Proof question: Prove that 2^(odd integer) + 5^(odd integer) + 2 is a multiple of 3, and 4^(any integer) + 1 can be expressed as 5n, 5n + 1 or 5n + 2. I had been working on the claim in the above question for sometime now. Statistically speaking, it works. For example:
$$\left(2^1 \right)+\left(5^1\right)+2=9,\\
\left(2^3\right)+\left(5^1\right)+2=15,\\ \ldots$$
It works with all the integers that I tested with. Can we give a simple proof for this one?
The second part is like this:
$$\left(4^1\right)+1=5,$$
$$\left(4^2\right)+1=17 = 15+2,$$
$$\left(4^3\right)+1=65,$$
$$\left(4^4\right)+1=257 = 255+2,\\ \ldots$$
So, the numbers are all expressed as $5n$, $5n+1$ or $5n+2$. Is there a generalised proof for this observation?
Thanks.

For questions of divisibility, it helps to think in terms of modular arithmetic. 
To show something is a multiple of $3$, we can check to see if it is $0 \pmod 3$:
$$2^{2m+1} + 5^{2n+1} + 2 \equiv -1 + 2 + 2 \pmod 3 \equiv 3 \pmod 3 \equiv 0 \pmod 3$$
Note: $5$ raised to any power always ends in $5$, and $5 \equiv 2 \pmod 3$. As well, $2 \equiv -1 \pmod 3$ and $-1$ raised to an odd power is always $-1$.
