Prove by induction that $3^{4n + 2} + 1$ is divisible by $10$ 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?
 A: 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 $10$
A: $N=3^{4n+2}+1=81^n\cdot9+1$. 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.
A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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$$
3^{4\pars{n + 1} + 2} + 1 = 3^{4n + 2}\overbrace{3^{4}}^{=\ 81} + 1
=
3^{4n + 2}80 + \pars{3^{4n + 2} + 1}
$$
A: $f(n): 3^{4n+2}+1$ 
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 $10$.
