Finding the solution of a biquadratic

Question

If $a$, $b$, $c$, $d$ are distinct integers such that $(x-a)(x-b)(x-c)(x-d)=9$ has an integral solution say $h$. Then find the value of $(a+b+c+d)/h$.

Answer 1

Since $9=-3\cdot-1\cdot 1\cdot 3$, we have \begin{align} x-a&=-3,\\ x-b&=-1,\\ x-c&=1,\\ x-d&=3. \end{align} Of course, we can switch the solutions, (e.g., $x-b=-3$ and $x-a=-1$), but the only possibility is to use $-3$, $-1$, $1$ and $3$ on the right hand side, as the integers have to be distinct. It follows that for $h\ne 0$, $$ \frac{a+b+c+d}{h}=\frac{(a-h)+(b-h)+(c-h)+(d-h)+4h}{h}=\frac{-3-1+1+3+4h}{h}=4. $$

Answer 2

Hint: $x-a$, $x-b$, $x-c$, $x-d$ are distinct factors of $9$. There aren't too many possibilities.