8.3. Given two numbers $a$ and $b$. It is known that out of the four numbers $a b, a+b, a-b, a / b$ exactly three are equal to each other ( $b$ is not equal to 0 ). Find $a$ and $b$.
Solution. Obviously, $a+b \neq a-b$, so exactly one of these two numbers is different from all the others. Then $a b=a / b$. If $a=0$, then the numbers from the condition are equal to 0, $b, -b, 0$ respectively, and the condition cannot be satisfied. Then $a \neq 0$, and it can be canceled, from which $b=1 / b$, which is equivalent to $b= \pm 1$.

If $b=1$, then the numbers have the form $a, a+1, a-1, a$. But the first number cannot be equal to either the second or the third, so this option is impossible.

If $b=-1$, then the numbers have the form $-a, a-1, a+1, -a$.

If $-a=a-1$, then $a=1 / 2$.

If $-a=a+1$, then $a=-1 / 2$.

From this, we get the answer: $a= \pm 1 / 2, b=-1$.

Criteria. Only a complete correct answer/answer with verification (and nothing else) - 1 point. Noted that $a+b \neq a-b-2$ points.

Proved that $b= \pm 1$ (or, equivalently, $b=1 / b$ or $b^{2}=1$) - 2 points. If the case $a=0$ is not considered, one point is deducted.

The case $b=1-1$ point is considered.

The case $b=-1-2$ points is considered. All points are summed ( $2+2+1+2=7$ )