Let k be a positive integer. Show that the function $a^5+a$ takes on every residue class mod $5^k$ for any $0 \le b \lt 5^k$, there exists an integer $a$ such that $a^5+a \equiv b$ mod $5^k$.
I used induction but stuck on the inductive step.
 A: If you are familiar with Hensel lifting, you will note that the solution below is a special case of lifting.
It is easy to verify that for any fixed integer $b$, the congruence $x^5+x\equiv b\pmod{5}$ has a solution. Suppose now that there is a solution modulo $5^k$. We show there is a solution modulo $5^{k+1}$.
Suppose that $a^5+a\equiv b\pmod{5^k}$. We produce a solution modulo $5^{k+1}$. Let $a^5+a=b+q5^k$.
We look for a solution of shape $a+t5^k$. So we want
$$(a+t5^k)^5+a+t5^k  \equiv b\pmod{5^{k+1}}.$$
Expand the left-hand side, and note that $(a+t5^k)^5\equiv a^5\pmod{5^{k+1}}$.
So we want
$$a^5+a+t5^k\equiv b\pmod{5^{k+1}},$$
or equivalently
$$q5^k+t5^k\equiv 0\pmod{5^{k+1}},$$
or equivalently
$$q+t\equiv 0\pmod{5}.$$ 
The above congruence has a solution. This completes the induction step.
Remark: Since we now know that we can take $t=-q$, we can avoid much of the calculation and just write down the lifted solution. But I wanted to give a proof that mirrors the general lifting argument.
