2
$\begingroup$

I'm having difficulty proving this statement.

For every positive integer $n$, $n^2-n+11$ is a prime

So I know the obvious, that a prime can only be divided by $1$ and itself, but I'm not sure how to turn that into some generic equation to compare to the one mentioned in the question. Could someone give me a start?

$\endgroup$
  • 6
    $\begingroup$ Hint: consider the case of $n=11$. $\endgroup$ – Regret Mar 10 '15 at 20:17
  • 1
    $\begingroup$ Ah, so a counter example to disprove it? $\endgroup$ – ADH Mar 10 '15 at 20:18
  • 1
    $\begingroup$ Yes, precisely! $\endgroup$ – Regret Mar 10 '15 at 20:18
  • 1
    $\begingroup$ Duh, thanks a lot! $\endgroup$ – ADH Mar 10 '15 at 20:18
  • 2
    $\begingroup$ @Gregory: What about the constant function $2$? $\endgroup$ – Regret Mar 10 '15 at 20:26
3
$\begingroup$

Hint $\ f(11n)\, =\, 11\,(11n^2\!-n+1)\,$ always prime implies $\,11n^2\!-n+1 = 1 $ for all $\,n>0,\,$ contra a nonzero quadratic has at most $2$ roots.

Remark $\ $ I presented the proof this way because exactly the same proof shows that any nonconstant polynomials cannot produce only primes. You may find it much more instructive to prove this more general statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.