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.

Yes. A0 is 1, B0 is 3. That would make (A0+B0)/2 = 2. I'll call the third one C0. The product of A0, B0, and C0 is 6. This is obviously a factor of 6^2008. So, if we multiply each number by 6^(2007/3) or 6^669, we get A = 6^669, C = 2*6^669, and B = 3*6^669, which would give a product of 6^2008.