Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.5 National olympiad Prove it

5.32 Given a polynomial with integer coefficients $a_{1}, a_{2}, \cdots, a_{n}$
$f(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n} .$

It is also known that there exist four distinct integers $a, b, c, d$ , such that
$f(a)=f(b)=f(c)=f(d)=5 \text {. }$

Prove that there is no integer $k$ , such that $f(k)=8$ .

Solution

[Proof] From the given conditions, the polynomial
$f(x)-5$

has four distinct integer roots $a, b, c, d$ . Therefore, we can assume
$f(x)-5=(x-a)(x-b)(x-c)(x-d) \cdot g(x)$

where $g(x)=x^{m}+b_{2} x^{m-1}+\cdots+b_{m}$ , and $b_{1}, b_{2}, \cdots, b_{m}$ are integers.
If there exists an integer $k$ such that $f(k)=8$ , then
$(k-a)(k-b)(k-c)(k-d) \cdot g(k)=3.$

Thus, among the four integers $k-a, k-b, k-c, k-d$ , at least three have an absolute value of 1, and therefore at least two of them are equal. This contradicts the condition that " $a, b, c, d$ are four distinct integers". Hence, the proposition is proved.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.