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If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.

I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.

I don't know how I could solve this, I don't have any ideas. I know that $4^n+2^n+1$ is obviously odd. Assume $m=0$, i.e. $n=1$. Then $3$ is prime. So that doesn't disprove what we're trying to show.

So assume $m>0$. Then $4^n+2^n+1>3$, so $4^n+2^n+1$ is either $3k+1$ or $3k+2$, where $k\in\mathbb N$. So either $4^n+2^n$ or $4^n+2^n-1$ is divisible by 3. I think it's not something that could help out at solving this. Just a little observation. I don't have any ideas.

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    $\begingroup$ Try to relate it to $2^{3n}-1$. $\endgroup$ Dec 28, 2013 at 0:55
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    $\begingroup$ Do you mean to relate $4^n+2^n+1$ to $2^{3n}-1$? How could they be related? I don't quite understand. $\endgroup$
    – user26486
    Dec 28, 2013 at 1:00
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    $\begingroup$ @mathh $4^n=(2^n)^2$. $\endgroup$ Dec 28, 2013 at 1:05
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    $\begingroup$ Do you know a formula for factoring the difference of two cubes? $\endgroup$ Dec 28, 2013 at 1:06
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    $\begingroup$ I do know it. $2^{3n}-1=(2^n-1)(2^{2n}+2^n+1)$. $\endgroup$
    – user26486
    Dec 28, 2013 at 1:11

4 Answers 4

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Hint. For prime $p\not=3$ we have $$(x^2+x+1)\mid (x^{2p}+x^p+1)$$ Proof.

\begin{align}x^{2p}+x^p+1&=\frac{x^{3p}-1}{x^p-1}\\&=\frac{(x^3-1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{x^p-1}\\&=\frac{(x^2+x+1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{(x^p-1)/(x-1)}\end{align} Now it suffices to prove \begin{align}\frac{x^p-1}{x-1}\big|\sum_{k=0}^{p-1}x^{3k}\end{align} Since $x^p\equiv 1\pmod{x^p-1}$, we see $x^{np+j}\equiv x^j\pmod{\frac{x^p-1}{x-1}}$. Now note that since $3\not\mid p$, for $k=0,1,\cdots,p-1$ the sequence $3k$ form a complete residue system of $p$. So \begin{align}\sum_{k=0}^{p-1}x^{3k}&\equiv\sum_{j=0}^{p-1}x^j\pmod{\frac{x^p-1}{x-1}}\\ & \equiv\frac{x^p-1}{x-1}\equiv 0\pmod{\frac{x^p-1}{x-1}}\end{align}

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    $\begingroup$ it is enough that, if $p$ is not divisible by $3,$ prime or not, $\omega = \frac{-1 + i \sqrt 3}{2}$ and $\omega^2 = \bar{\omega}$ are roots of $x^{2p} + x^p + 1,$ since $p \equiv 1 \pmod 3$ and $2p \equiv 2 \pmod 3,$ or else $p \equiv 2 \pmod 3$ and $2p \equiv 1 \pmod 3.$ $\endgroup$
    – Will Jagy
    Oct 15, 2016 at 17:57
  • $\begingroup$ This is a brilliant answer. Could you potentially prove that $3 \not\vert p$ implies that the sequence $3k$ for $k \in [0, p-1]$ forms a complete residue system? I think I'm being stupid but that doesn't come immediately to me. $\endgroup$
    – gowrath
    Oct 16, 2016 at 0:10
  • $\begingroup$ @WillJagy Thanks for your neat approach! $p$ is taken as prime just for convenient. In fact I never used its primeness in the proof. Anyway, your method is very elegant. $\endgroup$ Oct 16, 2016 at 3:22
  • $\begingroup$ @gowrath The sequence $3k~(k\in[0,p-1])$ are pairwise distinct mod $p$. To see this, check that $p\not|3(i-j)$ for $i,j\in[0,p-1],i\not=j$ since $\gcd(3,p)=1$ and $|i-j|<p$. Now, for $k\in[0,p-1]$ there are $p$ distinct numbers mod $p$, they must form a complete residue system. $\endgroup$ Oct 16, 2016 at 3:26
  • $\begingroup$ One last question. This proves that p must be a multiple of 3, but not that it must also be a power of 3. Or does that entail and I'm just not seeing it? (sorry I'm very very green in number theory). $\endgroup$
    – gowrath
    Oct 16, 2016 at 5:19
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If $n$ is not $3^m$, then $n=3^mr$ for some integer $r$, $r\gt1$, $\gcd(r,3)=1$.

$$(2^n-1)(4^n+2^n+1)=2^{3n}-1=2^{3^{m+1}r}-1=(2^{3^{m+1}}-1)q$$ for some integer $q\gt1$. Now $$\gcd(2^a-1,2^b-1)=2^c-1$$ where $c=\gcd(a,b)$, so $$\gcd(2^n-1,2^{3^{m+1}}-1)=2^{3^m}-1$$ Hence, $$\gcd(2^{3^{m+1}}-1,4^n+2^n+1)\gt1$$ But $2^{3^{m+1}}-1\lt4^n+2^n+1$, so $$\gcd(2^{3^{m+1}}-1,4^n+2^n+1)\lt4^n+2^n+1$$ Therefore, $4^n+2^n+1$ can't be prime.

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The case $n=1$ is trivial. Let $n\ge2$

By contradiction assume that $n=3^km$ where $3\not|m$, $m\ge2$ and let $$P_q(x)=x^{2q}+x^q+1\in\mathbb Z[x]$$ so $$P_n(x)=(x^{3^k})^{2m}+(x^{3^k})^{m}+1=P_m(x^{3^k})$$ We have $$P_m(j)=P_m(j^2)=j^{2m}+j^m+1=\frac{j^{3m}-1}{j^m-1}=0$$ hence $P_1(x)|P_m(x)$ in $\mathbb Z[x]$ so $$P_1(2^{3^k})|P_m(2^{3^k})=P_n(2)=4^n+2^n+1$$ and $$1<P_1(2^{3^k})<4^n+2^n+1$$ so $4^n+2^n+1$ isn't prime. Contradiction.

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  • $\begingroup$ $j=e^{2i\pi/3}$ is a cubic root of $1$ and this notation is standard. We have $j^2+j+1=0$. $\endgroup$
    – user63181
    Dec 28, 2013 at 2:42
  • $\begingroup$ This is simple just we have: $$j^3=\left(e^{2i\pi/3}\right)^3=e^{2i\pi}=1$$ $\endgroup$
    – user63181
    Dec 28, 2013 at 2:50
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Lemma: $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

Proof: Several proofs can be found here


$4^n+2^n+1 = \frac{8^n-1}{2^n-1}$. Let $n=3^t \times k$, where $k$ is not divisible by $3$. The numerator becomes $8^{3^t \times k}-1 = \left( {8^{3^t}} \right) ^k-1$, which has a factor $\left(8^{3^t}-1 \right)$.

The denominator becomes $2^{3^t \times k}-1$. By our lemma, $\gcd(8^{3^t}-1, 2^{3^t \times k}-1)$ $= \gcd(8^{3^t}-1, 8^{3^{t-1} \times k}-1)$ $ = 8^{\gcd(3^t, {3^{t-1} \times k})} - 1 = 8^{3^{t-1}}-1 = 2^{3^{t}}-1$, since $3 \nmid k$.

We now rewrite the fraction $\frac{8^n-1}{2^n-1}$as follows:

$$\frac{ \left( 2^{3^{t}}-1 \right) \left( {\left(2^{3^{t}}\right)}^2 + 2^{3^{t}} +1\right) \left( {\left(8^{3^{t}}\right)}^{k-1} + {\left(8^{3^{t}}\right)}^{k-2} + ... +1\right)}{\left( 2^{3^{t}}-1 \right) \left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right)}$$

$\gcd\left( \left( {\left(2^{3^{t}}\right)}^2 + 2^{3^{t}} +1\right),\left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right) \right) = 1$, so the only way the fraction cancels to a prime is when $\left( {\left(8^{3^{t}}\right)}^{k-1} + {\left(8^{3^{t}}\right)}^{k-2} + ... +1\right)$ $=\left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right)$. This only occurs when $k=1$. Hence, $n$ is of the form $3^t$.

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