Very hard!
If for natural number n ,
(2 +√3)^n +(2-√3)^n +1 is prime,
prove that there exist non-negative integer m such that n = 3^m

Question

Very hard!
If for natural number n , 
(2 +√3)^n +(2-√3)^n +1  is prime,
prove that there exist non-negative integer m such that  n = 3^m

anonymous · Answer

I have no idea!

anonymous · Answer

is it like this?$$(2+\sqrt{3})^n+(2-\sqrt{3})^n+1$$? because if you plug in n = 2, that isnt prime, its 15.

anonymous · Answer

@joemath314159
no its not claiming that its true for all n, its saying IF its true, then prove the second statement

when you expand it, all the odd powers of sqrt(3) cancel

anonymous · Answer

ah, i see.

anonymous · Answer

try expanding it

anonymous · Answer

this is hard...

anonymous · Answer

Yeah, I would try expanding it and writing it in a simpler form.

mani_jha · Answer

If you use the binomial expansion of each term, you'll get in the end:
$$2^n+2^{n-2}\sqrt{3}^{2}+2^{n-4}\sqrt{3}^{4}....2\sqrt{3}^{n-1}$$+1
Note that n is an odd number if the condition is satisfied(it is a power of 3)
I think we shall be able to sum up this series, which is a mix of two G.P.s
Or we could also note that:
$$1/(2+\sqrt{3})=2-\sqrt{3}$$
Substitute this in the second term and then expand.

kinggeorge · Answer

Let $a_n=(2 +\sqrt3)^n +(2-\sqrt3)^n +1$

Without having proven anything, but using wolfram to factor some of the results, it looks like we want to prove that if $n$ is a product of at least two distinct primes such as $$n=p^{e}k$$where $p$ is prime, and $\gcd(p, k)=1$, then $a_n$ has a divisor of $a_{p}$.

Additionally, if $n=p^r$ where $p\neq3$ then $a_n$ has a divisor of $p$.

kinggeorge · Answer

I can prove it partially. Let $n=p$ where $p\neq3$. Then look at $$a_p\equiv(2 +\sqrt3)^p +(2-\sqrt3)^p+1 \pmod p$$Because of weirdness mod p, this equals$$a_p\equiv(2 +\sqrt3 +2-\sqrt3)^p +1 \pmod p$$$$a_p\equiv4^p +1 \pmod p$$$$a_p\equiv 4+1 \pmod p$$$$a_p\equiv 5 \pmod p$$This means that if $p=5$ it's obvious that 5 is a divisor of $a_5$. However, since it's equivalent to 5 for ALL primes p, and it has a divisor of 5 for $p=5$, it must be divisible by 5 for $p\neq5$

experimentx · Answer

*

anonymous · Answer

._. ill pray for u

anonymous · Answer

@Mani_Jha i think you forgot $$\left(\begin{matrix}n \ r\end{matrix}\right)$$when you used binomial theorem