- Write down two linear congruences which do not have solutions modulo \(15\), but do have solutions modulo \(16\). (You do not have to solve them.)
- Write out the multiplication table for \(\mathbb{Z}_{11}\) completely, by hand.
- Show that Wilson's Theorem fails for \(p=10\) and check that it works for \(p=11\) by hand.
- Pick one, and really do some exploration and write about it.
- Do exploration to try to find a criterion for which primes \(p\) there are square roots of \(-1\). You will have to examine primes less than 10 by hand to make sure you are right!
- Do exploration to find out anything you can about how many square roots of \(1\) there are for a given \(n\).
- Find some conjecture/pattern to state about values of \(a^n\) mod (\(p\)), for \(p\) prime and \(0\leq n < p\) you discovered using the Sagelet which we did not talk about in class. This could be anything profounder than \[a^0\equiv 1\text{ mod }(p)\text{ or }1^n\equiv 1\text{ mod }(p)\] for all prime \(p\) and for all \(n\), but should at least be some pattern you tested for a number of values.
- Use Fermat's Little Theorem to help you calculate each of the following very quickly:
- \(512^{372}\) mod (13)
- \(3444^{3233}\) mod (17)
- \(123^{456}\) mod (23)
- Prove Fermat's Little Theorem using the steps above, or any way you would like.
- Prove that Wilson's Theorem always fails if the modulus is not prime. Hint: use the fact that the modulus \(n\) then has factors \(m\) other than \(1\) or \(n\).
- Keep on trying to figure out the pattern (if there is one) for square roots of negative one modulo a prime. We now know there can be at most two!
- Find solutions to \(3x-4\equiv 0\) mod (\(343\)) and \(x^2+8\equiv 0\) mod (\(121\)) using the method above (Hensel's Lemma).
- Solve \(f(x) = x^3-x-1\equiv 0\) mod (\(5^e\)) for \(e=1,2,3\). (Jones and Jones)
- Prove that it is impossible for \(p\mid x^2+1\) if a prime \(p\) has \(p\equiv 3\) mod (\(4\)) – that is, if \(p\) is of the form \(4n+3\). (Hint: look at everything modulo 4.)
- Prove that \(x^2+y^2=p\) has no (integer) solutions for prime \(p\) with that same form.
- Show that \(y^2=x^3+999\) has no (integer) solutions.