<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] The statement is true for $n=1$. Suppose it holds for $k$: $2^{4k}+5=21p$, so $2^{4k}=21p-5$, so
$$
2^{4(k+1)}+5=2^{4k}\cdot 16+5=
16(21p-5)+5=21\cdot 16p-80+5=21\cdot16p-75
$$
Ops! Something seems wrong. We have proved that

if $21\mid 2^{4k}+5$, then $21\nmid 2^{4(k+1)}+5$

Actually, $3\mid 2^{4n}+5$ (prove it), but in general $7\nmid 2^{4n}+5$; indeed,
$$
2^{4n}+5=16^n+5\equiv 2^n+5\equiv2^n-2\equiv2(2^{n-1}-1)\pmod{7}
$$
Now
\begin{align}
2^0-1&\equiv 0\pmod{7}\\
2^1-1&\equiv 1\pmod{7}\\
2^2-1&\equiv 3\pmod{7}\\
2^3-1&\equiv 0\pmod{7}
\end{align}
so $2^{4n}+5$ is divisible by $7$ if and only if $3\mid(n-1)$, therefore when $n=3m+1$, for some $m$.
</s>