I don't know the skill level of everyone on this forum, but this is a relatively easy olympiad problem for you to try:

Find all real valued functions f such that:
[f(x)+f(y)][f(u)+f(v)]=f(xu-yv)+f(xv+yu).

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2007-02-06 17:44
sounds difficult , maybe i will think about it for several days . HA!!

2007-02-06 19:00
What makes olympiad problems fun is that fact that you are free to explore and play around with the problem.

Usually, with functional equations, it is nice to play around, conjecture, and then prove.

2007-02-26 19:22
Okay, I'll start the proof.
First of all, letting x=y=u=v=0, we get that 4f(0)^2=2f(0). This means that f(0)=0 or f(0)=1/2.

If f(0)=1/2 then letting x=y=0 shows trivially that f(x) is constant and equals 1/2 for all x.

Now see what you can derive under the condition that f(0)=0.

2007-02-26 19:23
Actually I'm pretty confident this is not quite difficult. It was the first problem on a functional equations packet!

(Well, there were 2 warm-up problems beforehand.)

2007-03-07 18:54
Actually, I have known the answer already.
i. f(x.) = 0
ii. f(x) = 1/2
iii. f(x) = x^2

2007-03-07 19:33
by the way , I think there are many people in this website might take part in the AIME test .
So, everybody good luck!!

p.s. I think i will participate in the test in tainan , since the test won't be held in chia-yi!!

2007-03-07 19:37
to isomorphism : you will get a excellent grade on the AIME test because your math is so good. Today , I wrote a test sheet about the AIME test in 2002 . i only get fifty ! XDXD

My math still need to be improved !!!

2007-03-08 19:57
AIME test, it seems not easy for me.
I think I just had a great fortune on the test last year.
(ps. The questions were also easier last year.)

2007-03-11 10:29
Your solution to the functional equation is correct. Solving any IMO problem is already a great feat in itself; I admire your talent in mathematics.

About the AIME, last year's test indeed was easier. I'm aiming for a 14+ this year (even though a 7 will probably get me into USAMO).