After a little playing, and looking at the options, you might guess that the answer is likely to be a large number, and the only such option is (c). Looking at (c), it seems to be claiming that [MATH]3^{3^{333}}+1[/MATH] is a divisor of [MATH]3^{3^{334}}+1[/MATH]. Or maybe, in general, [MATH]3^{3^{n}}+1[/MATH] is a divisor of [MATH]3^{3^{n+1}}+1[/MATH]. Can we convince ourselves of this?
Again, playing with n = 0, 1, 2, you see that the claim is that [MATH]3^{1}+1[/MATH] is a divisor of [MATH]3^{3}+1[/MATH], [MATH]3^{3}+1[/MATH] is a divisor of [MATH]3^{9}+1[/MATH], and [MATH]3^{9}+1[/MATH] is a divisor of [MATH]3^{27}+1[/MATH].
In fact, that's true because, for example, [MATH]3^9+1 = (3^3+1)(3^6-3^3+1)[/MATH], which is how we factor a sum of cubes, and ... aha! Suddenly you can factor [MATH]3^{3^{n+1}}+1[/MATH], and one of the factors turns out to be ... [MATH]3^{3^{n}}+1[/MATH]!
There's the proof. Doing that last step will be good practice in algebra. Look up "sum of cubes" if you've forgotten it.
