I need help with this excercise.
prove that $2^{4n}+3n-1$ is divisible by 9 for all positive intergral values of n greater than 1.
I know that $n=k$: $2^{4k}+3k-1=9m$
then,
For $n=k+1$
$$2^{4(k+1)}+3(k+1)-1=2^{4}2^{4k}+3k+3-1$$ I don't know how to continue, maybe $$=(2^{4k}+3k-1)+(15 . 2^{4k}+3 )$$
Proving $2^{4n} + 3n \equiv 1 \pmod{9}$ is equivalent to your problem.
\begin{equation} 2^{4n} + 3n \equiv (-2)^n +3n \equiv 1 \pmod{9} \end{equation}
For $n \equiv 0 \pmod{3}, (-2)^n \equiv 1\pmod{9}$ and $3n \equiv 0\pmod{9}\Rightarrow (-2)^n +3n \equiv 1 \pmod{9}$
For $n \equiv 1 \pmod{3}, (-2)^n \equiv -2\pmod{9}$ and $3n \equiv 3\pmod{9}\Rightarrow (-2)^n +3n \equiv 1 \pmod{9}$
For $n \equiv 2 \pmod{3}, (-2)^n \equiv 4\pmod{9}$ and $3n \equiv -3\pmod{9}\Rightarrow (-2)^n +3n \equiv 1 \pmod{9}$
For all $n$, $(-2)^n +3n - 1 \equiv 0 \pmod{9} \Rightarrow 9 \mid 2^{4n} +3n - 1$
Everything $\pmod 9$
$$\begin{array} {c|c|c} n & 2^{4n}=7^n & 3n \\ \hline 1 & 7 & 3 \\ 2 & 4 & 6 \\ 3 & 1 & 0 \\ 4 & 7 & 3 \\ \end{array}$$
$$7 + 3 = 1$$ $$4 + 6 = 1$$ $$1 + 0 = 1$$
Other answers may be simpler, but I'll assume that you know no formal modular arithmetic and stick to proof by induction only.
We know that $2^{4n}+3n-1$ is a multiple of $9$ and
$$\begin{align}2^{4(n+1)}+3(n+1)-1-(2^{4n}+3n-1)&=16\times 2^{4n}+3n+3-1-(2^{4n}+3n-1)\\&=15\times 2^{4n}+3\\ 2^{4n+1}+3(n+1)+1&=(2^{4n}+3n-1)+(15\times 2^{4n}+3)\end{align}$$
so it follows that $2^{4(n+1)}+3(n+1)-1$ is a multiple of $9$ iff $15\times 2^{4n}+3$ is (for all $n$).
We can prove this new claim by induction: the inductive case is $$\begin{align}15\times 2^{4(n+1)}+3&=16\times 15\times 2^{4n}+3\\&=16(15\times 2^{4n}+3)-45\end{align}$$ and we know that both $15\times 2^{4n}+3$ and $45$ are multiples of $9$ (the former using the inductive hypothesis).
Now just the base case of $15\times 2^{4n}+3$ remains (what value of $n$ would that be?).
