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Please prove or disprove: If $n \in ℤ^{+}$, then $n^{2} + n + 41$ is prime.

I know that the above statement is not true because if you plug in 41 for $n$, the result is not a prime number. How can I write a formal proof (or disproof) showing that this is a false statement?

rtybase
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Joshua Vaughan
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    If I remember correctly, the statement is false, so you must show with a counterexample. – DonQuixote Sep 26 '15 at 21:02
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    You are already done. As DonQuixote said, you merely need to give a counterexample, which you have already done by letting $n=41$. – Tom Tseng Sep 26 '15 at 21:05
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    You have done it. Maybe you should indicate how you know that the number is not prime when $n=41$. Or maybe it is so obvious that it is not worth doing. By the way, the number is also composite at $n=40$, and is prime for $n=1$ to $39$. – André Nicolas Sep 26 '15 at 21:07

2 Answers2

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There is no polynomial with integer coefficients that is able to take only prime values over $\mathbb{Z}$ or $\mathbb{N}$.

The reason is simple: if $p(x)\in\mathbb{Z}[x]$, $a-b$ is a divisor of $p(a)-p(b)$ for every $a\neq b$ (that obviously holds for monomials, hence it holds for polynomials, too). Moreover, a polynomial cannot take the same value in more points than its degree. Assuming that $p(1)$ is some prime $q$, then $q$ divides $p(kq+1)-p(1)$, hence $p(kq+1)$, for every $k\in\mathbb{Z}^+$. If $n$ is the degree of $p(x)$, it follows that at least a number among $p(q+1),p(2q+1),\ldots,p((2n+1)q+1)$ cannot be a prime.

Jack D'Aurizio
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"An Introduction To The Theory Of Numbers" by Hardy (or http://matematica.cubaeduca.cu/medias/pdf/842.pdf), 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 $n$.

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