Integer Solutions to x^3+y^3+z^3=2: Proving Infinitely Many

[lfdahl]

Prove, that the equation:$x^3+y^3+z^3 = 2$- has infinitely many integer solutions.

 

[kaliprasad]

Prove, that the equation:$x^3+y^3+z^3 = 2$- has infinitely many integer solutions.

Spoiler

we have $(p+1)^3-(p-1)^3 = 6p^2 + 2$

so if we chose q such that $q^3 = 6p^2$

then $(p+1)^3 - (p-1)^3 - q^3 = 2$

so the 3 numbers are $(p+1, - p + 1, - q )$ and $p = 6m^3=>q=6m^2$ giving solution

so we have solution set $x = 6m^3+1, y = - 6m^3 + 1, z = -6m^2$ is the set for integer m >0 is the solution set

so there are infinite solutions

 

[lfdahl]

Spoiler

we have $(p+1)^3-(p-1)^3 = 6p^2 + 2$

so if we chose q such that $q^3 = 6p^2$

then $(p+1)^3 - (p-1)^3 - q^3 = 2$

so the 3 numbers are $(p+1, - p + 1, - q )$ and $p = 6m^3=>q=6m^2$ giving solution

so we have solution set $x = 6m^3+1, y = - 6m^3 + 1, z = -6m^2$ is the set for integer m >0 is the solution set

so there are infinite solutions

Very short and elegant, kaliprasad. Thankyou for your contribution!