<s>[INST] Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me?
 [/INST] $k$ is a huge even number, hence $k^2+2^k$ is for sure an even number. To compute the last digit of $k^2+2^k$, we just need to understand what $k^2+2^k\pmod{5}$ is. We have $k\equiv 0\pmod{4}$ and 
$$ k \equiv 3^2+4^{1004} \equiv -1+1\equiv 0\pmod{5}, $$
hence $k^2+2^k\equiv 0+1\equiv 1\pmod{5}$ by Fermat's little theorem, so the last digit of $k^2+2^k$ is $\color{red}{6}$.
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