Three digit number such that $A^2+B^2+C^2$ is divisible by $26$. Find all three digit natural numbers $ABC,(A \neq 0)$ such that $A^2+B^2+C^2$ is divisible by $26$.
Could someone give me some hint as how to approach this question as I am not able to initiate?
 A: $0^2, 1^2, 2^2,....,9^2 \equiv 0,1,4,9,-10,-1,10,-3,12,3$
Triplets that add to $0 \equiv 26$ are 
$(0,0,0)(0,1,-1),(0,10,-10)(0,3,-3)(1,9,-10)(4,-1,-3)$
So we can have any three digit numbers with the following sets of digits:
$0,0,0 \rightarrow $ no such numbers.
$0,1,5\rightarrow 150,105,510,501$.
$0,4,6\rightarrow 460,406,640,604$.
$0,7,9\rightarrow 790,709,970,907$.
$1,3,4\rightarrow 134,143,314,341,413,431$
$2,6,8 \rightarrow 268,286,628,683,826,862$
I do feel there should be a more direct way.  Maybe making use of $26 = 3^3 -1$ but I'm not seeing it.
A: As $26=2\cdot13$
and $x^2\equiv x\pmod2\implies A^2+B^2+C^2\equiv A+B+C\pmod2$
So, there should be even number of odd values among $\{A,B,C\}\ \ \ \  (1)$
Again, as $A,B,C$ are decimal digits, $$0\le A,B,C\le9\ \ \ \  (2)$$
Now for any integer $y, y^2\equiv0,\pm1,\pm4,\pm3\pmod{13}$
For $13\mid(A^2+B^2+C^2),$
we need $\{A^2,B^2,C^2\}\equiv$  to be one of $\{0,0,0\};\{0,\pm1\};\{0,\pm4\};\{0,\pm3\};\{4,-1,-3\};\{-4,1,3\}$
$\implies\{A,B,C\}\equiv$  to be one of
$\{0,0,0\};$
$\{0,\{1,-1\equiv12\text{(which is unacceptable as }>10\},\{5,-5\equiv8\}\};$
$\{0,2,3\};$
$\{0,\{4,-4\equiv9\},\{7,-7\equiv6\}\};$
$\{2,\{5,-5\equiv8\},\{7,-7\equiv6\}\};$
By $(1),(2);$ we can accept $$\{0,1,5\};\{0,4,6\};\{0,9,7\};\{2,5,7\};\{2,8,6\}$$
