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 $a+1$ .

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Yes.Corrected it.

lawrence Bush - 6 years ago

Ah. That makes more sense.

I've updated it to "any integer $a > 2$ " for clarity.

This is an interesting problem.

Calvin Lin Staff - 6 years ago

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 - 6 years ago

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 - 6 years ago

Another hint: Use the quadratic formula.

Joel Tan - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$