Science:Math Exam Resources/Courses/MATH312/December 2013/Question 02 (a)

MATH312 December 2013

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[hide] Question 02 (a)
Find all solutions to the congruence $\displaystyle x^{2}\equiv 1\mod {55}$

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.

[show] Hint 1
You might be tempted to just say that the solutions are $x\equiv \pm 1\mod {55}$ but you must be careful! The modulus 55 is not prime. There is more work to do here.

[show] Hint 2
Notice that reducing the given equation modulo 5 and 11 gives the system of congruences ${\begin{aligned}x^{2}&\equiv 1\mod {5}\\x^{2}&\equiv 1\mod {11}\end{aligned}}$ Formulating the question like this should be clear that the Chinese Remainder Theorem would be really useful here.

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.
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[show] 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 follow the solution as outlined in the hints. Notice that reducing the given equation modulo 5 and 11 gives the system of congruences ${\begin{aligned}x^{2}&\equiv 1\mod {5}\\x^{2}&\equiv 1\mod {11}\end{aligned}}$ Solving these gives ${\begin{aligned}x&\equiv 1,4\mod {5}\\x&\equiv 1,10\mod {11}\end{aligned}}$ As 5 and 11 are coprime, we may apply the Chinese remainder theorem. The above congruences can be solved as follows. For the case ${\begin{aligned}x\equiv 1\mod {5}\end{aligned}}$ we see that there is an integer m such that $x=1+5m$ . If we further suppose that ${\begin{aligned}x\equiv 1\mod {11}\end{aligned}}$ we can see by combining these two equations that $1\equiv x\equiv 1+5m\mod {11}$ and so we see that $m\equiv 0\mod {11}$ . Writing $m=11n$ for some integer n, we have that $x=1+5m=1+5(11n)=1+55n$ and reducing modulo 55 yields $x\equiv 1\mod {55}$ . If instead in the second congruence we suppose that ${\begin{aligned}x\equiv 10\mod {11}\end{aligned}}$ we have that $10\equiv x\equiv 1+5m\mod {11}$ and so we see that $5m\equiv 9\mod {11}$ . Multiplying both sides by -2 gives $m\equiv -10m=-18\equiv 4\mod {11}$ and so $m=4+11n$ for some integer n. Hence, we have that $x=1+5m=1+5(4+11n)=21+55n$ and reducing modulo 55 yields $x\equiv 21\mod {55}$ . Now we go back to the first congruence and use instead the congruence given by ${\begin{aligned}x\equiv 4\mod {5}\end{aligned}}$ we see that there is an integer m such that $x=4+5m$ . If we further suppose that ${\begin{aligned}x\equiv 1\mod {11}\end{aligned}}$ we can see by combining these two equations that $1\equiv x\equiv 4+5m\mod {11}$ and so we see that $5m\equiv 8\mod {11}$ . Multiplying by $-2$ on both sides yields $m\equiv -10m\equiv -16\equiv 6\mod {11}$ . Writing $m=6+11n$ for some integer n, we have that $x=4+5m=4+5(6+11n)=34+55n$ and reducing modulo 55 yields $x\equiv 34\mod {55}$ . If instead in the second congruence we suppose that ${\begin{aligned}x\equiv 10\mod {11}\end{aligned}}$ we have that $10\equiv x\equiv 4+5m\mod {11}$ and so we see that $5m\equiv 6\mod {11}$ . Multiplying both sides by -2 gives $m\equiv -10m=-12\equiv 10\mod {11}$ and so $m=10+11n$ for some integer n. Hence, we have that $x=4+5m=4+5(10+11n)=54+55n$ and reducing modulo 55 yields $x\equiv 21\mod {55}$ . Thus, the complete set of solutions is given by $x\equiv 1,21,34,54\mod {55}$