sgw wrote: Determine all pairs of real numbers $a$ and $b$, $b > 0$, such that the solutions to the two equations $$x^2 + ax + a = b \qquad \text{and} \qquad x^2 + ax + a = -b$$ are four consecutive integers. Sum of these 4 consecutive numbers is $-2a$ So the four numbers are $-\frac{a+3}2,-\frac{a+1}2,-\frac{a-1}2,-\frac{a-3}2$ So $(-\frac{a+3}2,-\frac{a-3}2)$ are roots of one of the two equations And $(-\frac{a+1}2,-\frac{a-1}2)$ are roots of the other So $(-\frac{a+3}2)(-\frac{a-3}2)+(-\frac{a+1}2)(-\frac{a-1}2)=(a-b)+(a+b)=2a$ Which is $a^2-4a-5=0$ and so $a\in\{-1,5\}$ $a=-1$ gives $(-1,0,1,2)$ and so $b=1$ $a=5$ gives $(-4,-3,-2,-1)$ and so $b=1$ again Hence the anwer $\boxed{(a,b)\in\{(-1,1),(5,1)\}}$