Proof by induction that $3^{4n + 1} + 5^{2n + 1}$ is divisble by $8$ This is a homework problem:

Prove that: $$ 3^{4n+1} + 5^{2n+1}$$ is divisible by $8$ for every natural number $n$.

Base case: $$n = 0$$
$$ 3^{0 + 1} + 5^{0 + 1} = 8$$
$$8\bmod8 = 0 $$ Base case verified.
Inductive Hypothesis: Assume for some $n$, $$ 3^{4n+1} + 5^{2n+1} =8k$$ where $k$ is any non-negative integer.
Prove that: $$ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$$
Indeed: $$ 3^{4n+4+1} + 5^{2n+2+1} = 3^4\cdot3^{4n+1} + 5^2\cdot5^{2n+1}$$
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (80+1)\cdot3^{4n+1} + (24+1)\cdot5^{2n+1} $$
$$ = 80\cdot3^{4n+1} + 24\cdot5^{2n+1} + 3^{4n+1} + 5^{2n+1} $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1}) + 8k $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1} + k) $$
Is this wrong in any particular way? My answer is different from the textbook, but I feel like I sufficiently proved that $ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$.
Edit: What the book did:
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (56+25)\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25(3^{4n+1} + 5^{2n+1}) $$
$$ = 56\cdot3^{4n+1}+25(8k) $$
$$ = 8(7\cdot3^{4n+1}+25k) $$

The two methods differ only in how they rearrange the summands to show dvisibility by $8$
$$\begin{eqnarray} 81a+25b &=&\,80a+24b+\color{#c00}{a+b} &=&\, (10a+3b+\color{#c00}k)\,\color{#c00}8\ \ {\rm by}\ \ \color{#c00}{a+b = 8k} \\ 81a+25b &=&\, 56a + 25(\color{#c00}{a+b}) &=&\, (7a+25\color{#c00}k)\,\color{#c00}8\\
 \end{eqnarray}$$
More simply: $\ {\rm mod}\ 8\!:\ 81\equiv 1\equiv 25\ \Rightarrow\ 81a + 25b\equiv \color{#c00}{a+b \equiv 0}$
