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

Skipping through the basis and onto the induction:

$4\cdot 4^{n+1}+5^2 \cdot 5^{2n-1}=21a $ for some integer $a$

The following steps were:

$4\cdot (21a-5^{2n-1})+ 125 \cdot5^{2n-1} \\= 84a-4 \cdot5^{2n-1} + 125 \cdot5^{2n-1} $

But I can't factor out a 21 from here... Any input?

Dimitri
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  • http://math.stackexchange.com/questions/227728/induction-prove-that-4n152-n-1-is-divisible-by-21-for-all-n-geq-1 and http://math.stackexchange.com/questions/460317/prove-that-5-is-factor-of-32n-2n1 – lab bhattacharjee Dec 07 '13 at 05:10

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$$4^{n+2}+5^{2n+1}=4\cdot 4^{n+1}+25\cdot 5^{2n-1}=4\cdot 4^{n+1}+(21+4)\cdot 5^{2n-1}$$

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

Could you continue from here?

Salech Alhasov
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    I reviewed my answer, but for some reason my head said that $5^2 = 125$ instead of $25$... By checking this, I got to the same place as you. Thank you.. – Dimitri Dec 07 '13 at 04:32