Science:Math Exam Resources/Courses/MATH312/December 2012/Question 02 (c)

MATH312 December 2012

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Question 02 (c)
State whether the following are true or false with full justification. Each question carries 4 marks. The number 25 passes Miller's test for the base 7.

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 7 passes Miller's test if $\displaystyle 7^{d}\equiv 1\mod {25}$ or the following holds for some r $\displaystyle 7^{2^{r}d}\equiv -1\mod {25}$ where $\displaystyle 25-1=24=2^{s}d$ for $s=3$ and $d=3$, and $\displaystyle 0\leq r\leq s-1$. If 7 passes the test, then 25 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

The answer is true. Recall that 7 passes Miller's test if $\displaystyle 7^{d}\equiv 1\mod {25}$ or the following holds for some r $\displaystyle 7^{2^{r}d}\equiv -1\mod {25}$ where $\displaystyle 25-1=24=2^{s}d=2^{3}\cdot 3$ and $\displaystyle 0\leq r\leq 3-1$. If 7 passes the test, then 25 is a probable prime. If it fails the test, then the number is not prime. So we check manually: $\displaystyle {\begin{aligned}7^{3}&\equiv 49\cdot 7\equiv -1\cdot 7\not \equiv 1\mod {25}\\7^{2^{1}\cdot 3}&\equiv 49^{3}\equiv (-1)^{3}\equiv -1\mod {25}\end{aligned}}$ Thus, 7 passes Miller's test. (Note that 25 is not prime but still 7 passes Miller's test).

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