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A sequence is defined by the nth term: $$n^3 -21n^2 +99n +121$$

What primes does the sequence contain if continued to infinity?

(Question given by maths teacher in stretch and challenge workshop)

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  • $\begingroup$ Think of some way to factorize the expression $\endgroup$
    – PNT
    May 7 at 20:32
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    $\begingroup$ $n^3 - 21n^2 + 99n + 121 = (n+1)(n-11)^2$ $\endgroup$
    – Alan
    May 7 at 20:33

1 Answer 1

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Note that $$a_n=n^3−21n^2+99n+121=\left(n+1\right)\left(n-11\right)^2$$ And for $n\ge 13$ both $n+1$ and $(n-11)^2$ are $>1 $ so $a_n$ is not a prime for $n\ge 13$.

Now you just want to check $n\le 12$ and see what primes the sequence gives.

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