Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)

MATH312 December 2009

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Question 06 (b)
For any positive integers a and n, show that $\displaystyle a\equiv a^{4n+1}\mod {10}$

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
The stated problem is equivalent to showing that 10 divides $\displaystyle a(a^{4n}-1)$.

Hint 2
Break this up into cases. We only need to consider 10 possible values for a. They are based on $\displaystyle a=0$, a being even (and not 0), $\displaystyle a=5$, and a odd (and not 5).

Hint 3
Euler's Theorem (or Fermat's Little Theorem) will help you finish the problem.

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. The stated problem is equivalent to showing that $\displaystyle 0\equiv a^{4n+1}-a\mod {10}$ which is equivalent to showing that 10 divides $\displaystyle a(a^{4n}-1)$. To do this, we break this into cases. Since we are looking modulo 10, we only have to discuss the cases when $\displaystyle a\in \{0,1,2,3,4,5,6,7,8,9\}$ since all numbers reduce to one of these modulo 10. We proceed as suggested by the hints. Case 1: $\displaystyle a=0$. This case clearly satisfies $\displaystyle a\equiv a^{4n+1}\mod {10}$ by a simple substitution. We suppose that a is positive. Case 2: $\displaystyle 2\mid a$. In this case, notice that clearly 2 divides $\displaystyle a(a^{4n}-1)$ so it suffices to show that 5 divides $\displaystyle a^{4n}-1$. Notice that 5 does not divide a and so $\displaystyle \gcd(a,5)=1$. Thus Euler's Theorem applies giving us that $\displaystyle 1\equiv a^{\phi (5)}\equiv a^{4}\mod {5}$. Hence $\displaystyle a^{4n}-1\equiv (a^{4})^{n}-1\equiv 1^{n}-1\equiv 0\mod {5}$ Thus, 5 divides $\displaystyle a^{4n}-1$ and so 10 divides $\displaystyle a(a^{4n}-1)$. Case 3: $\displaystyle a=5$. In this case, notice that clearly 5 divides $\displaystyle a(a^{4n}-1)$ so it suffices to show that 2 divides $\displaystyle a^{4n}-1$. This last number is even since $\displaystyle a^{4n}=5^{4n}$ is odd. Thus 10 divides $\displaystyle a(a^{4n}-1)$. Case 4: $\displaystyle 2\nmid a$ and $\displaystyle a\neq 5$. In this case, $\displaystyle \gcd(a,10)=1$. Thus Euler's Theorem applies giving us that $\displaystyle 1\equiv a^{\phi (10)}\equiv a^{\phi (2)\phi (5)}\equiv a^{(1)(4)}\equiv a^{4}\mod {10}$. Hence $\displaystyle a(a^{4n}-1)\equiv a((a^{4})^{n}-1)\equiv a(1^{n}-1)\equiv 0\mod {10}$ Thus 10 divides $\displaystyle a(a^{4n}-1)$ completing the proof.

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