Question

If n € N such that -

$4^n + 2^n + 1$ is a prime then prove that $n = 3^m$
for some m € N ​

Answer 1

Question :-

If n € N such that -

$4^n + 2^n + 1$ is a prime then prove that $n = 3^m$

for some m € N

Answer:-

If nn is not 3m3m, then n=3mrn=3mr for some integer rr, r>1r>1, gcd(r,3)=1gcd(r,3)=1.

If nn is not 3m3m, then n=3mrn=3mr for some integer rr, r>1r>1, gcd(r,3)=1gcd(r,3)=1.(2n−1)(4n+2n+1)=23n−1=23m+1r−1=(23m+1−1)q

If nn is not 3m3m, then n=3mrn=3mr for some integer rr, r>1r>1, gcd(r,3)=1gcd(r,3)=1.(2n−1)(4n+2n+1)=23n−1=23m+1r−1=(23m+1−1)q(2n−1)(4n+2n+1)=23n−1=23m+1r−1=(23m+1−1)q

for some integer q>1q>1. Now

for some integer q>1q>1. Nowgcd(2a−1,2b−1)=2c−1

for some integer q>1q>1. Nowgcd(2a−1,2b−1)=2c−1gcd(2a−1,2b−1)=2c−1

where c=gcd(a,b)c=gcd(a,b), so

where c=gcd(a,b)c=gcd(a,b), sogcd(2n−1,23m+1−1)=23m−1

where c=gcd(a,b)c=gcd(a,b), sogcd(2n−1,23m+1−1)=23m−1gcd(2n−1,23m+1−1)=23m−1

Hence,

gcd(23m+1−1,4n+2n+1)>1

gcd(23m+1−1,4n+2n+1)>1gcd(23m+1−1,4n+2n+1)>1

gcd(23m+1−1,4n+2n+1)>1gcd(23m+1−1,4n+2n+1)>1But 23m+1−1<4n+2n+123m+1−1<4n+2n+1, so

gcd(23m+1−1,4n+2n+1)>1gcd(23m+1−1,4n+2n+1)>1But 23m+1−1<4n+2n+123m+1−1<4n+2n+1, sogcd(23m+1−1,4n+2n+1)<4n+2n+1

gcd(23m+1−1,4n+2n+1)>1gcd(23m+1−1,4n+2n+1)>1But 23m+1−1<4n+2n+123m+1−1<4n+2n+1, sogcd(23m+1−1,4n+2n+1)<4n+2n+1gcd(23m+1−1,4n+2n+1)<4n+2n+1

gcd(23m+1−1,4n+2n+1)>1gcd(23m+1−1,4n+2n+1)>1But 23m+1−1<4n+2n+123m+1−1<4n+2n+1, sogcd(23m+1−1,4n+2n+1)<4n+2n+1gcd(23m+1−1,4n+2n+1)<4n+2n+1Therefore, 4n+2n+14n+2n+1 can't be prime.

Step-by-step explanation:

hopes it helps...

Answer 2

In Case n= 1 is Not Important.

$let \: n ≥2$

Let us Assume that

$n = {3}^{k}m$

And

$3≠m$ , $m≥2$

and Let

$p_{q}(x) ={x}^{2q} + {x}^{q} + 1∈ \: Z$

So,

$p_{n}x = { ({x}^{3k}) }^{2m} + { ({x}^{3k}) }^{m} + 1 = p_{m} {(x)}^{3k}$

Now,

$p_{m}(j) = p_{m} {(j)}^{2} = {j}^{2m} + {j}^{m} + 1 = \frac{ {j}^{3m} - 1}{ {j}^{m} - 1} = 0$

Hence,

$p _{1}(x) |p _{m}(x) \: in \: Z[x]$

So,

$p_{1} {(2)}^{3k} |p_{m} {(2)}^{3k} =p_{n}(2) = {4}^{n} + {2}^{n} + 1$

And

$1 < p_{1} {(2)}^{3k} < {4}^{n} + {2}^{n} + 1$

$So, {4}^{n} + {2}^{n} + 1 \: isn't \: prime$