Puzzle: $(\Box @)+(\Box @) = (\Box\bigstar\Box$)

You have tried to solve it more algebraically than normal people would do. (I usually do the same thing - too scientific an approach - which occasionally makes us less efficient in similar puzzles.)

Spoilers are found below. Please stop reading if you want to solve it yourself.

Each of the two 2-digit expressions are smaller than $b^2$, so their sum is smaller than $2b^2$. Because this is a 3-digit expression, its first digit has to be $1$. So the square is equal to 1. So we have $$(1@) + (1@) = (1\bigstar 1)$$ Note that it ends with a $1$ - odd digit - even though the left-hand side is even - twice $(1@)$. It can only happen if the base $b$ is odd.

At any rate, the equation simplifies to $$2(b+@) = b^2+1 + b\bigstar$$ Moving everything non-negative to the right hand side, we have $$ 2@ = (b-1)^2 + b \bigstar.$$ The first term is a square of an integer, i.e. $4,16,36,64,\dots $ for $b=3,5,7,9,\dots$. Note that the identity above implies that $2@$ can't be smaller than $(b-1)^2$ but $2@<2b$ and if $b\geq 5$, $2b$ is clearly smaller than $(b-1)^2$. So the only chance to find a solution is $b=3$. Then the equation simplifies to $$ 2@ = 4+3 \bigstar .$$ The left hand side is even, so the right hand side must also be even. Therefore $\bigstar$ is either $0$ or $2$. For $\bigstar=2$ we would get $@=5$ which is too high so the only other option is $\bigstar=0$ and $@=2$ which works: $$ (12) + (12) = (101).$$ This trinary equation is translated to decimal base as $5+5=10$. Note that those ETs have 3 fingers in total, so if they have an even number of hands, it follows that different hands have different numbers of fingers, for example 1 left finger and 2 right fingers.

I would rather hide this behind a spoiler so that others can have fun, too. If there is a way of doing it, @-message me, please.

The only carry to the third postion that we can get in the sum of two double digit numbers is 1. Therefore `square'='1'. The sum is obviously an even number, so as the sum ends with a '1' = odd digit, the base 'b' must be an odd number. The double digit integer $x=$ 'square @' is $<2b$. Because of the carry we get that $2x\ge b^2$, so $4b>b^2$, and therefore $b<4$. The only alternative for the base is thus $b=3$. $y=$'@' must satisfy $2y>b$, because otherwise the least significant digit couldn't overflow, so '@' must be $=2$. The only remaining possibility is $$ 12_3+12_3=101_3, $$ which checks out.