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

MATH437 December 2011

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Question 07 (i)
Let $\displaystyle a\in \mathbb {Z}$ such that $\displaystyle a\equiv 5\mod {8}$ and let $\displaystyle \alpha \in \mathbb {N}$. (i) Prove that the order of a modulo $\displaystyle 2^{\alpha +2}$ is $\displaystyle 2^{\alpha }$

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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
First, using Euler's theorem, argue that it suffices to show that $\displaystyle a^{2^{\alpha -1}}\not \equiv 1\mod {2^{\alpha }}$ and $\displaystyle a^{2^{\alpha }}\equiv 1\mod {2^{\alpha }}$.

Hint 2
The proof should use mathematical induction.

Hint 3
Try to classify solutions to the equation $\displaystyle x^{2}\equiv 1\mod {2^{e}}$. This general statement will be useful.

Hint 4
Lemma The solutions to $\displaystyle x^{2}\equiv 1\mod {2^{e}}$ for e a positive integer are given as follows: 1) If $\displaystyle e=1$, then $\displaystyle x=1$. 2) If $\displaystyle e=2$, then $\displaystyle x=\pm 1$. 3) If $\displaystyle e\geq 3$, then $\displaystyle x\equiv \pm 1\mod {2^{e}}$ or $\displaystyle x=\pm 1+2^{e-1}\mod {2^{e}}$.

Hint 5

With the lemma and the induction hypothesis, what we can do is notice that $\displaystyle 2^{\alpha -2}\not \equiv 1\mod {2^{\alpha +1}}$ and $\displaystyle 2^{\alpha -1}\equiv 1\mod {2^{\alpha +1}}$ gives rise to a nontrivial solution to $\displaystyle x^{2}\equiv 1\mod {2^{\alpha +1}}$. We proceed by contradiction assuming (post induction hypothesis) that $\displaystyle 2^{\alpha -1}\equiv 1\mod {2^{\alpha +2}}$ and thus giving $\displaystyle 2^{\alpha -2}\not \equiv 1\mod {2^{\alpha +2}}$ which is inconsistent with the above.

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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. We first make some observations before beginning the formal proof. Firstly, recall that the order of an odd element in $\displaystyle \mathbb {Z} /2^{k}\mathbb {Z}$ divides $\displaystyle \phi (2^{k})=2^{k-1}$ for any $\displaystyle k\geq 1$. This follows from Euler's theorem and using the Euclidean algorithm (for any element if $\displaystyle d^{b}\equiv 1\mod {m}$ and $\displaystyle d^{c}\equiv 1\mod {m}$ then $\displaystyle d^{\gcd(b,c)}\equiv 1\mod {m}$). Thus the order of a must be a power of 2. Suppose that the order of a is the value r. The above shows that r is a power of 2. Moreover, $\displaystyle (a^{r})^{2}\equiv 1\mod {2^{\alpha +2}}$ Thus, to show that r is $\displaystyle 2^{\alpha }$, it suffices to show that $\displaystyle a^{2^{\alpha -1}}\not \equiv 1\mod {2^{\alpha }}$ and $\displaystyle a^{2^{\alpha }}\equiv 1\mod {2^{\alpha }}$. Next, we will make use of this small lemma. Lemma The solutions to $\displaystyle x^{2}\equiv 1\mod {2^{e}}$ for e a positive integer are given as follows: 1) If $\displaystyle e=1$, then $\displaystyle x=1$. 2) If $\displaystyle e=2$, then $\displaystyle x=\pm 1$. 3) If $\displaystyle e\geq 3$, then $\displaystyle x\equiv \pm 1\mod {2^{e}}$ or $\displaystyle x=\pm 1+2^{e-1}\mod {2^{e}}$. Proof of lemma The first two cases are clear so let's suppose that e is at least 3. Clearly the four x values do give solutions so it suffices to show that these are the only ones. Factoring shows that $\displaystyle (x-1)(x+1)\equiv 0\mod {2^{e}}$. Notice that $\displaystyle \gcd(x+1,x-1)=\gcd(x+1,2)=2$. Thus, we must have that one of the following is true $\displaystyle x-1\equiv 0\mod {2^{e-1}}$ or $\displaystyle x+1\equiv 0\mod {2^{e-1}}$ These give one of $\displaystyle x=\pm 1+2^{e-1}k$ for some integer k. We break into cases. Case 1: $\displaystyle k=2m+1$ odd. Then, we see that $\displaystyle x=\pm 1+2^{e-1}(2m+1)=\pm 1+2^{e-1}+2^{e}m\equiv \pm 1+2^{e-1}\mod {2^{e}}$ Case 2: $\displaystyle k=2m$ even. Then, we see that $\displaystyle x=\pm 1+2^{e-1}(2m)=\pm 1+2^{e}m\equiv \pm 1\mod {2^{e}}$ and this is what was to be shown completing the proof of the lemma. After all this we can begin the proof of the problem. The proof proceeds by induction. Notice that the claim is trivial for $\displaystyle \alpha =0$ or $\displaystyle \alpha =1$. For $\displaystyle \alpha =2$, we have that $\displaystyle a\equiv 5,13\mod {16}$ and both of these elements can be checked to have order 4. Thus, we suppose that $\displaystyle \alpha \geq 3$ and that the claim holds for $\displaystyle \alpha -1$. To prove the claim for $\displaystyle \alpha$, we proceed in two parts as suggested by the preliminary discussion. First we show that $\displaystyle a^{2^{\alpha -1}}\not \equiv 1\mod {2^{\alpha +2}}$. Notice that the induction hypothesis gives us that $\displaystyle a^{2^{\alpha -1}}\equiv 1\mod {2^{\alpha +1}}$ and $\displaystyle a^{2^{\alpha -2}}\not \equiv 1\mod {2^{\alpha +1}}$ Notice that the element $\displaystyle a^{2^{\alpha -2}}$ is a nontrivial solution to $\displaystyle x^{2}\equiv 1\mod {2^{\alpha +1}}$. Thus, the lemma states that $\displaystyle a^{2^{\alpha -2}}\equiv \pm 1+2^{\alpha }\mod {2^{\alpha +1}}$ or $\displaystyle a^{2^{\alpha -2}}\equiv -1\mod {2^{\alpha +1}}$ (since the solution 1 is not consistent with the above information). Notice that for the two negative solutions, If we look at the equation modulo 4 (recalling that we assumed that $\displaystyle \alpha \geq 3$), we get that $\displaystyle 1\equiv a^{2^{\alpha -2}}\equiv -1\mod {4}$ where we used the fact that $\displaystyle a\equiv 5\mod {8}$. This gives a contradiction and thus $\displaystyle a^{2^{\alpha -2}}\equiv \pm 1+2^{\alpha }\mod {2^{\alpha +1}}$. Next, we assume that $\displaystyle a^{2^{\alpha -1}}\equiv 1\mod {2^{\alpha +2}}$ and try to reach a contradiction. From above, we saw that $\displaystyle a^{2^{\alpha -2}}\not \equiv 1\mod {2^{\alpha +1}}$ and thus, $\displaystyle a^{2^{\alpha -2}}\not \equiv 1\mod {2^{\alpha +2}}$. Once again, using the lemma and similar logic to the above, we see that $\displaystyle a^{2^{\alpha -2}}\equiv 1+2^{\alpha +1}\mod {2^{\alpha +2}}$. This implies that $\displaystyle a^{2^{\alpha -2}}\equiv 1\mod {2^{\alpha +1}}$ which contradicts the above. Secondly, we wish to show that $\displaystyle a^{2^{\alpha }}\equiv 1\mod {2^{\alpha +2}}$. Thankfully, this is much more straightforward. Using the induction hypothesis given by $\displaystyle a^{2^{\alpha -1}}\equiv 1\mod {2^{\alpha +1}}$ we recast this information as $\displaystyle a^{2^{\alpha -1}}=1+2^{\alpha +1}k$ for some integer k. Squaring both sides yields $\displaystyle a^{2^{\alpha }}=(a^{2^{\alpha -1}})^{2}=(1+2^{\alpha +1}k)^{2}=1+2^{\alpha +2}k+2^{2\alpha +2}k^{2}\equiv 1\mod {2^{\alpha +2}}$ which completes what was required to show.

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