Consider three positive integers x < y < z in arithmetic sequence, and determine all possible triplets (x,y,z) that satisfy this equation:

x⁴ + x²y² = z⁴ – 76

I If (x-a)^2*x^2+76 = (x+a)^4 -(x-a)^4, then

II x(8a^3-a^2 x+10ax^2-x^3)=76, with x={1,2,4,19,38,76},

when {a=1,x=2} is the only solution in the positive integers.

Posted by broll on 2011-08-24 06:39:28