Prove that $2^{2k-1}+2^{k}+1$ is not divisible by $7$ for any $k$ natural number

Question

I am trying to prove this, but I really can't seem to get anywhere with it.. I tried transforming this into something else, but no transformation yields in any useful expression whatsoever.. As always, I'm searching only for a little hint, just to start going, I dont want a solution posted here. I will post a solution after I get it so that someone else can find it is a full ;) Thanks.

Answer 1

As Daniel said, double, which does not change the question of divisibility by 7. And $$ x^2 + 2 x + 2 $$ is never divisible by 7 for integer $x.$ So, what is $x?$

Answer 2

You can actually say more: The number is not divisible by any prime of the form $4n+3$.

Here is the proof; Let $$ x=2^k $$ Then the expression is $$ h=x^2/2 + x + 1 = \frac{x^2+2x+2}{2} = \frac{(x+1)^2+1}{2}$$

Now a prime $p$ divides $h$ if and only if $p$ divides $(x+1)^2+1$, only if $-1$ is a quadratic residue of $p$. Since $-1$ is not a quadratic residue of primes of the form $4n + 3$ such prime cannot divide $h$

Answer 3

There are many, many ways to approach this problem. Here's a hint for one approach: What can you say about the smallest possible counterexample $k$?

Answer 4

Hint $ $ Note $\, 7\mid 2^{2k-1}+ 2\,2^{k-1}+ 1\,\overset{\rm Lemma}\Rightarrow\, {\rm mod}\ 7\!:\ \color{#c00}{{-}1} \equiv \color{#0a0}{n^2}\,\overset{\rm cube}\Rightarrow -1\equiv n^6\, $ contra little Fermat.

Lemma $\ \ \ m\mid b^{2k-1} + 2\,b^{k-1} + a\,\ (=:\, c)\ $ implies that $\ \color{#c00}{1-ab}\ $ is a $\rm\color{#0a0}{square}$ mod $m$

Proof $\ {\rm mod}\ m\!:\,\ 0\equiv bc = b^{2k} + 2b^k +ab \equiv (b^k+1)^2 + ab-1\,\Rightarrow\, \color{#c00}{1-ab} \equiv \color{#0a0}{(b^k+1)^2}$