smc

Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$, and the distance between them is $2k+1$. Let this be equivalent to $3d$, so that the one-third of the distance between the squares is equivalent to $d$. The numbers $abc$ and $acb$ are one-third and two-thirds of the way between $k^2$ and $(k+1{)}^2$. Therefore, the distance between these two numbers is also one-third the distance between the squares, or $d$. Setting these equal to each other, we have $\frac{2k+1}{3}=9(c-b)$ $\Rightarrow 2k+1=27(c-b)$. Notice that since $c$ and $b$ are digits, their difference is at most $9$ and at least $0$. Also notice that since $acb$ is greater than $abc$, $c>b$. Representing this as an inequality, we have $27\le 27(c-b)\le 243$. Substituting $2k+1$, we have $27\le 2k+1\le 243$ $\Rightarrow 13\le k\le 121$. However, we know that $abc$ is a $3$-digit number, and since $k^2$ is less than $abc$, $k^2$ must be at most $961$, or $31^2$. Therefore $k\le 31$. Plugging this back into our inequality, we have $13\le k\le 31$ $\Rightarrow 27\le 2k+1\le 63$ $\Rightarrow 27\le 27(c-b)\le 63$ $\Rightarrow 1\le (c-b)\le \frac{7}{3}$. But (c-b) must be an integer, so now we have $1\le (c-b)\le 2$ $\Rightarrow 27\le 27(c-b)\le 54$ $\Rightarrow 27\le 2k+1\le 54$ $\Rightarrow 13\le k\le \frac{53}{2}$ $k$ is also an integer, so now we have $\Rightarrow 13\le k\le 26$ $\Rightarrow 27\le 2k+1\le 53$ $\Rightarrow 27\le 27(c-b)\le 53$ $\Rightarrow 1\le (c-b)\le \frac{53}{27}$. Once again, $(c-b)$ must be an integer, so we have $1\le (c-b)\le 1$ $\Rightarrow (c-b)=1$ $\Rightarrow 27(c-b)=27$ $\Rightarrow 2k+1=27$ $\Rightarrow k=13$ The two squares are $13^2$ and $14^2$, or $169$ and $196$. A third of the distance between them is $9$, and $169+9=178$. $1+7+8=16\Rightarrow \boxed{16}$.