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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 .

Vinyl_cape_jawa
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Majd
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  • the hint here is to let $2^{4^n}+5=21k$ for some $k\in\Bbb Z$. now, rearrange the expression to get $2^{4^n}$ as the subject. substitute it into the inductive step argument. what do you get? – thesmallprint Nov 04 '19 at 21:34
  • $2^{4^{n+1}}=(2^{4^n})^4\equiv(-5)^4\equiv-5\mod21$ – J. W. Tanner Nov 04 '19 at 22:19

4 Answers4

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).

Milten
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  • how you got that $2^{{4}^{n}}+5 = 21k$ ? and what is k ? and why we use this method here ? – Majd Nov 04 '19 at 21:39
  • $2^{4^n}+5$ is divisible by $21$, which is the same as saying that there exists a $k$ so $2^{4^n}+5=21k$. This method is exactly equivalent to the other answers. My reason for doing it is that to get from $n$ to $n+1$ is the same as raising $2^{4^n}$ to the fourth power. Therefore I would like to have everything else on the other side of the equation, so that I can raise everything to the fourth power. – Milten Nov 04 '19 at 21:59
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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

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

user
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1

$2^{4^n}+5$ 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.

lhf
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0

In $\mod (25)$

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

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

Mohammad Riazi-Kermani
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