Three digit numbers $abc$ such that $26\mid a^2+b^2+c^2 $

Question

Find all three digit numbers $abc$ such that $26\mid a^2+b^2+c^2.$

I found this in a Olympiad question paper and was unable to solve it.

Answer 1

Compute the squares of $0 \to 9$ mod $26$. You will get the list $0,1,4,9,16,25,10,23,12,3$. Now, you need three of these numbers to sum to $26$ or more. You will immediately get some answers:

$1 + 9+16=26$, giving $134,143,314,341,413,431$.

$1+0+25=26$, giving $105,150,501,510$.

$4+25+23=52$, giving $257,275,527,572,725,752$.

$0+16+10=26$, giving $406,460,604,640$.

$4+10+12 = 26$, giving $268,286,628,682,826,862$.

$23+3+0= 26$, giving $709,790,907,970$.

That completes the list of numbers.

Answer 2

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16 \equiv -10 \mod 26$

$5^2 \equiv -1 \mod 26$

$6^2 \equiv 10 \mod 26$

$7^2 \equiv -3 \mod 26$

$8^2 \equiv 12 \mod 26$

$9^2 \equiv 3 \mod 26$

$0^2 = 0$.

$0 + x + (-x) \equiv 0 \mod 26$ so any number with the digits $(1,5,0),(4,6,0),(7,9,0)$ so the accounts for 12 answers.

Let's see what else...

$1 + 9 -10 = 4-1 - 3$ So all numbers with digits $(1,3,4)$ or $(2,5,7)$. That accounts for 24 more answers.

I think that's all.