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

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.

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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. 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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