MATH312 December 2012
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 • Q4 • Q5 •
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!
For the last part, use the fact that
where and where we recall that means that fully divides n (that is, this is the highest power of p that divides n).
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.
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gives the required definitions.
For the last part, we proceed by a contradiction and suppose that for some integer n.
As stated in the hint, we have
where . Now every odd prime present in the factorization of n contributes a power of two to due to the factor. Thus, there can be at most one odd prime dividing n, say .
If occurs to a power of two, then because it divides the left hand side, it must divide 14 and so must be 7. The prime cannot occur to a third power or higher since otherwise there are two factors of on the left and at most one on the right. However and so the power must be 1. Thus for some . Notice that as the phi-function is multiplicative, we have that
(or we have that if no power of 2 divides n)
Thus, as is even, we have that (or n is odd) and further that which is not a prime. This is a contradiction and hence this equation has no solution.
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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Divisibility, Pages using DynamicPageList parser function, Pages using DynamicPageList parser tag