Recent content by elimqiu

Please take a look at http://math.elinkage.net/showthread.php?tid=102&pid=698#pid698 Let me know what do you think. Thanks

Suppose 1\le d \mid \gcd(a+b,\frac{a^p+b^p}{a+b}), then we have the following 1\le d \mid a+b\implies b \equiv -a\ (\text{mod }d)\implies \sum_{k=0}^{p-1}(-1)^k a^{p-1-k}b^k \equiv pa^{p-1}(\text{mod }d) \frac{a^p+b^p}{a+b} \equiv pa^{p-1}(\text{mod }d). Now since \gcd(d,a)=1, this means...

Show that if a,b\in\mathbb{N}^+,\ \gcd(a,b) = 1 and p is an odd prime, then \gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)\in \{1,p\}

http://elinkage.net/math/viewtopic.php?f=5&t=519"

No one interested in a proof of such a pretty formula?

It's a tool to prove f(x)=a_1\sin x+\cdots+a_n\sin nx,\quad |f(x)|\le |\sin x|\quad (\forall x\in\mathbb{R})\implies |a_1+\cdots+a_n|\le 1 It's not fit for homework in any math course I guess:)

Thanks micromass, not see real advantage yet...geometric sequence cannot be handled easily with double summation...

Show that \displaystyle{\sum_{k=1}^{n-1}\sin\frac{km\pi}{n}\cot\frac{k\pi}{2n} = n-m}\quad\quad(m,n\in\mathbb{N}^+,\ m\le n)

It's hard to show that that's the only possible n

Show that there is only one integer n ( > 1) such that http://latex.codecogs.com/gif.latex?1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6} is a perfect square

Find smallest d=2a^2=3b^3+2=5c^5+3 where a,b,c are integers. This problem is very hard. Either no solution or d is toooooooo big