# Fermat Little Theorem [closed]

The problem I am trying to solve is how to use Fermat Little theorem to prove that the number 66013 is not prime. I found this problem on another website (Question Cove). The student who had to solve this problem didn't know how to proceed. I offered to help. I read the following lecture notes Princeton lecture notes on cryptography and thought I would follow the same procedure. I wrote $66013 = 257^2 + 36 = 256^2 + 513 + 36 = 2^{16} + 2^9 + 37$ but I am stuck after that. What is the correct approach?

## closed as off-topic by Adrian Keister, Mostafa Ayaz, Isaac Browne, José Carlos Santos, NamasteJul 31 '18 at 0:06

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Fermat's little theorem is an incredibly inefficient method for proving primality. In this case the standard test for divisibility by 3 (by adding the digits and checking whether the sum is a multiple of 3) shows that $66213$ is a multiple of $3$.

• 66213 = 9* 7357 – Thierry Kauffmann Jul 29 '18 at 20:27
• Yes, indeed the quotient is easy to find. I was just pointing out that 66213 falls to the very simplest rule of thumb for checking primality. – Rob Arthan Jul 29 '18 at 20:31

Pick $a:=3$. We have $$a^{66212}=3^{66212}\not\equiv1\pmod{66213}$$ because $3\mid 66213$. Well, there, I used Fermat's Little Theorem!

Edit: Originally, the OP's question was to disprove the primality of $66213$.

For the corrected question (with number $66013$), choose $a:=15781$. Then, $$a^2=249039961\equiv 38925\pmod{66013}\,.$$ That is, $$a^4\equiv 38925^2=1515155625\equiv25249\pmod{66013}\,.$$ Finally, $$a^5=a\cdot a^4\equiv 15781\cdot 25249=398454469\equiv 1\pmod{66013}\,.$$ That is, $$a^{66013-1}=a^2\,\left(a^5\right)^{13202} \equiv a^2\cdot 1^{13202}\equiv a^2\equiv 38925\pmod{66013}\,.$$ Thus, $66013$ is not prime.

• If $66213$ were prime, any $a$ between $1$ to $66212$ could be used. The point is, because $a=3$ couldn't be used, $66213$ is not prime. – Batominovski Jul 29 '18 at 20:20
• But you can't claim $3|66213$ unless you know that $66213$ is not prime in the first place. This argument is circular. – fleablood Jul 29 '18 at 20:21
• No, I didn't have to use circular logic to verify that $3\mid 66213$. I can just divide it. This problem is a joke. – Batominovski Jul 29 '18 at 20:23
• I agree with Batominovski, it is a form of torture to set students tasks like this. And to placate Anurag A and fleablood, we are entitled to assume for a contradiction that $a = 3$ and $66213$ are relatively prime (since if they aren't then the prime $3$ divides $66213$). – Rob Arthan Jul 29 '18 at 20:25
• @Batominovski May I ask how you picked a:=15781.? – Thierry Kauffmann Jul 30 '18 at 13:07