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 ?

**$$ \color{red} {2^{12}=4096=21*195+1}$$
$$2^{12n+4}+5\equiv\\ 2^4 .2^{12n}+5 \equiv\\16.4096^n+5  \equiv\\16 (4095+1)^n+5\equiv\\16.(21.195+1)^n+5 \equiv \\ 16(1)^n+5 \equiv \\21\equiv0$$ so $$21|2^{12n+4}+5\\n=0,1,2,3,4,5,...$$
