Solutions $a,b,c \mid 1+a+b+c $ over positive integers algebraically (avoiding brute force)?
Because of symmetry, you can assume that $a\leq b\leq c$, then we have $$ck= 1+a+b+c \implies k\leq 4$$
- If $\boxed{k=1}$ we have $c=1+a+b+c$ a contradiction
If $\boxed{k=2}$ we have $2c=1+a+b+c$ so $c= 1+a+b$. Then $b\mid 2+2a+2b$ so $b\mid 2+2a$ so $mb = 2a+2\leq 4b\implies m\leq 4$ and now is easy to check all 4 subscases...
- If $m=1$ we get $b=2a+2\implies a\mid 2+2a+4a+4\implies a\mid 6$...
- If $m=2$ we get $b=a+1\implies a\mid 2+2a+2a+2\implies a\mid 4$...
- If $m=3$ we get $3b=2a+2\leq 2b+2\implies b\leq 2$ so $b=2$ and $a=2$.
- If $m=4$ we get $4b=2a+2 \leq 2b+2\implies b=1$
If $\boxed{k=3}$ we have $3c=1+a+b+c$ so $2c=1+a+b$.
If $b\leq c-1$ then $2c\leq 1+2c-2 = 2c-1$ a contradiction.
If $b= c$ then $c=1+a$ and now we have $a\mid 3+3a \implies a\mid 3$ so $a=1$ and $b=c=2$ or $a=3$ and $b=c=4$.
- If $\boxed{k=4}$ we have $4c = 1+a+b+c$ so $a=b=c=1$.