Science:Math Exam Resources/Courses/MATH312/December 2009/Question 02 (a)

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MATH312 December 2009

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Question 02 (a)
Determine whether 209 passes Miller's test to the base 2.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that 2 passes Miller's test if $\displaystyle 2^{d}\equiv 1\mod {209}$ or the following holds for some r $\displaystyle 2^{2^{r}d}\equiv -1\mod {209}$ where $\displaystyle 209-1=208=2^{s}d$ and $\displaystyle 0\leq r\leq s-1$. If 2 passes the test, then 209 is a probable prime. If it fails the test, then the number is not prime.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Recall that 2 passes Miller's test if $\displaystyle 2^{d}\equiv 1\mod {209}$ or for some value r $\displaystyle 2^{2^{r}d}\equiv -1\mod {209}$ where $\displaystyle 209-1=208=2^{s}d$ and $\displaystyle 0\leq r\leq s-1$. If 2 fails the test, then 209 is not prime. For us, we see that $\displaystyle 208=2^{3}(26)=2^{4}\cdot 13$ and so d is 13 and r is 4. We compute manually. $\displaystyle 2^{13}\equiv 2^{8}\cdot 2^{5}\equiv (256)(32)\equiv (47)(32)\equiv 1504\equiv 41\not \equiv \pm 1\mod {209}$ and for the powers of 2, $\displaystyle 2^{2\cdot 13}\equiv 41^{2}\equiv 1681\equiv 9\not \equiv \pm 1\mod {209}$ $\displaystyle 2^{4\cdot 13}\equiv 9^{2}\equiv 81\not \equiv \pm 1\mod {209}$ $\displaystyle 2^{8\cdot 13}\equiv 81^{2}\equiv (2\cdot 41-1)^{2}\equiv 4\cdot 41^{2}-4\cdot 41+1\equiv 4\cdot 9-164+1\equiv 36+45+1\equiv 82\not \equiv \pm 1\mod {209}$ $\displaystyle 2^{16\cdot 13}\equiv 82^{2}\equiv (81+1)^{2}\equiv 81^{2}+162+1\equiv 82+162+1\equiv 36\not \equiv \pm 1\mod {209}$ and so the number 209 is not prime and 2 fails Miller's test. Thus 209 is composite. In fact $\displaystyle 209=11\cdot 19$.

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