# Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

Find the number of positive divisors of $$(2008^3 + (3 * 2008 * 2009) + 1)^2$$

What I Tried: I have no idea of an elegant solution to this. I cannot seem to guess or figure out the factors as there is a $$(+1)$$ added to it, which seems to bother me. As the number is a perfect square I can guess it will have even number of positive divisors or so.

I decided to solve it by hand. I calculated the expression to be $$(8108486729)^2$$ , and turns out , the prime factorization of $$8108486729$$ is amazingly $$7^6 * 41^3$$ .

My first question is, how is that factorization coming? It looks a bit like magic to me, can someone explain that?

So from here $$(8108486729)^2$$ will have prime factorization as $$7^{12} * 41^6$$ , so total number of factors will be $$(13)(7) = 91$$ , problem solved. This is however, not an elegant solution, and I can't seem to find any.

Can anyone help?

• Hint: $2009 = 7^2 \times 41$ Dec 23, 2020 at 7:30
• I figured that out, but what more can you say with the terms $2008^3$ and the $(+1)$ ? Dec 23, 2020 at 7:32

Hint: $${(2008)}^3+3\cdot 2008\cdot 2009+1={(2008+1)}^3={(2009)}^3$$

• Oh that's it, I feel pretty dumb. Dec 23, 2020 at 7:33

$$z^3+3z^2+3z+1=(z+1)^3$$

$$z=2008$$

$$z+1=2009$$

$$2009 = 7^2 * 41$$

$$2009^3 = 7^6 * 41^3$$

• This is more or less the same answer as the other one, I already figured it out. Dec 23, 2020 at 7:39
• @Anonymous maybe the OP was already typing out the answer while i posted so (s)he didnt see that i had posted before Dec 23, 2020 at 7:41