1. ## prime question

The question is does n^2 +2 or n^2 + 3n +2 give infinitely many primes....

I just want to see if my answer is sufficient....

1st equation yields the following results.
n=0, yeilds 2
n=1, yields 3
n=2, yields 6
n=3, yields 11
n=4, yields 18
n=5, yields 27

Therefore, the equation n^2 + 2 does not yield infinitely many primes....

Same for the 2nd equation results are as follows....
2
6
12
20
30
42

Therefore the equation n^2 + 3n +2 doesn't yield infinitely many primes...

Is this sufficient?

2. Forget the post i made before, i misinterpreted the question

I should take TPH's advice and not drink and derive

3. Hello, duggaboy!

Does $n^2 +2\,\text{ or }\,n^2 + 3n +2$ give infinitely many primes?

I just want to see if my answer is sufficient....

1st equation yields the following results.
$n=0$ yields 2.
$n=1$ yields 3
$n=2$ yields 6 . . . . You can stop here

Therefore, $n^2 + 2$ does not yield infinitely many primes.

Same for the 2nd equation, results are as follows:
2
6 . . . . You can stop here

Therefore, $n^2 + 3n +2$ does not yield infinitely many primes.
I would have explained them like this . . .

For $n^2 + 2$

If $n$ is even, $n = 2k$, then we have:
. . $n^2 + 2 \:=\:(2k)^2 + 2 \:=\:4k^2 + 2 \:=\:2(2k^2+1)$ . . . an even number

Therefore, $n^2+2$ does not generate infinite primes.

For $n^2 + 3n + 2$

Note that it factors: . $n^2 + 3n + 2 \:=\:(n+1)(n+2)$

Therefore, for $n \geq 1$, it never produces a prime.

4. ## Awesome

HI.....

Thank you, your expansion makes more sense..Thank you again!!

5. I don't think duggaboy's answers were sufficient. your's are. at least for the second one. for the first, i have a question
Originally Posted by Soroban

For $n^2 + 2$

If $n$ is even, $n = 2k$, then we have:
. . $n^2 + 2 \:=\2k)^2 + 2 \:=\:4k^2 + 2 \:=\:2(2k^2+1)" alt="n^2 + 2 \:=\2k)^2 + 2 \:=\:4k^2 + 2 \:=\:2(2k^2+1)" /> . . . an even number

Therefore, $n^2+2$ does not generate infinite primes.
how do we know it doesn't generate infinitely many primes for odd n's?

6. Originally Posted by Soroban
Hello, duggaboy!

I would have explained them like this . . .

For $n^2 + 2$

If $n$ is even, $n = 2k$, then we have:
. . $n^2 + 2 \:=\2k)^2 + 2 \:=\:4k^2 + 2 \:=\:2(2k^2+1)" alt="n^2 + 2 \:=\2k)^2 + 2 \:=\:4k^2 + 2 \:=\:2(2k^2+1)" /> . . . an even number

Therefore, $n^2+2$ does not generate infinite primes.

For $n^2 + 3n + 2$

Note that it factors: . $n^2 + 3n + 2 \:=\n+1)(n+2)" alt="n^2 + 3n + 2 \:=\n+1)(n+2)" />

Therefore, for $n \geq 1$, it never produces a prime.
Hmmmm ..... Just because n^2 + 2 produces some numbers that aren't prime, that's not a proof that the equation does not produce infinitely many primes. All it proves is that the equation does not always produce a prime ......

To show that n^2 + 2 does not produce an infinite number of primes, it's needed to show that it produces only a finite number of primes. Soroban has done half the work, since if n is even you always get an even number => no primes.

So you need to show that when n is odd, n = 2m + 1 say, there are only a finite number of primes. In other words there are NOT an infinite number of primes of the form (2m + 1)^2 + 2 = 4m^2 + 4m + 3 ......

7. Originally Posted by mr fantastic
Hmmmm ..... Just because n^2 + 2 produces some numbers that aren't prime, that's not a proof that the equation does not produce infinitely many primes. All it proves is that the equation does not always produce a prime ......

To show that n^2 + 2 does not produce an infinite number of primes, it's needed to show that it produces only a finite number of primes. Soroban has done half the work, since if n is even you always get an even number => no primes.

So you need to show that when n is odd, n = 2m + 1 say, there are only a finite number of primes. In other words there are NOT an infinite number of primes of the form (2m + 1)^2 + 2 = 4m^2 + 4m + 3 ......
my point exactly...but so much more articulate

8. Oh okay, so adress the point that it does produce finite many primes just not infinitely many primes?
That does look more complete but then what arguing form would I use? n^2 +2 only produces finite many primes and n^2 +3n +2 produces only finite many primes...
It makes sense that we can't show it producing infinitely many primes so therefore, it would produce finite many....I'm just not strong enough in the "proofs" yet to conclude this properly...

Thank you so much for input and direction!!

9. I think the problem of whether there are infinitely many prime of form n^2+1 is an unsolved problem, I would assume we have a similar situation with n^2+2.

10. but couldn't you say that the equation doesn't produce infinitely many primes only finite many primes? since it doesn't work for every number?

11. Originally Posted by duggaboy
but couldn't you say that the equation doesn't produce infinitely many primes only finite many primes? since it doesn't work for every number?
No. It can happen that it alternates. The first one is prime the second is not, the third is prime, the fourth is not. And so on. It can happen the first 1000 are not prime but after that all are prime. And in all these cases you still have infinitely many of them. So it is not good enough to say that is does not produce primes for some of them.