Prove by induction that $21 | 4^{n+1} + 5^{2n-1}$

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The problem that I have is: Prove by induction that 21 divides 4ⁿ⁺¹ + 5^2n-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

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2 Answers

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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.

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Consider this:

$$4^{n+2} + 5^{2n+1} = 4\cdot 4^{n+1} + (21+4)\cdot 5^{2n-1} = 21\cdot5^{2n-1} + 4\cdot (4^{n+1}+5^{2n-1})$$

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