Prove by induction that $2^{4^n}+5$ is divisible by 21 .

Question

After I did the first two steps in the induction , I am stuck in the last step ; to prove for $n+1$ that $2^{4^n}+5$ is divisible by $21$ , so I know that $2^{4^n}+5$ is divisible by $21$ is true . I want to prove for $n+1$ ( $n$ is natural ): $$2^{4^{n+1}}+5=$$ $$=2^{4^{n}\cdot 4}+5=$$ $$=2^{4^{n}\cdot 4}+5=$$ $$=2^{{4}^{{4}^{n}}}+5=$$ $$=16^{{4}^{n}}+5.$$ this is the last step I reached , I would be happy to get some recommendation on how to continue from here .

Answer 1

Hint:

The statement is equivalent to $2^{4^n} = -5 + 21k$ for some $k\in\mathbb Z$. Then raise both sides to the $4$'th power. (Looks cleaner with modular arithmetic if you know it).

Answer 2

We have that $2^{4^n}+5\equiv 0 \mod 21 \implies 2^{4^n}\equiv -5 \mod 21$ then for the induction step use that

$$2^{4^{n+1}}+5\equiv(2^{4^{n}})^4+5 \equiv(-5)^4+5\equiv (25)^2+5 \equiv 16+5\equiv0 \mod 21$$

Answer 3

In $\mod (25)$

$$2^{4^n}\equiv -5 \implies 2^{4^{n+1}}\equiv (-5)^4= 625 $$

$$625+5 = 630 = 21(30)$$

Answer 4

$2^{4^n}+5$ is divisible by $21$ iff $2^{4^n} \equiv -5 \equiv 16 = 2^4 \bmod 21$.

Let $a_n = 2^{4^n}$. Then $a_{n+1} = a_n^4$.

Thus, to make the induction work, it is enough to prove the $(2^4)^4 \equiv 2^4 \bmod 21$, which is immediate.