Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (e)

MATH312 December 2008

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[hide] Question 01 (e)
For each of the following statements, indicate if it holds for every positive integers $\displaystyle a,b,c$ (if so, a simple true without a proof suffices, if not, a false together with a counterexample is expected). If $\displaystyle \gcd(a,b)=1$ , then $\displaystyle a^{b-1}\equiv 1\mod {b}$ .

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!

[show] Hint
This should remind you of a mixture of Fermat's Little Theorem and Euler's Theorem. Can you use these to guide your solution?

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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
The statement is false We seek a counter example and we take the smallest composite number for b, namely $\displaystyle b=4$ . We then take $\displaystyle a=3$ . Then $\displaystyle 3^{4-1}\equiv 3^{3}\equiv 27\equiv 3\not \equiv 1\mod {4}$ . For a more enlightening proof, recall that by Euler's Theorem $\displaystyle a^{\phi (b)}\equiv 1\mod {b}$ Thus if we find a value of b where $\displaystyle \gcd(\phi (b),b-1)=1$ we would be able to construct a counter example since in this case there are integers s and t such that $\displaystyle s\phi (b)+(b-1)t=1$ and so choosing an a coprime to b between 2 and b-1 would give us that $\displaystyle a\equiv a^{s\phi (b)+(b-1)t}=(a^{\phi (b)})^{s}(a^{b-1})^{t}\equiv 1\mod {b}$ which would be a contradiction. This is what we did above.