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$3^{3n}-2^{3n}=27^n-8^n=(27-8)(27^{n-1}+27^{n-2}\cdot 8+...+27^1\cdot8^{n-2}+8^{n-1})$

If $n$ is even,

$3^{3n}+2^{3n}=27^n+8^n=(27+8)(27^{n-1}-27^{n-2}\cdot 8+...-27^1\cdot8^{n-2}+8^{n-1})$

If $n$ is even and a power of $2$, $3^{3n}+2^{3n}$ can't be factorized.

If $n$ is even, $n=m\cdot 2^k,m>1,k>0$ and $m$ is odd $\Rightarrow$ $$27^n+8^n=(27^{2^k}+8^{2^k})\sum_{i=1}^m 27^{(m-i)2^k}(-b^{2^k})^{i-1}$$

How to check divisibility using these cases?

Reference

user300048
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3 Answers3

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if n is odd $3^{3n} + 2^{3n} = (27+8)(27^{n-1} - 27^{n-2}8 +\cdots + 8^{n-1})$ and $35|(3^{3n} + 2^{3n})$

if $n$ is even, $n = 2k:$

$(3^{3n}-2^{3n}) = (3^{n}-2^{n})(3^{2n} + 3^n2^n + 2^{2n})$

$(3^{n}-2^{n}) = (3^{2k} - 2^{2k}) = (3^2 - 2^2)(3^{2k-1} +\cdots + 2^{2k-1})\\ 5|(3^{2k}-2^{2k})$

What is left? showing that $7|(3^{6k} - 2^{6k})$

Lets use Fermat's little theorem on this one when $p$ is prime and $p$ does not divide $a,$ $a^{p-1} \equiv 1\pmod p$

$(3^{6k} - 2^{6k}) \equiv 1-1 \equiv 0\pmod 7$

We probably should have brought in the modular arithmetic earlier, but you had already started by factoring.

Doug M
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Hint $\ {\rm mod}\ 35\!:\,\ 27\equiv -8\,\ $ so $\,\ \overbrace{27^{\large n}}^{\Large 3^{\Large 3n}}\equiv (-8)^{\large n}\equiv \pm\! \overbrace{8^{\large n}}^{\Large 2^{\Large 3n}}\,$ depending on parity of $n.\ $ QED.

Bill Dubuque
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HINT.-Modulo $5$ you have $$3^3\equiv 2\pmod5\\2^3\equiv-2\pmod5$$and módulo $7$ you have $$3^3\equiv -1\pmod7\\2^3\equiv1\pmod7$$ You can deduce easily the result for $3^{3n}\pm2^{3n}$

Piquito
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  • That's because $,3^{\large 3}\equiv -2^{\large 3}\pmod{35},,$ so there's not need to break it down modulo $5$ and $7$, e.g. see my answer. – Bill Dubuque Nov 23 '16 at 23:18
  • That's right but I used the "standard" way I mean when it is not apparent the divisibility by both factors. Thank you for your comment. Regards. – Piquito Nov 24 '16 at 11:05
  • The point is that here the divisibility is apparent mod the product $5\cdot 7$ since it is equivalent to $\ 27^n\equiv (-8)^n\equiv \pm 8^n\pmod{35}.,$ There is no need to prove it first mod $5$ and $7$ since it is already clear mod $35\ .$ – Bill Dubuque Nov 24 '16 at 14:55