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I would appreciate if somebody could help me with the following problem:

Q: Show that for any integer $n\geq 1$, all the numbers $(3 n + 1)^5 + 5$ are composite (i.e. not prime).

I expand the formula $$(3 n + 1)^5 + 5=243 n^5+405 n^4+270 n^3+90 n^2+15 n+6$$ and .....

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    $\begingroup$ … can you find a common factor of the coefficients? $\endgroup$ – Daniel Fischer Mar 9 '16 at 11:18
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    $\begingroup$ It is readily seen that all such numbers are divisible by $3$. $\endgroup$ – Crostul Mar 9 '16 at 11:19
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Another way to solve this (besides expanding) is noting that

$$ (3n+1)^5 + 5 \equiv 1^5 + 5 \equiv 0 \mod 3$$

so $(3n+1)^5+5$ is divisible by 3. Since $(3n+1)^5+5 >3$, it is composite.

In fact for all positive integers $k$, $(3n+1)^k + 5$ is divisible by 3 by the same reasoning. (And thus is also not composite)

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