help

There are 3 different integers bigger than 0 :
"a", "b" and "(a+b)/2".

Is it possible the multiplication a*b*(a+b)/2 equal x^2008 where x is integers bigger than 0 too.

>>13
What about a0 = 4, b0 = 6, (a0+b0)/2 = 5?

Then a0*b0*(a0+b0)/2 = 120, which is obviously a factor of 120^2008, etc...

This seems a litte bit too trivial for my liking, all we have to do is find two integers with integral average.

Perhaps OP is asking, GIVEN an integer x>0, can we find a and b s.t. (a+b)/2 is an integer, and a*b*(a+b)/2 = x^2008 ?