Diophantine Equation $x^2+y^2+z^2=c$ 
$x^2+y^2+z^2=c$
Find the smallest integer $c$ that gives this equation one solution in
  natural numbers.
Find the smallest integer $c$ that gives this equation two distinct
  solutions in natural numbers.
Find the smallest integer $c$ that gives this equation three distinct
  solutions in natural numbers.

Clearly the first answer is c = 3. I know how to do linear diophantine equations, but I am stumped on this one.
By distinct solutions, I am looking for different threesomes (unordered triples)
Can you help?

Since the equation is symmetric: if $(x,y,z)$ is solution than every permutation of$(x,y,z)$ is the solution too. So it's impossible for this equation to have 2 solutions (because if for example $x\neq y$ than we have at least three distinct permutations).
The smallest $c$ for three distinct solutions is $6$: $(2,1,1), (1,2,1), (1,1,2)$.
