<s>[INST] Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction 
$$ 21 \mid (2^{4n}+5) $$
So I assume:

$ 2^{4k}+5= 21p$

to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ = 

$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 16 \cdot 21p - 80+5 = 16 \cdot 21p - 75 $

But its not divisible by 21. Whats I doing wrong ?
 [/INST] for  $n=2$
$2^{4n}+5=256+5=261$
and
$261=21\times12+9$
so your identity is not true for $n=2$.
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