http://kvant.mirror1.mccme.ru/1991/09/p70.htm ]]>

1. All 1998-problems were NOT mine. I just came to USA 1 day before

Olympiad took place :-))). I think that all 1998 problems were taken (by

Rachel Kamisnky) from some other sources, so they were not original

ones. You can refer to the problem as “Streamline School Olympiad

Problem”.

2. Some of 1999 problems were suggested by Rachel Kaminsky (she took

them from some other sources), others were suggested by myself (and some

of them were indeed original ones)

The first 1999 problem you mentioned was suggested by Rachel, the second

one was created by myself, and the third one was suggested (either by

Rachel or by myself, I cannot tell now for sure), it was not an original

one.

3. The 2000 problem you mentioned was created by myself.

4. At the end of the first paragraph it should be “Kharkov (Ukraine)

Olympiad” instead of “Ukraine Olympiad”

5. I can provide a simple solution for 1998 problem in the blog (of

course, it would be better to say “real numbers” instead of just

“numbers”, but “other source” did not say that, and I did not have a

chance to review the problems before the Olympiad :-))) )

Oleg

]]>We know a=(b-c)^2 and b=(a-c)^2.

Subtract the second equation from the first, and factor to obtain

a-b = (a-b)(2c-a-b).

So either a=b, or a+b=2c.

The same conclusion holds if we permute a,b,c. But if all permutations of a+b=2c hold, then it’s easy to see that a=b=c. So without loss of generality, a=b must hold, and c=0. Also, a=(b-0)^2=a^2, so the only possible solutions are a=1 or a=0.

]]>Definitely a keeper.

]]>That one took me a while too. {0,0,0} and {0,1,1} both seem to work and if you restrict the possibilities to real numbers, I believe you can pretty easily prove those are the only solutions. I haven’t tried to figure out it any solutions exist for imaginary or complex numbers. ]]>