Christmas Streak 16/88: Hell in 1997

Consider expressing $B$ as: $\{(x,~x)|P(x)Q(x)=0\}.$

The number of elements of $B$ would be maximized if $P(x)=0$ and $Q(x)=0$ have distinct roots, which means if $P(a)=0,$ $Q(a)\neq0$ and vise versa.

However, the answer cannot be $23.$

This is because the problem has stated that $A=\{(x,~y)|P(x)Q(y)=0~\text{and}~P(y)Q(x)=0\}$ is an infinite set.

If $P(x)=0$ and $Q(x)=0$ don't have any common roots, then any root of $P(x)=0,~\alpha,$ would satisfy $P(\alpha)Q(y)=0.$

However, since $Q(\alpha)\neq0,$ we can see that $P(y)=0$ is mandatory. But since $P(x)=0$ has finite roots, $A$ would be a finite set.

This is a contradiction, so we can think of a situation where $P(x)=0$ and $P(y)=0$ has exactly one common root, $\beta.$

Then not only $P(\beta)Q(y)=0$ but $P(y)Q(\beta)=0$ as well. Therefore, $y$ can be any reals, thus making $A$ an infinite set.

Therefore the maximum of the number of elements of $B$ is $7\text{(distinct roots of P)}+14\text{(distinct roots of Q)}+1\text{(common root)}=\boxed{22}.$

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P.S. When this problem was given in the Korean SAT, only 0.08% of the students got it right. (Majority of the students wrote 23 as the answer.)