Recent content by xax

Yes CRGreathouse, that's why I said that only one can be negative. My question was are there other solutions when one number is negative besides (t,1,-t)?

Thanks a lot to all of you and expecialy to CRGreathouse. I've proven that a<=3, b<=4 and found all the solutions(there are 3 in total). Edit: If one of them is negative the solution is (t,1,-t). Do you think there are more?

D H, I thought I was in the right forum. What I did so far: b+c = (a-1)*t and b*c = a*t and (c-1)*(b-1) = t+1. I'm stuck because I get too many posibilities.

Need to find all the posible solutions (a,b,c) for 1/a + 1/b + 1/c = 1, a,b,c integers.

Nevermind guys, I figured it out. Thanks for your input.

I thought it was clear what the question was(since the others understood): prove that this is true for every n prime, n!= 2. Using the Wilson theorem, I didn't go far: ((n-1)/2)!*((n+1)/2)*((n+3)/2)*...*(n-1) = -1 (mod n).

p = n

p is prime, not 2. Thanks in advance

You make perfect sense rodigee and this is a nicer demonstation than mine. Thank you.

You all have good points and I thought of all of them and the reason I've posted this is because of the square numbers as rodigee said. Well I think if n = k*k then k is one of the numbers between 1 and n-1. since n=k*k then k divides (n-k) and this is smaller the n-1 which means n divides (n-1)!.

I need to prove this for any n natural, n>= 5, n not prime.

Thanks again robert.

Hi robert. I was thinking instead of working on 8x^3 - 6x - 1 to substitute 2x with y and so the equivalent equation y^3 - 3y - 1 will be easier to prove it has no rational solutions( r can be only +/- 1). What do you think?

I've started to work on the it, just tell me if I'm on the right track. cos (45-30) = (\sqrt{3} + 1) / 2\sqrt{2} so cos 15 is irrational. cos3x = 4cos^3x - 3cos x \Rightarrow cos 5 is irrational cos 4x... cos 20 If this is a bad way, maybe someone knows a better one. Here is what I think...

you've saved me again robert, thanks.