Let $a,b,c,d$ be distinct integers such that the equation $(x-a)(x-b)(x-c)(x-d)-9=0$ has an integer root $r$,then find the value of $a+b+c+d-4r.$

Question

Let $a,b,c,d$ be distinct integers such that the equation $(x-a)(x-b)(x-c)(x-d)-9=0$ has an integer root $r$,then find the value of $a+b+c+d-4r.$

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As $r$ is the integer root of the equation $(x-a)(x-b)(x-c)(x-d)-9=0$,so $(r-a)(r-b)(r-c)(r-d)=9$

or $(a-r)(b-r)(c-r)(d-r)=9$

We need to find $(a-r)+(b-r)+(c-r)+(d-r)$ but i do not know how to find that.

Answer 1

$a,b,c$ and $d$ are distinct integers.

So $r-a,r-b,r-c$ and $r-d$ are all distinct integers.

Again, $9=3^2$

Hence the only possible factorisation: $$(r-a)(r-b)(r-c)(r-d)=3\cdot 3\cdot 1 \cdot 1$$ or $$(a-r)(r-b)(c-r)(r-d)=3\cdot 3\cdot 1 \cdot 1$$

So we have $a=r+3$, $b=r-3$, $c=r+1$, $d=r-1$.

Thus $\color{blue}{a+b+c+d-4r} = \color{red}{0}$