Science:Math Exam Resources/Courses/MATH437/December 2011/Question 07 (ii)

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MATH437 December 2011

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Question 07 (ii)
Let $\displaystyle a\in \mathbb {Z}$ such that $\displaystyle a\equiv 5\mod {8}$ and let $\displaystyle \alpha \in \mathbb {N}$. (ii) For each odd integer x, prove that there exists a unique $\displaystyle i\in \{0,1\}$ and a unique $\displaystyle j\in \{0,1,2,...,2^{\alpha }-1\}$ such that $\displaystyle x\equiv (-1)^{i}a^{j}\mod {2^{\alpha +2}}$

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

Part (i) shows us that the order of a modulo $\displaystyle 2^{\alpha +2}$ is equal to $\displaystyle 2^{\alpha }$. Thus each of $\displaystyle a^{j}\mod {2^{\alpha +2}}$ are distinct odd numbers and thus each of the $\displaystyle (-1)a^{j}\mod {2^{\alpha +2}}$ are also distinct odd numbers. Hence it suffices to show that $\displaystyle (-1)a^{k}\not \equiv a^{j}\mod {2^{\alpha +2}}$ for any $\displaystyle 0\leq j,k\leq 2^{\alpha }-1$ and finish off the proof with a counting argument.

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

Repeating the hint, part (i) shows us that the order of a modulo $\displaystyle 2^{\alpha +2}$ is equal to $\displaystyle 2^{\alpha }$. Thus each of $\displaystyle a^{j}\mod {2^{\alpha +2}}$ are distinct odd numbers between 0 and $\displaystyle 2^{\alpha +2}$ for $\displaystyle 0\leq j\leq 2^{\alpha }-1$ and thus each of the $\displaystyle (-1)a^{j}\mod {2^{\alpha +2}}$ are also distinct odd numbers between 0 and $\displaystyle 2^{\alpha +2}$. As there are $\displaystyle 2^{\alpha -1}$ total odd numbers between 0 and $\displaystyle 2^{\alpha +2}$, we know that provided all of the above numbers are distinct, we have found $\displaystyle 2^{\alpha }+2^{\alpha }=2^{\alpha +1}$ total odd numbers and hence we can form a bijection. Hence it suffices to show that $\displaystyle (-1)a^{k}\not \equiv a^{j}\mod {2^{\alpha +2}}$ for any $\displaystyle 0\leq j,k\leq 2^{\alpha }-1$. This shows the claim for all odd numbers since all odd numbers are equivalent modulo $\displaystyle 2^{\alpha +2}$ to an odd number between 0 and $\displaystyle 2^{\alpha +2}$. Assume towards a contradiction that we found two such numbers: $\displaystyle (-1)a^{k}\equiv a^{j}\mod {2^{\alpha +2}}$ Without loss of generality, suppose that $\displaystyle k\geq j$. Then, $\displaystyle a^{k-j}\equiv -1\mod {2^{\alpha +2}}$ Once again, modulo 4 arguments (recalling that $\displaystyle a\equiv 5\mod {8}$) show that the left hand side above is congruent to 1 modulo 4 and the right hand side is still congruent to -1 modulo 4. This is a contradiction. Thus any odd number can be expressed uniquely of the form $\displaystyle (-1)^{i}a^{j}$ with $\displaystyle i\in \{0,1\}$ and $\displaystyle 0\leq j\leq 2^{\alpha }-1$ as required.

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