This is the home page for Math 290.

OK, here's the best of the class pictures...
Link to image...
but it's way to dark in Explorer and Netscape to show on this page. It looks ok in other viewers, and maybe I'll find a way to lighten it up.

Syllabus - Course description, contact information.

Current schedule - includes homework assignments. Tentative.

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Obsolete Notes:

4/21/03 (updated 4/22):Here is ne.html, a table of all the n's and e's I have so far. (If you have sent me your key and it doesn't appear on the table, please be patient...) These are public keys, so you can use them to send anyone on the list a message. But beware, 7-digit n's aren't really secure.

Here is NTcalc.xls, the spreadsheet mentioned in class 4/14. It's a (relatively primitive) tool for computing inverses mod m and powers mod m. (EXCEL DOCUMENT)

Here is the homework handout from 4/14 (The RSA exercise, maybe due 4/21). (WORD DOCUMENT)

4/7/03. Here's a link to the prime number result by Goldston and Yildirim (thanks to Jessica Dodge). The link is to the Am. Inst. of Math's home page, so if they add new headlines you'll have to scroll down to the prime number result.

Here's what they proved: There are infinitely many primes p such that the gap between p and the next prime is less than (log p)^(8/9). That's a big improvement over previous results. The twin primes conjecture says that there are infinitely many primes p such that the gap between p and the next prime is 2.

4/6/03. As you noticed, I didn't get around to posting a new quiz last week. So, everybody scores 3/3 for free. There won't be a quiz next week because of the exam, so the only remaining quizzes are April 17 and April 24.

Here are the questions that should have been on the quiz. They seem like good questions for the exam, too:

• Given a and p (where p is an odd prime and a isn't divisible by p) decide whether a is a quadratic residue mod p. That is, evaluate the Legendre symbol with a and p (+1 or -1). Examples: Legendre(40, 53) or Legendre(-4, 1003). (Hint: one of those is +1, one is -1.) Key facts:
  • You can reduce a mod p;
  • Legendre(ab,p) = Legendre(a,p) times Legendre(b,p);
  • Legendre(any square, p) = +1;
  • Legendre(2,p) = +1 only if p is congruent to 1 or 7 (mod 8);
  • Quadratic reciprocity theorem.
• Be able to grind out Pythagorean triples at will. Enough to know that, at least for primitive triples, z itself is the sum of two squares and one of x,y is the difference of the same two squares. (If you choose the squares badly then the resulting triple might not be primitive, but it will still be Pythagorean.)

Also 4/6/03: Did you notice that the text's solution to problem 13.1#9 (the one about solving x^2 + 3y^2 = z^2) is way incomplete? This problem was too hard for the author and all the editors, and it was too hard for me until today, but you can do it anyway.

NO CLASS WEDNESDAY 3/19. Don't forget the quiz (Thursday noon or sooner). See you Monday.

Self-scheduled exam available about 2/19, due back 2/24 (despite the snow day)---
Covers Appendices A and B, sections 1.2 (induction), 1.4 (divisibility), 2.1 (just binary <-> decimal), 3.1 through 3.5, plus everything else we have done, but not anything from Chapter 4

2/19/03. About quiz 4:
1. What is ( 3 * 5^3 * 7^2 , 2^2 * 5 * 7 * 11 ) ? The gcd is 5 * 7 = 35.
2. What is [ 3 * 5^3 * 7^2 , 2^2 * 5 * 7 * 11 ] ? The lcm is 2^2 * 3 * 5^3 * 7^2 * 11.
3. If (323, 2975) = 17, what is [323, 2975] ? 323 * 2975 / 17.
4. Are there infinitely many primes of the form 6k+4, where k is an integer? No, because 2 is a factor of all of these...
5. Are there infinitely many primes of the form 4k+7, where k is an integer? Yes, by Dirichlet's theorem. Also, numbers of this form are also of the form 4k+3, and we proved that there are infinitely many 4k+3 primes, by showing that if there were finitely many, then there would be no 4k+3 factors available for their product minus 1.

2/10/03. Next Wednesday (2/19) would be a good day to start a self-scheduled exam (ending 2/24).

2/10/03. Warning! There was a typo in the extra credit problem. It was supposed to be this:
If n is in Z+, show that (n-2) and (n^2+n+1) can't both be perfect cubes. (Note that n^2 means n with superscript 2; that is, n squared.)
(NOT (n-3)) (This is corrected on the schedule, now.)

2/10/03. About quiz 3:
1. gcd = 10093
2. Yes, you can write 3 = 6*41 - 3*81. I know that 3 isn't the gcd of 41 and 81, but it is a multiple of the gcd, and that's all it takes.
3. Since 323 doesn't divide 2975, and obviously 323 doesn't divide 228 either, it seems that the wish theorem doesn't apply to 323. So, 323 must not be a prime. Some of you noted that 323 = 17 * 19, and that each of those factors divides one of 2975 and 228.

2/3/03. About quiz 2:You all showed you're on top of things, but then, the questions were easy. They'll get harder. For quiz 3 I'll ask for the gcd of a couple of 5- or 10-digit numbers.

Question 3 was really dumb. Sorry. Most people thought I was asking about the numbers 1 and 2, but I meant the numbers in the problems 1 and 2---that is, the numbers 99 and 86. Do you remember the TV show Get Smart? It featured Agent 86 and Agent 99. Like I said, it was a dumb question.

1/23/03. Follow this link: http://www.thinkgeek.com/tshirts/frustrations/5aa9/

1/26/03. How to do problem 2a (Appendix A):
Theorem: If a is an integer, then (-1)*a = -a. (Sorry, can't type raised dot; * means times)
Proof:By definition, -a is the solution to the equation a+x=0. So, if we can prove that (-1)*a is a solution to that equation, then we can conclude that by definition, (-1)*a is -a.
So, let's see whether (-1)*a satisfies a+x=0:
a + (-1)*a = 1*a + (-1)*a (because a=1*a by an axiom)
  =(1+(-1))*a (distributivity axiom)
  =0*a (since 1 + (-1) is zero by axiom)
  =0 (since 0*anything is 0 by example 1 in Appendix A)
so we're done.
That's how to approach problem 2b, too.