Cool Problem (BOMC-2)

[sourav das]

(Bulgarian Math Olympiad 1981, #4) Prove that, if $1 + 2^n + 4^n$ is prime, then $n = 3^k$ for some integer $k$.
Hint

A little known trick just kill the problem

[FahimFerdous]

Asolei josh!

[sourav das]

Show your solution, (Feel really cool when you solve that.... )

[sourav das]

Fahim has solved it in same way as mine. Too bad he couldn't post it from his mobile. Here is the sketch of his solution

We can write it in form of $1+x^n+x^{2n}$. Now remind that $1+\omega + \omega ^2 =0$, So if some $k$ not equal to 3 divide $n$ then write it in form of $1+y^k+y^{2k}$ and show that $1+y+y^2$ divides $1+y^k+y^{2k}$

[Nadim Ul Abrar]

hehehehehehe

[itsnafi]

@ Nadim. Check out your solution.What is the case of $n=3l$ but $3l\neq 3^k$ ?

[*Mahi*]

@ Nadim. Check out your solution.What is the case of $n=3l$ but $3l\neq 3^k$ ?

Then prove $l$ is also of the form $3i$

[Nadim Ul Abrar]

if $n\neq 3^k$
then $n$ will be of form $3l+1$,$3p+2$,$3^m(3l+1)$or $3^m(3p+2)$ ; $(l+1,p \geq0)$

Case 1 : $n$ has the form $3l+1$

$1+2^n+4^n \equiv 1+2+4 \equiv 0 (mod7)$

Case 2 : $n$ has the form $3p+2$

$1+2^n+4^n \equiv 1+4+16 \equiv 0 (mod7)$

Case 3 : $n$ has the form $3^m(3l+1)$

$1+2^n+4^n \equiv 1+2^{3l+1}+4^{3l+1} \equiv 1+2+4 \equiv 0 (mod7)$

Case4 : $n$ has the form $3^m(3p+2)$

$1+2^n+4^n \equiv 1+2^{3p+2}+4^{3p+2} \equiv 1+4+16 \equiv 0 (mod7)$

[*Mahi*]

Show your solution, (Feel really cool when you solve that.... )

Complex numbers and number theory

[Nadim Ul Abrar]

@ Nadim. Check out your solution.What is the case of $n=3l$ but $3l\neq 3^k$ ?

Then prove $l$ is also of the form $3i$

hehehe that was an excito .

I became so much excited when i found $(mod7)$ .. hehe .. sorry