Let $p>2$ are prime number and $2^p - 1 $ prime number. Prove that exists $n \in \mathbb N$ such that $n^2 + n + 1$ divisible by $2^p - 1$.
I need an elementary proof of this fact
Prove that $\exists n \in \mathbb N: 2^p - 1|n^2 + n + 1$
3
$\begingroup$
prime-numbers
integers
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1How elementary, exactly? Can we use the totient function? Fermat's little theorem? (Not sure if those are useful here, but what is the level?) – 2017-01-21
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0level is middle school – 2017-01-21
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0Can we use Fermat's little theorem – 2017-01-21
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0So no use of quadratic properties modulo primes ? – 2017-01-21
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0no use of quadratic properties modulo primes – 2017-01-21
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0Good luck. :) I'm not even sure I see a *non*-elementary proof of that fact. But then, I haven't gone as deep into number theory as I might have. – 2017-01-21
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0@Wildcard An elementary proof does exist, but given the nature of its depth, I am not sure if it can be converted to middle school elementary. – 2017-01-21
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0I ask not to close this problem. I need it very much! – 2017-01-21
1 Answers
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Let $q=2^p-1$. As $\Bbb Z/q\Bbb Z$ is a finite field, it's multiplicative group is cyclic of order $q-1$. Let $a+q\Bbb Z$ be a generator.
Since $p$ is odd, $q=2^p-1\equiv 1\pmod 3$, so $q=3m+1$ for some $m$.
Then $(a^m)^3=a^{3m}=a^{q-1}\equiv 1\pmod q$, but $a^m\not\equiv 1\pmod q$ (because $0 We conclude that for $n:=a^m$,
$$q\mid\frac{n^3-1}{n-1}=n^2+n+1.$$
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0Why $a^m \not \equiv 1 (\bmod q)$. If $m$ even and $a=q-1$ then $$a^m \equiv 1 (\bmod q)$$ – 2017-01-22