Seventeenth Stars of Mathematics Competition

The required integers are $n=1$ and $n=2$. In the former case, $(a_1,b_1)=(1,-2)$ is the unique pair of integers satisfying the conditions in the statement; and in the latter, only $(a_1,b_1,a_2,b_2)=(2,0,-1,-2)$ and $(a_1,b_1,a_2,b_2)=(-1,-2,2,0)$ fit the bill. Verification is routine.

The degree $2n$ monic polynomial $f={\prod }_{i=1}^n(X^2+a_{i}X+b_i)$ vanishes at each $a_k$ and at each $b_k$. Since these $2n$ numbers are pairwise distinct and $\mathrm{deg}f=2n$, they form the root set of $f$. Moreover, since the root set of each factor $f_i=X^2+a_{i}X+b_i$ has size at most 2, and the union of these $n$ root sets has size $2n$, they must be pairwise disjoint sets of size 2 each. Recall now that $f$ is monic to write $f={\prod }_{k=1}^n(X-a_k)(X-b_k)$. Consequently, ${\prod }_{k=1}^n(X^2+a_{k}X+b_k)={\prod }_{k=1}^n(X-a_k)(X-b_k)$.

If no $b_k$ is zero, identification of constant terms of both sides yields $a_1\cdots a_n=1$. This is impossible for pairwise distinct integers unless $n=1$, in which case $a_1=1$, and $f=X^2+X+b_1$. This polynomial must vanish at $a_1=1$, so $b_1=-2$. Consequently, $f=X^2+X-2=(X-1)(X+2)=(X-a_1)(X-b_1)$, as required.

If some $b_k=0$, say, $b_1=0$, then $a_1\ne 0$, so $f_1=X^2+a_{1}X$ does not vanish at $a_1$. Since the root set of $f_1$ has size 2, this forces $n\ge 2$. The polynomial equality at the end of the second paragraph now reads $(X+a_1){\prod }_{k=2}^n(X^2+a_{k}X+b_k)=(X-a_1){\prod }_{k=2}^n(X-a_k)(X-b_k)$.

Since $a_1,b_2,\dots ,b_n$ are all non-zero, identification of constant terms of both sides above yields $a_2\cdots a_n=-1$. This is impossible for pairwise distinct integers unless $n=2$ or $n=3$.

If $n=2$, then $a_2=-1$, and $f_1=X^2+a_{1}X$ and $f_2=X^2-X+b_2$. Recall that the root sets of $f_1$ and $f_2$ are disjoint and both have size 2. Since $f_2(b_1)=f_2(0)=b_2\ne b_1=0$ and $f_2(b_2)=b_2^2\ne 0$, the roots of $f_2$ must be $a_1$ and $a_2$, so the other root of $f_1$ must be $b_2$. The condition $f_2(a_2)=0$ yields $b_2=-2$, and the condition $f_1(b_2)=0$ then yields $a_1=2$. Consequently, $(a_1,b_1,a_2,b_2)=(2,0,-1,-2)$.

The other quadruple, $(a_1,b_1,a_2,b_2)=(-1,-2,2,0)$, corresponds to the case where $b_2=0$.

Finally, to rule out the case $n=3$, we may and will assume that $a_2=-1$ and $a_3=1$. Then $f_1=X^2+a_{1}X$, $f_2=X^2-X+b_2$, $f_3=X^2+X+b_3$, and $f=f_1{f}_2{f}_3$ has the (pairwise distinct) roots $a_1,a_2=-1,a_3=1,b_1=0,b_2$ and $b_3$. Since $f_1(-1)f_1(1)=(1-a_1)(1+a_1)\ne 0$, it follows that $-1$ and $1$ are roots of $f_2{f}_3$. Notice that $f(1)=b_2\ne 0$ and $f_3(-1)=b_3\ne 0$, so $2+b_2=f_2(-1)=0=f_3(1)=2+b_3$, i. e., $b_2=b_3$. This contradiction settles the case.