1. ## Need Help Please (Prime or Not Prime)

I need help solving the following question:

Prove or disprove the following:

There exists an integer n such that (8n^3 + 12n^2 + 6n + 1) is a prime number.

2. Originally Posted by mathnb
I need help solving the following question:

Prove or disprove the following:

There exists an integer n such that (8n^3 + 12n^2 + 6n + 1) is a prime number.
Hint: We can factor this!

$\displaystyle 8n^3 + 12n^2 + 6n + 1 = 8n^3 + 1 + 12n^2 + 6n$ ........now factor by grouping

$\displaystyle = (2n + 1)(4n^2 - 2n + 1) + 6n(2n + 1)$ ................the first pair is the sum of two cubes...

$\displaystyle = (2n + 1)(4n^2 + 4n + 1)$

$\displaystyle = (2n + 1)^3$

Now what can you say? We can factor this, if we assume it is prime, what must happen? Could it work?

3. Originally Posted by Jhevon
Hint: We can factor this!

$\displaystyle 8n^3 + 12n^2 + 6n + 1 = 8n^3 + 1 + 12n^2 + 6n$ ........now factor by grouping

$\displaystyle = (2n + 1)(4n^2 - 2n + 1) + 6n(2n + 1)$ ................the first pair is the sum of two cubes...

$\displaystyle = (2n + 1)(4n^2 + 4n + 1)$

$\displaystyle = (2n + 1)^3$

Now what can you say? We can factor this, if we assume it is prime, what must happen? Could it work?

We can disprove the statement, i.e. prove that $\displaystyle 8n^3 + 12n^2 + 6n + 1 = (2n + 1)^3$ is a composite number for all integers n.
But to prove that it is a non-prime number we must show that it can be written as a product of two integers (x*y, where both x and y are different from 1)?

4. Originally Posted by mathnb

We can disprove the statement, i.e. prove that $\displaystyle 8n^3 + 12n^2 + 6n + 1 = (2n + 1)^3$ is a composite number for all integers n.
But to prove that it is a non-prime number we must show that it can be written as a product of two integers (x*y, where both x and y are different from 1)?
You are thinking along the right lines, but we won't need to go in that direction. If you like the x*y thing, you can stop at the second to last line, set each factor equal to 1, and solve for n. None will work.

1 is not considered a prime number last I checked. And so, in light of this, no cube can be prime. How do we know this?

5. Got it. Thanks!