, but $35=5*7$

Thanks!

Answer 1

For a number to have at least 3 prime factors we can just consider numbers which have 3 factors. If the factors are not prime then they are the product of several prime numbers and so you are guaranteed to have at least 3 prime factors.

As shown in the comments this can be achieved in many ways. Without further details it is hard to tell what exactly you are after.

Possible answers (including those from the comments) are:

$$P(n)=24=2\cdot2\cdot2\cdot3=5^2-1$$ $$P(n)=4n(n+1)=2\cdot2\cdot n\cdot(n+1)=(2n+1)^2-1$$ $$P(n)=n(n+1)(n+2)(n+3)=(n^2+3n+1)^2−1$$ $$P(n)=3(2n+1)(6n+1)=(6n+2)^2-1 \text{ (this one has even $m$)}$$ $$P(n)=(2k-1)(2k+1)(4kn-2n+1)(4kn+2n+1)=4((4k^2-1)n+k)^2-1, k\in\mathbb{Z}$$ (this one also has even $m$ and will let you build as many $P(n)$ as you like)

Answer 2

What about $P(n)=n^8-1$? Then $P(n)=(n-1)(n+1)(n^2+1)(n^4+1)$ $\forall$ $n$. And $P(n)=(n^4)^2-1$ so it's of the form you want.