7. For four different integers, all their pairwise sums and pairwise products were calculated and written on the board. What is the smallest number of different numbers that could have appeared on the board? (I. Rubanov)
Answer: 6. Solution: If we take the numbers $-1,0,1,2$, it is easy to verify that each of the numbers written on the board will be equal to -2, -1, 0, 1, 2, or 3 - a total of 6 different values. We will show that fewer than six different numbers could not appear on the board. Let the taken numbers be $ac+d$. If $ac+d$. Variant for completing the solution. Let $u$ and $v$ be the two largest numbers in absolute value, with $|u| \leq|v|$. If $|u| \geq 2$, then $|uv| \geq 2|v|$, which is greater than any sum. If $|u| \leq 1$, then among the original numbers there must be $-1,0$, 1. If $v>0$, at least 6 different numbers are written on the board: $-1,0,1, v,-v, v+1$. The case $v<0$ is handled similarly.

Criteria. Only the answer - 0 points.

A correct example for 6 - 2 points.

Proved that there are no fewer than five values, with no further substantial progress - 1 point.

Proved that there are no fewer than 6 different numbers, but no example - 3 points.

A correct example for 6 and proof that there are no fewer than five values, without further substantial progress - 3 points.

The same plus substantial, but incomplete progress in proving the existence of a sixth value - 4 points.