TITLE: Prove by induction that $21 | 4^{n+1} + 5^{2n-1}$
QUESTION [5 upvotes]: The problem that I have is: Prove by induction that 21 divides 4n+1 + 52n-1
So far I have:
Base Case:
$n = 1$
$4^{1+1} + 5^{2-1} = 4^2 + 5^1 = 16 + 5 = 21$
Inductive Step:
Assume: $4^{k+1} + 5^{2k-1} = 3m$
$4^{(k+1)+1} + 5^{2(k+1)-1} = 3m$
$4^{k+2} + 5^{2k+1} = 3m$
I'm pretty sure I am far off on this one and not sure where to go. 
Thanks

REPLY [5 votes]: You can proceed with the inductive step as follows:
Assume that $$21 \mid 4^{k+1}+5^{2k-1}$$which implies 
$$21\mid 25(4^{k+1}+5^{2k-1})$$ $$\Longrightarrow21\mid 25(4^{k+1})+5^{2k+1}$$
We can subtract a multiple of $21$ on the right side to obtain
$$21\mid 25(4^{k+1})+5^{2k+1}-21(4^{k+1})$$
$$\Longrightarrow 21\mid4(4^{k+1})+5^{2k+1}$$
$$\Longrightarrow 21\mid4^{k+2}+5^{2k+1}$$
$$\Longrightarrow 21\mid4^{(k+1)+1}+5^{2(k+1)-1}$$
This completes the proof.