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The number of prime values of the polynomial $n^3 − 10n^2 − 84n + 840$ where $n$ is an integer is..?

How do we do this? Is there some sort of specific method or formula that I can learn, if yes what is it called?

E. Joseph
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user408113
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    Hint: $(n-10)(n^2-84)$ leaves darn few choices for $n$. No magic formula, just the need to think. Basically you must hope that something like this factorization exists, or a congruence argument proves that the values always have a certain factor. Otherwise a problem of this type will be a nightmare. – Jyrki Lahtonen Jan 26 '17 at 14:05
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    See also this answer to "Polynomials representing primes". – Bill Dubuque Jan 26 '17 at 14:16

1 Answers1

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Taking Jyrki's hint:

$P(n) := n^3 − 10n^2 − 84n + 840=(n-10)(n^2-84)$

For $P(n)$ to be prime, therefore, we need the factors to be both positive or both negative, and one of them to be equal to $\pm1$. $|n^2-84|$ is never equal to one, so we only need to consider $n=9$ and $n=11$. $P(9)=3$ and $P(11)= 37$, so there are two values of $n$ with $P(n)$ prime.

Joffan
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