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!
$\displaystyle \displaystyle 8n^3 + 12n^2 + 6n + 1 = 8n^3 + 1 + 12n^2 + 6n$ ........now factor by grouping
$\displaystyle \displaystyle = (2n + 1)(4n^2 - 2n + 1) + 6n(2n + 1)$ ................the first pair is the sum of two cubes...
$\displaystyle \displaystyle = (2n + 1)(4n^2 + 4n + 1)$
$\displaystyle \displaystyle = (2n + 1)^3$
Now what can you say? We can factor this, if we assume it is prime, what must happen? Could it work?
Thanks for the quick reply
We can disprove the statement, i.e. prove that $\displaystyle \displaystyle 8n^3 + 12n^2 + 6n + 1 = (2n + 1)^3$ 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?