Recently, I've got stuck at this problem
Find natural nonzero number $a$ knowing that it exists $n$ such as $a^{n+1} + 2^{n+1} + 1$ is divisible with $a^n + 2^n + 1$
All I was thinking was Fermat's theorem and I have to admit that I don't know how to approach this type of problem.
A hint/an advice would be great. Thanks!
Step 0: check manually $a=1$ and $a=2$ and show that it never works. Now we may safely assume that $a\ge3$.
OK, so the greater number should be a multiple of the smaller one. Multiply the smaller number by $a$; that would be another multiple of it. The difference of these two multiples must be also a multiple. BTW, it equals $$a\cdot(a^n + 2^n + 1)-(a^{n+1} + 2^{n+1} + 1)=2^n(a-2)+a-1$$
Now, if $n$ is at least 3, then $$a^n=a\cdot a^{n-1}=a\cdot2^{n-1}\cdot\left({a\over2}\right)^{n-1}\ge a\cdot2^{n-1}\cdot\left({3\over2}\right)^2=a\cdot2^{n-1}\cdot\left(2+{1\over4}\right)\ge2^n\cdot a+a$$ therefore the thing above is smaller than $a^n+2^n+1$, and hence cannot be divisible by it.
Now let's check $n=2$. We need $5a-9$ to be divisible by $a^2+5$, which obviously never happens for the same reason.
What remains is $n=1$, and it proves fruitful (well, to some extent). Plugging $n=1$ into the original expression, we see that $a^2+5$ must be divisible by $a+3$. But wait, $a^2-9=(a-3)(a+3)$ is definitely divisible by it, and so must be the difference between these two, which is 14. Now, what are the divisors of 14?