It's showed that X^3+Y^3+Z^3-2XYZ = 19 has infinitely many integral solutions on MathStackExchange website[1]. We show that X^3+Y^3+Z^3-nXYZ = (n-3)(n^2+3n+9) has infinitely many integral solutions for arbitrary n.1.IntroductionX^3+Y^3+Z^3-nXYZ = (n-3)(n^2+3n+9) has infinitely many integral solutions as follows. (m)^3+(n-m)^3+(-3)^3-n(m)(n-m)(-3) = (n-3)(n^2+3n+9) m,n are arbitrary. Proof. X^3+Y^3+Z^3-nXYZ....................................................(1) Put X=t+a, Y=-t+b, Z=c..............................................(2) Substitute (2) to equation (1), and simplifying equation (1), we obtain (nc+3a+3b)t^2+(-(-na+nb)c+3a^2-3b^2)t+a^3-nabc+b^3+c^3=0.............(3) Equating to zero the coefficient of t and t^2, then we obtain b = -1/3nc-a. Substitute b = -1/3nc-a to equation (1) and let c=-3 and m=t+a, then we obtain equation (4). X^3+Y^3+Z^3-nXYZ = (n-3)(n^2+3n+9)..................................(4) (X,Y,Z) = (m, n-m, -3). m,n are arbitrary. Hence equation (4) has infinitely many integral solutions. Q.E.D.@ @@2.TheoremX^3+Y^3+Z^3-XYZ = ±26 has infinitely many integral solutions. X^3+Y^3+Z^3-2XYZ = ±19 " X^3+Y^3+Z^3-3XYZ = 0 " X^3+Y^3+Z^3-4XYZ = ±37 "3.Examples[1]: StackExchange: For what N does N=x^3+y^3+z^3-2xyz have infinite integer solutions?4.Reference

HOME