Problem 1270

Solution. We will assume that $a+b>0$, otherwise the proof is already complete. Since the trinomial has integer roots, its discriminant $(ab)^2-4(a+b)$ is a perfect square, i.e., $(ab)^2-4(a+b)=k^2$ for some non-negative integer $k$. Then the numbers $ab$ and $k$ have the same parity. From the relation $00$ and even, so $|ab| \geq k+2$. Therefore, $4(a+b) \geq (ab)^2-(|ab|-2)^2=4|ab|-4$, and thus $|ab|-a-b+1 \leq 2$. If $a, b \geq 0$, then $(a-1)(b-1) \leq 2$. In the case where both brackets are positive, one of the numbers $a$ and $b$ equals 2, and the other is no more than 3, so their sum does not exceed 6. If one bracket is zero (let's say $a=1$ for definiteness), then $k^2=b^2-4(b+1)=(b-2)^2-8$. This is possible only if $b-2=3$, since the numbers $b-2$ and $k$ have the same parity, and $(m+2)^2-m^2=4m+4>8$ for $m>1$. Thus, $a+b=6$. If one of the brackets is negative (let's say the first one for definiteness), then $a \leq 0$ and $b>0$. Therefore, $2 \geq |ab|-a-b+1=-ab-a-b+1$ and, thus, $(a+1)(b+1) \geq 0$, which is impossible (if $a=0$ this is obvious, and if $a<0$, then $a<-1$ and the first factor is negative, while the second is positive).