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$S=1+x+x^2+x^3+...x^{n-1}=\dfrac{x^n-1}{x-1}$

Now I wanted to prove it using induction. Using induction for $n=1$, gives us $S=1$ which is prime. OK!

Now assume for $n=m$, it is true.

Then $\dfrac{x^n-1}{x-1}$ is prime $\implies m=prime$

Now if $m\ne2$ or $m\ne1$, then $\notin m+1$ such that $m+1$ is $ \space prime$

Thus w/o loss of generality, let $m+k$ where $k\in \mathbb{N}$ and $k<m$ such that $m+k$ is prime and also assume $m+k$ is the next prime after $m$

Now required to prove $\dfrac{x^{m+k}-1}{x-1}$ is prime

We see here that $\dfrac{x^{m+k}-1}{x-1}=1+x+x^2+.....x^{m+k-1}=1+x+x^2+...x^m+x^{m+1}+...x^{m+k-1}$

Now we know that $\underbrace{1+x+x^2+...x^{m-1}}_{\text{is prime by hypothesis}}+x^{m}+...x^{m+k-1}=\mathbb{P}+x^m(1+x+x^2+...x^{k-1})$

Now as $k<m$, we can say that $(1+x+x^2+...x^{k-1})$ is also prime if we use strong induction.

So now we get $\mathbb{P}+\underbrace{x^m\underbrace{(1+x+x^2+...x^{k-1})}_{\text{is prime by using strong induction}}}_{\text{is composite}}\implies \mathbb{P}+\text{a composite number}$ which is a possible candidate for prime.

But how can I say that if I add it, I must get a prime again?

Or Is the approach entirely wrong?Please suggest with reasoning.

Ok so by popular demand let us use contradiction here

Assume $n$ is not prime.

Then what to do next?

Saradamani
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    It is usually very hard to prove things about primes using induction. What I'd suggest to do here is to prove the contrapositive: assuming that $n$ is not a prime, prove that $1+x+...+x^n$ is not a prime. – idok Apr 29 '18 at 10:40
  • $1$ is not thought to be prime in most current definitions – Henry Apr 29 '18 at 10:42
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    $x^{ab}-1$ is divisible by $x^{a}-1$ and by $x^{b}-1$ – Henry Apr 29 '18 at 10:44
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    cyclotomic polynomials? – Angina Seng Apr 29 '18 at 10:51
  • @LordSharktheUnknown Yes you are right.https://en.wikipedia.org/wiki/Cyclotomic_polynomial – Saradamani Apr 29 '18 at 10:53
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    Hint: for all natural numbers $a$ and $b$ and for all real numbers $t$ it holds that, $$t^{ab-1}+t^{ab-2}+\ldots +t+1=\left(t^{a-1}+t^{a-2}+\ldots +t +1\right)\left(t^{(b-1)a}+t^{(b-2)a}+\ldots +t^a+1\right)_.$$ Set $n=ab$ and $x=t$. – Git Gud Apr 29 '18 at 10:59
  • @GitGud Is this a variation of Chebyshev inequality(looking like it)? – Saradamani Apr 29 '18 at 11:01
  • I have no idea what that is and this is an equality, it's really just simple arithmetical manipulation, there's nothing deep here, even if it's not obvious. – Git Gud Apr 29 '18 at 11:02

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