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Suppose I had a statement that said For all positive integers of n, ${n^2 + 21n + 1}$ is prime.

Attempt:

The first thing that I decided to do was to try and factor it. I immediately saw that it wasn't going to happen. This meant that the GCF for each term was one.

I then checked using the discriminant method and my result was ${\sqrt{437}}$ Though 437 is not prime since it is divisible by 19 and 23

My question is, is this enough to say that for all integers of this polynomial, ${n^2 + 21n + 1}$ is prime. If not, how else can I efficiently check to see that their exists at least one case that makes this not true.

Martin Sleziak
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Deathslice
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    Can you use Wolfram Alpha? If so, try $n = 18$. – JimmyK4542 Jun 17 '15 at 05:10
  • Also, the 2nd sentence here might be helpful. – JimmyK4542 Jun 17 '15 at 05:13
  • I would not expect any closed form expression to always generate a prime. I think you are mistaken about factoring. If the quadratic is factorable then certainly there exists integers which will give a non-prime (i.e. a composite number), however, just because it's not factorable, doesn't mean it cannot produce composite numbers. I mean, every quadratic is "factorable" if we're allowed non-integer factors--your reasoning would suggest that this cannot create an integer value (which it obviously can). – Jared Jun 17 '15 at 05:18
  • @Jared Well I didn't really make any final conclusions in regards to whether this polynomial was always prime. That is why I did 2 steps and then asked if this was enough and if not, what other steps could i employ to prove that this proposition was true or false. – Deathslice Jun 17 '15 at 05:23
  • As far as I know polynom can not generate primes (for all arguments, of course) – Michael Galuza Jun 17 '15 at 05:24
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    Yes, no single variable polynomial with integer coefficients can generate primes. See http://math.stackexchange.com/questions/304330/proof-of-lack-of-pure-prime-producing-polynomials – Shailesh Jun 17 '15 at 05:29
  • @MichaelGaluza: but a polynomial in more variables can, if you just take th positive values it achieves. See https://en.wikipedia.org/wiki/Formula_for_primes#Formula_based_on_a_system_of_Diophantine_equations – Ross Millikan Jun 17 '15 at 05:31

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When $n=1$, our polynomial is $23$. Now evaluate it at $n=1+23$. From "failure," success!

Remark: The same basic idea can be used to show that no non-constant polynomial $P(n)$ with integer coefficients can be prime for all natural numbers $n$.

André Nicolas
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  • Well besides the rudimentary plug and chug, what other techniques can I employ? – Deathslice Jun 17 '15 at 05:19
  • Didn't plug and chug. After noting that $P(1)=23$, I immediately went to $n=1+23$. – André Nicolas Jun 17 '15 at 05:21
  • Oh I see. Thanks for the heads up. – Deathslice Jun 17 '15 at 05:26
  • @Deathslice I will add that you don't even have to evaluate. Since $24\equiv1\pmod{23}$, you also have $P(24)\equiv P(1)\pmod{23}$ for any polynomial with integer coefficients. In this case you get $P(24)\equiv0\pmod{23}$, which means that $23\mid P(24)$. (And it is easy to see that $P(24)>23$.) – Martin Sleziak Jun 17 '15 at 05:58
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I would say $n^2+21n+1=(n+1)^2+19n$, so if $n+1$ has a common factor with $19$, the expression will be divisible by $19$. In fact, $18^2+21\cdot 18+1=703=19\cdot 37$

Ross Millikan
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André Nicholas illustrates a general observation which is easy and which works more generally.

If you wanted to continue your own observation further, one way would be:

$$f(n)=n^2+21n+1$$

$$4f(n)=(2n+21)^2-437$$

So $f(n)$ would have a factor $437$ whenever $2n+21$ is a multiple of $437$.

And since $437=441-4=21^2-2^2=19\times 23$, $f(n)$ will have a factor $19$ whenever $2n+21$ is divisible by $19$, and $23$ whenever $2n+21$ is divisible by 23.

You just need to confirm that the relevant multiple is greater than $1\times 437, 19, 23$ most of the time.

Mark Bennet
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