Recent content by mafendee

thanks for your reply and sorry for late response. I did a lot of reading on this. My further question would be, is it safe to say that the invariant properties is very much related to the isometric properties, and in case of euclidean geometry such that: f(ab)=f(a).f(b) where a and b are...

Dear all, i'm trying to understand geometry by studying the subject myself. i came across idea that I'm very much confuse of. it say's that 'geometry is a studies of geometric properties that is invariant under transformation' such as distance for euclidean geometry. my question is: why do...

tiny-tim, i will try to work on that simple proof you said. thank you.

tiny-tim, was my answer in previous post correct?

here we go again, for even power, (a + b)2n, let n=1, we have (a + b)2 1 2 1 we have gcd(2ab, a+b) here gcd=1/2 let n=2, 1 4 6 4 1 we have gcd(2ab(2a2+3ab+2b2), a+b) = gcd (2ab((2a+2b)(a+b) -ab), a+b) = gcd (2a2b2, a+b) here gcd=1/2 let n=3, 1 6 15 20 15 6 1 we have...

tiny-tim, for even power, (a + b)2n, let n=1, 1 2 1 we have gcd(2ab, a+b) here gcd=1/2 let n=2, 1 4 6 4 1 we have gcd(2ab(2a2+3ab+2b2), a+b) here gcd=1/2 let n=3, 1 6 15 20 15 6 1 we have gcd(ab(6a4+15a3b+20a2b2+15ab3+6b4), a+b) here gcd=1 let n=4, 1 8 28 56 70 56 28...

tiny-tim, you are right. from gcd(2ab(2a2 + 3ab + 2b2),a+b), we have = gcd(2ab(2a+2b)(a+b) -4ab + 3ab),a+b) = gcd(2ab(ab), a+b)) if p|2a2b2 then p|2 or p|a2 (p|a) or p|b2 (p|b) but p does not divide both a and b at the same time so p does not divide a+b. So we have gcd(2, a+b) = 1 or...

thank you to both of you. i think Дьявол has proved the cases of +a,-b and -a,-b. "For n = 4, for example, you would start with gcd(2ab(2a2 + 3ab + 2b2),a+b) " Here is what I have: Let say p|2ab(2a2 + 3ab + 2b2), this could be the cases where case1: p|2 or p|a but not b, and p does...

you're right. I had a thinking that a and b should be even and odd which is not divisible by 2. But then it could also be the case where both of them are odds, hence a+b=even which is divisible by 2. so, the gcd = 2. is it correct to say that gcd(a^n+b^n, a+b) = n? (i haven't tried to...

thank you tiny-tim, "if p divides 2ab, then either", p|2 and/or p|a (p not divide b) or p|2 and/or p|b. (p not divide a) p>1 cannot divide both a,b since (a,b)=1. so gcd(2ab, a+b)=1. is this correct?

Homework Statement Given gcd(a,b)=1, what is the gcd(a^2+b^2, a+b) where ^=square. Homework Equations The Attempt at a Solution gcd(a^2+b^2, a+b) = gcd ( (a+b)^2 -2ab, a+b) which i think, we can reduce to gcd(2ab, a+b). Here is where I stucked. I am not sure how to proceed. Please...