Thanks, I looked at the article as well as a proof that all primes have at least one primitive root, so now it makes perfect sense. I don't think I would have been able to come up with that on my own without more of a background in number theory, so it's definitely appreciated. I'll explain it...
Okay, I have a step in the right direction. If p - 1 does not divide k, use the argument above to show that p is odd and p - 1 is even. So, there are an even number of terms in the sum.
Suppose k is odd. Then, consider a^{k}, where a\leq(p-1)/2. Consider its additive inverse, (p-a)^{k}. Using...
The problem I have understanding how to do that is...suppose I consider n^k and m^k for distinct m and n. We know that mm^{k} is congruent to m^{r} by using Fermat's little theorem again. I'm trying to show that then m^{r}\neqn^{r}, correct? But, that's not always the case. Suppose we have p =...
If the values are distinct, then each of 1, 2, ..., p-1 appears once, so their sum is (p^{2}-p)/2. If p = 2, then p - 1 = 1 which divides p, so the first case applies. So, p\geq3. Then, p must be odd, so p-1 is even, and (p-1)/2 is an integer. Therefore, (p^{2}-p)/2 = p(p-1)/2 = p*n for some n...
I'm trying to help a friend solve a problem but as I've never studied number theory, I'm having a bit of trouble myself figuring out how to do it.
We need to find the sum of 1^{k}+2^{k}+...+(p-1)^{k} (mod p), where p is prime. By writing a program that created a table from test cases...
That's what I originally figured, that angles were merely a convenient notation to express a rotation in a given direction. In R^2, it's obvious what angle t refers to due to the fact that there are only two possible directions for Euclidean rotation, clockwise and counterclockwise. However, in...
Okay, I've proved that it holds true for the trivial case of R^1.
I've also done so in R^2, as that much better serves to illustrate the mechanisms of what's occurring.
With that done, how would I prove inductively that this will work for any R^n? Showing that it will work for translation is...
Good points. I did say that my geometry was rusty. When I studied modern geometry some years ago, our instructor confined most of the teaching to Euclidean and absolute geometry, at a level almost dumbed down to high school, and did not include affine geometry. I also ordinarily only teach...
Should I be using householder transformations to prove this? I've tried a few approaches so far, for the base case only (trying R^1 and R^2), and have met with little success. If I were to try to prove for R^2, should I be using 2x3 orthogonal matrices rather than2x2, so that I'm reflecting...
I took another shot at it, so let me know if I'm wrong on any details or just going about this the wrong way. I tried to prove the base case in both R^1 and R^2. For R^1, since any transformation matrix of a reflection must have a determinant of -1, [-1] is the only allowable matrix, which...
It's been a long time since I've studied geometry, and had an extremely poor instructor at the time, so I'm having difficulty remembering how to prove certain theorems. My linear algebra is a bit rusty as well, though I'm more well versed in that than in geometry. One of my students needs help...