Basic bases

$\begin{aligned} 123_a &= 146_b\\ \Leftrightarrow a^2 + 2a + 3 &= b^2 + 4b + 6\\ \Leftrightarrow (a+1)^2 + 2 &= (b+2)^2 + 2\\ \Leftrightarrow (a+1)^2 &= (b+2)^2\\ \Leftrightarrow a+1 &= b+2 \;\; ,\;\; \left(\{ a,b \}\in \mathbb{Z}^+\right)\\ \Leftrightarrow a &= b+1 \Rightarrow a+b=2b+1. \end{aligned}$

The smallest possible value of $b$ is $6+1=7$ , so the smallest possible value of $a+b$ is $2\times7 +1 =\boxed{15}$ .

Since 123 in base a equals 136 in base b, we must have a > b. We also know that b >= 7. Therefore a+b >= 15.

Luckily, 123 in base 8 is equal to 136 in base 7, so 15 is both the lower bound and achievable.