Recent content by MrJB

I wrote a little search program. So you've got a list of all the points in your vector space. Put the first one into your set, try putting in the next point, and go down the list putting points in your set if you can do so without having three on a line. That'll give you a solution, likely not...

Daniel: You're right. For n=3 there is a 9 point set and for n=4 there is a 20 point set satisfying the problem.

Pick any two elements of F3. Let A be the set of 2^n vectors whose coordinates are those two elements of F3. No three points in A is collinear. I don't know if this is maximal though. For example let's call the elements of F3 0, 1, and 2. In the n=2 case, A is {{0,0},{0,1},{1,0},{1,1}} No...

You have Cos(2*stuff) so use a double angle trig identity to get something squared in the sqrt.

I've seen this, and would also recommend it.

How about science journalism?

No. For example, along the real line the imaginary part of the zeta function is zero, but the real part is certainly not always zero. Look at this page on mathworld: http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html There's a graph of the curves where the real parts are zero and...

There aren't any more. An even perfect number (other than 6) is not divisible by three since it is the product of a power of 2 and a mersenne prime. Therefore one of P-1 or P+1 must be divisible by 3, and thus not prime.

Assume that two distinct solutions exist, then derive a contradiction. It's difficult to say further without more information about your problem.

I agree there is no peak age. A physics professor once told me that even though he was older he was still doing research as fast as when he was younger. He said his mind wasn't as quick, but because of his years of experience, he would investigate in the right direction more often and not waste...

Rewrite (1,3,6,8)(4,6) as one permutation.

Do you mean a number with 100 digits? What's Cr mean?

This is a sketch of how I think you can prove that the only possible p(x) are the two constant solutions. There are some important details missing at the end. Let c be one of the roots of x^2 - x + 1 = 0. Let's see what happens at x=c in p(x)^2-1 = p(x^2+1). p(c)^2 -1 = p(c^2+1), but...

If you simplify your function: f(n) = n^2 - (n-1)^2 = n^2 - ( n^2 - 2n + 1) = 2n - 1 You can see that it produces all the odd numbers. So, your function will produce all the primes (except for 2) and it will give you many false positives since not every odd number is prime.

The Elliptic Curve Method (ECM) seems appropriate for what you're doing. There's a nice applet here: http://www.alpertron.com.ar/ECM.HTM