Prove that following is always divisble by 5

Question

I am not getting how to prove that $1+ 2^{2^{4n-2}}$ is always divisible by 5 for every natural $n > 2$.

Answer 1

By Fermat's Little Theorem $2^4\equiv 1 \pmod 5$ and so $2^{4m}=(2^4)^m\equiv 1^m\equiv 1 \pmod 5$.

If $4n-2 \ge 2$ (that is if $n \ge 1$) then $2^{4n-2}$ is divisible by $4$. So $2^{2^{4n-1}}\equiv 1\pmod 5$ and $1 + 2^{2^{4n-1}}\equiv 2 \pmod 5$ and so this is never divisible by $5$ (If $n\ge 1$).

I think maybe you meant $2^{2^{4n-2}} - 1$ is always divisible by $5$, which is true as $2^{2^{4n-2}} - 1 \equiv 1-1 \equiv 0 \pmod 5$.

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An alternative proof of the latter is

$2^{4k} - 1 = (2^2 + 1)(2^{4k-2} - 2^{4k - 4} + 2^{4k -6}...... +2^6-2^4+2^2 - 1)$

So $2^2 + 1 =5$ will divide and $2^{4k} - 1$ (e.g. $16-1, 256-1, 4096-1$). And $2^{4n-2} = 4\cdot 2^{4n-4}$.

Answer 2

It's not true, since for $n=1$:

$$1 + 2^{2^2}=1+2^4=1+16=17 \equiv 2 \pmod{5}$$

Edit: I don't think it's true for $n>2$. Correct me if I'm wrong, but if for $n$ we have $2^{2^{4n-2}} \equiv-1 \pmod{5}$, then for $n+1$ we have $2^{2^{4(n + 1)-2}} = \left(2^{2^{4n-2}}\right)^{16} \equiv (-1)^{16} \equiv 1 \pmod{5}$

Answer 3

Nope, it's not true for any natural number $n$. Sorry.

For all natural numbers $n$ from $1$ onwards, $2^{4n-2}\ge 4$ and therefore this must have the form $4m$ for some natural number $m$. We then have

$2^{4m}=(2^m)^4\equiv 1\bmod 5$ by Fermat's Little Theorem.

We should have made the constant term $-1$ instead of $+1$. Watch your signs!