A polynomial with integer coefficients that attains the value $5$ at four distinct points

Question

There is a polynomial $f$ of integer coefficients such that $\deg(f) \geq 4$. Let's assume that there are four integers $a,b,c,d$ for which $f(a)=f(b)=f(c)=f(d)=5$. Prove that there is no integer $k$ for which $f(k)=8$ - I'd like to have a full explanation to solution.

Answer 1

The polynomial $g(x)=f(x)-5$ has roots at $a$, $b$, $c$, and $d$. That means that the polynomial $x-a$ divides $g(x)$, and so on. We therefore get that

$$f(x)=h(x)\cdot (x-a)(x-b)(x-c)(x-d)+5$$

for some polynomial $h(x)$ (which may be a constant). Since $\mathrm{deg}(f)\ge 4$, $h(x)$ must be non-zero. If we now substitute into $f(k)=8$ we get

$$8=h(k)\cdot (k-a)(k-b)(k-c)(k-d)+5$$ $$3=h(k)\cdot (k-a)(k-b)(k-c)(k-d)$$

Note that $k-a$, etc. are distinct integers. We then have $3$ as the product of four or more integers, with at least four of them distinct. Three is prime, so the most number of distinct integers we can factor it into is three ($-3\cdot -1 \cdot 1$).

This is a contradiction, so we do not have $f(k)=8$.

Answer 2

You can write the polynomial as $(x-a)(x-b)(x-c)(x-d)f'+5$. We just have to prove $(x-a)(x-b)(x-c)(x-d)f'$ is never $3$, In fact it is never prime because for every $k$ at least two of $(k-a),(k-b),(k-c),(k-d)$ are not $1$ or $-1$ (In the case none of them are $0$)