I recently stumbled across this question in a test.
Paul says that "$n^2+n+1$ is prime $\forall\:n\in \mathbb{N}$".
- Paul is correct, because…
- Paul is wrong, because…
The answers sheet says that "all answers providing a valid counterexample, e.g. $n=4$, are valid; however, I'm interested in understanding why – is there some sort of theorem which can predict whether a polynomial is prime or not?
What I tried was:
$n^2+n+1$ can be factorized as $(n+\frac{1+\sqrt{3}i}{2})(n+\frac{1-\sqrt{3}i}{2})$. Neither of these factors is a suitable factor (i.e. is natural, different from $1$, and different from $n^2+n+1$) for any $n\in\mathbb{N}$, therefore $n^2+n+1$ must be prime $\forall\:n\in\mathbb{N}$.
Instead, $n^2-1$ can be factorized as $(n+1)(n-1)$. The factor $(n+1)$ is a suitable factor (i.e. natural, different from $1$, and different from $n^2-1$) for $n>2$; therefore, $n^2-1$ is not prime for $n>2$.
However, the above reasoning is clearly wrong, because "$n^2+n+1$ is prime $\forall\:n\in\mathbb{N}$" doesn't hold for the case $n=4$.
What is wrong in the reasoning? Can one tell whether a polynomial $p(n)$ is prime for all $n$ a priori?
Best Answer
Hint $\ \ f(1)=3\,\Rightarrow\, 3\mid f(1\!+\!3n),\ $ e.g. $\,3\mid f(4)=21$
Remark $\ $ Above we used the Polynomial Congruence Theorem
$\ {\rm mod}\ 3\!:\ \color{}{1\!+\!3n\equiv 1}\,\Rightarrow\, f(\color{}{1\!+\!3n})\equiv f(\color{}1)\equiv 0,\ $ for all $\,f\in\Bbb Z[x]$
Alternatively $\,\ 3n\mid f(1\!+\!3n)-f(1)\ $ by the Factor Theorem $\,x-y\mid f(x)-f(y)$
The idea generalizes to a proof that any nonconstant polynomial $\in\Bbb Z[x]\,$ has a non-prime value.