Does there exist a pair of integer solutions for a^3 + b^3 = c^3

Aside from (1,0,1), (0, 1, 1), (-1, 0, -1), (0, -1, -1), (1,0,1), (0,1,1)?

There are an infinite number of integer solutions to this equation.
Let a = 0, & b = c
Take all the pairs you want!
a = 0
b = 666
c = 666

a^3 + 666^3 = 666^3
In number theory, Fermat's Last Theorem states that no three POSITIVE integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two. Zero isn't positive.
Hope that helps:)

Sure, as long as you're not assuming a,b,c are all positive, then there are almost trivially an infinite number of solutions.

(a,b,c) = (z, -z, 0), (z, 0, z), (0, z, z)
for any integer z. That's the full list of solutions.

To see that all the solutions have that form, there are essentially two cases to consider:

(1) a,b negative and c positive; or a,b positive and c negative
For this case, it's easy to see that no such solution exists

(2) Everything else.
For these, you can use Fermat's Last theorem for exponent 3. This is _not_ easy to show otherwise (it wasn't shown until Euler did so).

You've repeated (1,0,1) and (0,1,1)

Other solutions:
a = 0, b = c -----> (0, 1, 1), (0, 2, 2), (0, −5, −5), ...
b = 0, a = c -----> (1, 0, 1), (2, 0, 2), (−5, 0, −5), ...
c = 0, b = −a -----> (1, −1, 0), (−1, 1, 0), (10, −10, 0), (−π, π, 0)

Mαthmφm

(0,0,0) only, and no other because in fact we cannot have a scenario where all of the numbers are 1 1 1 or greater, we can have less.

By fermats last thereom saying x^n + y^n is not equal to z^n where n is greater than 2

(0, 0, 0)