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Prime odd problem

Prove that for any integer \(a>2\), there exists a prime number p such that p divides \(a^3+1\) but not \(a+1\).

Note by Lawrence Bush
2 years, 6 months ago

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I've managed to reduce down the problem to showing that the solutions of \(a^2-a+1 = 3^k\) in integers satisfy \(a \le 2\), but can't proceed anywhere afterwards.

Note that since \(a^3+1 = (a+1)(a^2-a+1)\), the prime number we're looking for cannot divide \(a+1\) and hence must divide \(a^2-a+1\). Thus it suffices to find a prime number \(p\) that divides \(a^2-a+1\) but not \(a+1\).

However, \(\gcd(a^2-a+1, a+1) = \gcd(3, a+1)\), thus it suffices to find some prime factor \(p\) of \(a^2-a+1\) that is not \(3\). If this is found, \(p\) cannot divide \(a+1\), since if it does then \(p\) divides their GCD, but the GCD is either \(1\) or \(3\), impossible to be divisible by \(p\) whatever it is.

Also note that \(a^2-a+1 > 1\) for \(a > 2\), so it suffices to show that \(a^2-a+1\) is not in the form \(3^k\) for \(a > 2\), or in other words the only solutions of \(a^2-a+1 = 3^k\) in integers satisfy \(a \le 2\).

Ivan Koswara - 2 years, 6 months ago

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Another hint: Use the quadratic formula.

Joel Tan - 2 years, 6 months ago

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Great! You are 1 step away from a complete solution.

Hopefully that is a sufficient hint (and you shouldn't look down further).

If not,

Hint: If \( \gcd (3, a+1) = 3 \), what does that tell us about \(a\)?

Calvin Lin Staff - 2 years, 6 months ago

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Yes.Corrected it.

Lawrence Bush - 2 years, 6 months ago

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Ah. That makes more sense.

I've updated it to "any integer \( a > 2 \)" for clarity.

This is an interesting problem.

Calvin Lin Staff - 2 years, 6 months ago

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