Science:Math Exam Resources/Courses/MATH220/December 2011/Question 08

MATH220 December 2011

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q2 (a) • Q2 (b) • Q3 • Q4 • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 •

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[hide] Question 08
Prove that if p is a prime number and p > 4, then $\displaystyle p^{2}\equiv 1{\pmod {6}}$

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
Note: this hint was giving on the exam itself. Think about the possible remainders when p is divided by 6.

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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. By saying that we look at primes larger than 4, what we really mean is we look at all the primes except 2 and 3 (which makes sense since 2² = 4 and 3² = 9 which is 3 mod 6; so these two don't work anyway). Now any prime p other than 2 and 3 will then clearly NOT be a multiple of two and NOT be a multiple of 3. This means that $\displaystyle p\equiv 1{\text{ or }}5{\pmod {6}}$ since integers that are 0, 2 or 4 mod 6 are all even and since an integer that is 3 mod 6 is a multiple of 3. But since $\displaystyle 1^{2}\equiv 1{\pmod {6}}$ and $\displaystyle 5^{2}=25\equiv 1{\pmod {6}}$ we can conclude that if p is a prime larger than 4, then its square is always 1 mod 6.