# Math Help - False Existential

1. ## False Existential

Rusty returning student stuck on this one proof:

Prove false:
There exists an integer n such that $6n^2 + 27$ is prime.

I know to do this, I have to prove the universal negation true.
For all integers n such that $6n^2 + 27$ is not prime.

2. Originally Posted by MBoogie
Rusty returning student stuck on this one proof:

Prove false:
There exists an integer n such that $6n^2 + 27$ is prime.

I know to do this, I have to prove the universal negation true.
For all integers n such that $6n^2 + 27$ is not prime.
yes.

you can factor out a 3. what does that tell you?

3. oh no -

I had $3(2n^2 +9)$ written down so many times, and kept trying to wrong things with it. Is it as simple as this?:

$(2n^2 +9)$ is a sum of two squares which is prime. And because a composite is the product of two primes, $3(2n^2 +9)$ is composite?

4. Originally Posted by MBoogie
oh no -

I had $3(2n^2 +9)$ written down so many times, and kept trying to wrong things with it. Is it as simple as this?:

$(2n^2 +9)$ is a sum of two squares which is prime. And because a composite is the product of two primes, $3(2n^2 +9)$ is composite?
First $2n^2 + 9$ is not the sum of two squares. $2n^2$ is not a perfect square unless n = 0.

Second, $6n^2 + 27 = 3(2n^2 + 9)$

This means that, for any n, $6n^2 + 27$ is divisible by 3 and is thus a composite number. (A composite number is simply the product of two numbers, neither of which are 1, not necessarily two primes.)

-Dan

5. Originally Posted by MBoogie
oh no -

I had $3(2n^2 +9)$ written down so many times, and kept trying to wrong things with it. Is it as simple as this?:

$(2n^2 +9)$ is a sum of two squares which is prime. And because a composite is the product of two primes, $3(2n^2 +9)$ is composite?
it is, as topsquark says, even easier than that. you have an integer that is three times another integer. it means that your integer is divisible by 3 (an integer n is divisible by 3 if it can be written in the form $n = 3k$ for some integer $k$), and so is not prime.

6. Originally Posted by topsquark
This means that, for any n, $6n^2 + 27$ is divisible by 3 and is thus a composite number.
Originally Posted by Jhevon
it is, as topsquark says, even easier than that. you have an integer that is three times another integer. it means that your integer is divisible by 3 (an integer n is divisible by 3 if it can be written in the form $n = 3k$ for some integer $k$), and so is not prime.
Not true! 3 is divisible by 3 but 3 is not a composite number!

A positive integer divisible by 3 is not prime only if it’s not equal to 3 itself. It this particular problem, $6n^2+27$ is not prime because 3 divides it and $2n^2+9$ is not equal to 1 for any integer n.

7. Originally Posted by JaneBennet
...and $2n^2+9$ is not equal to 1 for any integer n.
of course. i should have seen that. thanks for the catch, Jane

that's a big "and," but i think i deserved it

8. wow - yeah I have a problem of getting ahead of myself. Answer was right in front of me the whole time. Thx Jhev and top

9. Originally Posted by MBoogie
wow - yeah I have a problem of getting ahead of myself. Answer was right in front of me the whole time. Thx Jhev and top
...and Jane

10. Originally Posted by Jhevon
of course. i should have seen that. thanks for the catch, Jane

that's a big "and," but i think i deserved it
I had to learn the hard way too. Once I wanted to prove that, while twin primes exist, you could never have “triple primes” (three consecutive odd primes). I thought I had cool proof in just one line: “Any sequence of three consecutive odd integers must contain a multiple of 3 and so that number in the sequence cannot be prime.” Then someone knocked me off my pedestal by showing me that one set of triple primes does exist: 3, 5, 7. I have learnt my lesson since then!