Prove by induction: $3^{(4n+2)} + 1$ is divisible by $10$.
My basic step: $3^{(4n+2)} + 1$, where $n = 1$ gives me $3^6 + 1 = 730$, which is divisible by $10$. However, then I have to do the induction hypothesis and I am kind of stuck because I do not have an equality. How do I finish proving this by induction?
Many thanks.
Edit: I am thinking of creating a formula which involves $10n$? Would this be correct?
Let $f(n): 3^{4n+2}+1$ be divisible by $10$
Clearly, $f(n)$ holds true for $n=1$
Let $f(n)$ holds true for $n=m$ i.e., $ 3^{4m+2}+1\equiv0\pmod{10}$
Now, $\displaystyle 3^{4(m+1)+2}+1=3^{4m+2}\cdot3^4+1\equiv 3^{4m+2}+1\pmod{10}$ as $3^4\equiv1\pmod {10}$
But by inductive hypothesis, $3^{4m+2}+1$ be divisible by $10$
$f(n): 3^{4n+2}+1$
STEP-$1$:
$f(1): 3^{4+2}+1 = 730$, which is divisible by $10$. Hence $f(1)$ holds true.
STEP-$2$:
Now let $n=k$, i.e., $f(n)=f(k)$ hold true .Hence, $f(k) = 3^{4k+2}+1$ is divisible by $10$.
Now we just need to prove that the criteria is satisfied for $n=k+1$.
STEP-$3$:
$$f(k+1) = 3^{4(k+1)+2}+1$$ $$ = 3^{4k+2}.3^{4}+1$$ $$ = 3^{4k+2}.(80+1)+1$$ $$ = (3^{4k+2}.80+3^{4k+2}.1)+1$$ $$ = 3^{4k+2}.80+(3^{4k+2}.1+1)$$
The first term is clearly divisible by 10. The second and third term are together divisible by 10 (from our assumption in step-2). So $f(n)=f(k+1)$ holds true.
Hence by induction $3^{4n+2}+1$ be divisible by $10$.
$N=3^{4n+2}+1=81^n\cdot9+1$. Using Newton's Binomial Theorem for $81^n=(80+1)^n$ we notice that it is of the form $100k+1$, meaning $N=900k+10=10\cdot(90k+1)$. Similar questions have been asked here and here.
Suppose $3^{4k+2} +1$ is divided by 10, then we need to show $3^{4(k+1)+2} +1$ is also divided by 10. Note that:
\begin{align*} \ 3^{4(k+1)+2} +1 &= 3^{4k+2+4}+(81-80) \\ &= 81\cdot3^{4k+2} + 81 - 80 \\ &= 81\cdot(3^{4k+2}+1) - 80 \\ &= 81\cdot10m -80\ldots\ldots (\text{where}~3^{4k+2}+1 = 10m ~~\text{for some integer}~m) \\ &= 10(81m-8) \end{align*}
Hence, $3^{4(k+1)+2} +1$ is also divided by 10, which complete the proof.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ 3^{4\pars{n + 1} + 2} + 1 = 3^{4n + 2}\overbrace{3^{4}}^{=\ 81} + 1 = 3^{4n + 2}80 + \pars{3^{4n + 2} + 1} $$
