MATH312 December 2009
• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) •
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 $\displaystyle 2091=208=2^{s}d$ and $\displaystyle 0\leq r\leq s1$. If 2 passes the test, then 209 is a probable prime. If it fails the test, then the number is not prime.

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Solution

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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 $\displaystyle 2091=208=2^{s}d$ and $\displaystyle 0\leq r\leq s1$. 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,
$\displaystyle 2^{2\cdot 13}\equiv 41^{2}\equiv 1681\equiv 9\not \equiv \pm 1\mod {209}$
$\displaystyle 2^{4\cdot 13}\equiv 9^{2}\equiv 81\not \equiv \pm 1\mod {209}$
$\displaystyle 2^{8\cdot 13}\equiv 81^{2}\equiv (2\cdot 411)^{2}\equiv 4\cdot 41^{2}4\cdot 41+1\equiv 4\cdot 9164+1\equiv 36+45+1\equiv 82\not \equiv \pm 1\mod {209}$
$\displaystyle 2^{16\cdot 13}\equiv 82^{2}\equiv (81+1)^{2}\equiv 81^{2}+162+1\equiv 82+162+1\equiv 36\not \equiv \pm 1\mod {209}$
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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