Science:Math Exam Resources/Courses/MATH437/December 2006/Question 08

MATH437 December 2006

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Question 08
Suppose that the positive integer n satisfies $\displaystyle 2^{n}\equiv -1\mod {n}$. Let p be the smallest prime divisor of n. Show that $\displaystyle p=3$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Firstly, note that $\displaystyle 2^{n}+1$ is odd for all values of n so the smallest prime factor must be odd.

Hint 2
Try using Fermat's Little Theorem.

Hint 3
Also, include some facts about the order of a number modulo a prime to take you home.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Note that $\displaystyle 2^{n}+1$ is odd for all values of n so the smallest prime factor must be odd. Then further, $\displaystyle 2^{p-1}\equiv 1\mod {p}$ by Fermat's Little Theorem. From the given, we know that $\displaystyle 2^{n}\equiv -1\mod {p}$ and thus $\displaystyle 2^{2n}\equiv (2^{n})^{2}\equiv (-1)^{2}\equiv 1\mod {p}$ Hence the order of 2 must be a divisor of $\displaystyle d=\gcd(2n,p-1)$ (you can see this by using the Euclidean algorithm to find an a and a b so that $\displaystyle a(p-1)+bn=d$ and then noting that $\displaystyle 2^{d}=2^{a(p-1)+bn}\equiv 1\mod {p}$). As p is the smallest prime factor of n and as noted it is odd, we see that $\displaystyle \gcd(2n,p-1)=2$. Since clearly $\displaystyle 2\equiv 1\mod {p}$ is false, we see that $\displaystyle 4=2^{2}\equiv 1\mod {p}$ which happens only when $\displaystyle p=3$ as required.

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