For every positive integer $n$, $n^2 + n +19$ is prime

Question

I'm trying to prove that for every positive integer $n, n^2 + n +19$ is prime.

I tried to disapprove it saying that is is not prime.

If it's not prime, then $n^2 + n +19$ has to have at least two factors which are greater than $1$. I'm stuck here.

I'm not sure if I'm going in the right direction.

If $n=19$, then $19\mid n^2+n+19 > 19$, so $n^2+n+19$ is composite if $n=19$. — user236182, Mar 17 '16 at 20:31
No polynomial with integer coefficients produces only primes. — Bruno Joyal, Mar 17 '16 at 20:33
Consider trying to prove that the all of the non-trivial zeros of the Riemann Zeta Function have real part equal to one half. This will help you with your understanding of the distribution of prime numbers. — Edward Evans, Mar 17 '16 at 20:34
Nooooooo, the correct theorem says that if for every positive integer $n$, $n^2 + n + 19$ is prime, then Santa exists. — Count Iblis, Mar 17 '16 at 20:41
@CameronWilliams It's very easy. Imitate the first commenter's idea! :) — Bruno Joyal, Mar 17 '16 at 20:41
I think you are trying to find $n\in \Bbb N$ for which $n^2+n+19$ is prime. Because, as @user236182, pointed out, it not true for a lot of positive integers. — , Mar 17 '16 at 20:43
If you expect something in the similar form, Euler found out that $n^2 + n + 41$ is prime for any $0 \leq n \leq 40$. — SiXUlm, Mar 17 '16 at 21:15

score 6 · Accepted Answer · answered Mar 17 '16 at 20:42

6

Let's try some values:

$n=1$ $$1^2+1+19=21=3*7$$

$n=2$ $$2^2+2+19=25=5*5$$

$n=5$ $$5^2+5+19=49=7*7$$

Even one counter example was enough to prove that the statement is false.

answered Mar 17 '16 at 20:42

Jens Renders

4,334

score 1 · Answer 2 · answered Mar 17 '16 at 21:31

Are you sure you didn't forget to mention some other condition that makes this a much more difficult problem, such as for example that $n \not\equiv 1 \pmod 3$?

If $n \equiv 1 \pmod 3$ then $n^2 \equiv 1 \pmod 3$, and since $19$ also satisfies that congruence, $n^2 + n + 19$ is a multiple of $3$.

Spend a few minutes looking at this page from FactorDB.com.

score 1 · Answer 3 · edited Apr 13 '17 at 12:19

1

I answered a similar question once.

"An Introduction To The Theory Of Numbers" by Hardy, Theorem 21, page 18.

THEOREM 21. No polynomial $f(n)$ with integral coefficients, not a constant, can be prime for all $n$, or for all sufficiently large $n$.

edited Apr 13 '17 at 12:19

Community

1

answered Mar 17 '16 at 21:55

rtybase

16,907

score 1 · Answer 4 · answered Mar 18 '16 at 16:08

This proposition is not true, and more generally it is not true if 19 is replaced by any positive integer $k$. For $k>1$ we can obtain a counterexample by putting $n=k-1$ so that: $$n^2 + n + k = (k-1)^2+(k-1)+k= (k^2-2k+1)+(k-1)+k=k^2$$ For $k=1$ a counterexample is $n=4$ implying $n^2+n+1 = 21 = 3*7$.

For every positive integer $n$, $n^2 + n +19$ is prime

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