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I've recently encountered this question: \begin{align} & \text{Prove that there is no positive integer $k$ such} \\ & \text{that } \forall x\in \text{ positive integers, } x^2+5x+k\text{ is prime}. \end{align} After playing around with it for a while, I decided to use the quadratic formula to solve and found some intricacies, namely, the discriminant (where $k$ is located) can only have values where $k < 6$; any value greater than that will not yield a real solution.

As it turns out, only $k = 4$ has an integer solution to $x$ and it is a negative number. At this point, I'm quite lost as to how this would be utilized to prove the statement and whether this would be a valid proof. Any ideas?

Thank you

Brandon

Michael Hardy
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Brandon S
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    Take $x=k$ (for $k>1$). – player3236 Oct 22 '20 at 18:06
  • @Äres Certainly you mean "for $k=1$, take $x=2$" – player3236 Oct 22 '20 at 18:09
  • @player3236 $15$ is not prime, for $x=k=1$ we have $7$. – Alessio K Oct 22 '20 at 18:10
  • I suggest proving a stronger statement: there is no non-constant polynomial with integer coefficients that returns only primes for natural number arguments. If the constant term is $\neq \pm 1$ this is obvious (since $k,|,f(nk)$ for any integer $n$). It's a little harder to cover the cases where the constant term is $\pm 1$ but you can dodge that in your case (just handle $k=1$ separately). – lulu Oct 22 '20 at 18:11
  • @Äres We want "not prime". – player3236 Oct 22 '20 at 18:11
  • @player3236 $x^2+5x+k$ is prime. – Alessio K Oct 22 '20 at 18:12
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    @Äres $\neg \exists k \forall x (x^2+5x+k \text{ prime}) \equiv \forall k \exists x \neg( x^2+5x+k \text{ prime})$ – player3236 Oct 22 '20 at 18:14
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    It seems from the comments that there is some confusion as regards the question. I think it is clear that it means: "show that, for all $k\in \mathbb N$, there exists some $n\in \mathbb N$ such that $f(n)$ is not prime." – lulu Oct 22 '20 at 18:15
  • But what is with discriminant? We are not solving for roots of the equation. $x$ is already a positive integer and so is $k$. The equation will never get to zero. – Math Lover Oct 22 '20 at 18:18
  • @lulu note the original problem https://math.stackexchange.com/questions/289338/is-the-notorious-n2-n-41-prime-generator-the-last-of-its-type/289357#289357 which happens if we just take $x=t-2$ and replca $k$ by $k-6$ – Will Jagy Oct 22 '20 at 18:18
  • @lulu Regarding the stronger result, I'm pretty sure I've seen it somewhere before (especially the case $k=\pm1$) Where can I find the result? It seems that it is due to Goldbach/Legendre. – player3236 Oct 22 '20 at 18:20
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    @player3236 it suffices to say that the values are eventually larger than $1;$ and strictly increasing. if we have $f(n_1) = p$ is prime, then $f(n_1 + p) $ is divisble by $p$ therefore composite – Will Jagy Oct 22 '20 at 18:29
  • This is true for any polynomial. – rtybase Oct 22 '20 at 18:38

2 Answers2

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Note that

$$x^2+5x+k=(x-1)(x+6)+(k+6)$$

Taking $x=(k+6)n+1$ for any $n$, we have $x-1=(k+6)n$, in which case

$$x^2+5x+k=(k+6)(n(x+6)+1)$$

which is a factorization of $x^2+5x+k$, with each factor greater than $1$ if $n\gt0$.

Barry Cipra
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Observe that $f(x+1)-f(x)=2x+6$, in other words it is strictly even.

First, we suppose that $k=2j$ i.e. even. Then, immediately we have $f(1)=2j+6$ (since $f(0)=2j$) which is composite and hence contradicts the original claim.

Now suppose $k=2j+1$ i.e. odd. Then, $f(1)=2j+7, f(2)=2j+15, f(3)=2j+25$.

See if you can spot why one of these must be a multiple of $3$. Once you've shown this, it contradicts the claim for all odd $k$ that $f(x)$ is prime $\forall x$, and so combining the two parts you have a proof that there is no such $k\in \Bbb N$

Rhys Hughes
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