For $f(x)$ a polynomial with integer coefficients, if $f(a)=f(b)=f(c)=f(d)=1991$, then $f(h) \neq 1993$.

Question

Could someone help me with this problem and explain the solution

Let $f(x)$ be a polynomial with integral coefficients. Suppose that $a$, $b$, $c$, and $d$ are distinct integers such that $$f(a)=f(b)=f(c)=f(d)=1991.$$ Show that $f(h)=1993$ cannot exist for any integer $h$.

Now in the solution it basically took a polynomial $$g(x)=f(x)-1991.$$ Hence $(x-a)$, $(x-b)$, $(x-c)$, $(x-d)$ are factors of $g(x)$. So it took $$g(x)=p(x)(x-a)(x-b)(x-c)(x-d),$$ where $p(x)$ has integral coefficients. It explained that, for $g(h)=2$, the values of $h-a$, $h-b$, $h-c$, $h-d$ have to be $-1$, $-2$, $1$, $2$, and so we get that $$g(h)=4p(h)=2.$$ This cannot be true.

This was the solution given, but I cannot understand why $p(h)$ cannot be $1/2$. Also, why does it have integral coefficients?

Could somebody explain?

Answer 1

Since $g$ has integer coefficients, then also $p$ has integer coefficients, provided $a,b,c,d$ are integer.

If they are distinct, then $h-a$, $h-b$, $h-c$, $h-d$ are four distinct divisors of $2$, so their product is $4$.

Thus you get the impossible relation $4p(h)=2$.

It’s impossible because, for integer $h$, $p(h)$ is integer.

Note 1. I had to add the conditions that $a,b,c,d$ are distinct integers and that $h$ is integer. Otherwise the statement is false, in general.

Note 2. If a polynomial with integer coefficients $g(x)$ is divided by the polynomial $x-a$ ($a\in\mathbb{Z}$), then the quotient has integer coefficient as well: just think to the long division algorithm, or prove it by induction on the degree of $g(x)$. The fact is obvious if $g(x)$ is constant. Suppose it has degree $n>0$ and that the result is true for polynomials of degree $<n$. Then $g(x)=cx^{n}+h(x)$, where $h$ has degree $<n$. Then we can write $$ g(x)=cx^n-ca^n+h(x)+ca^n=(x-a)(cx^{n-1}+cax^{n-2}+\dots+ca^{n-1})+(x-a)h_1(x) $$ by the well known formula about $x^n-a^n$ and the induction hypothesis on $h(x)+ca^n$ which has degree $<n$ and so $h_1(x)$ has integer coefficients.

Answer 2

Let $p(x)=q(x)+r(x)$, where $q(x)$ has integer coefficients and $r(x)$ contains the fractional parts. Then $$(x-a)(r_1x^k+r_2x^{k-1}+\ldots) \\ =r_1x^{k+1}+(r_2-ar_1)x^k+\ldots $$ so the leading coefficient of $(x-a)r(x)$ is the same as the leading coefficient of $r(x)$. So $(x-a)r(x)$, and then $(x-a)p(x)$, has a coefficient with a fraction. Repeat with $(x-b),(x-c)$ and $(x-d)$, to see that $f(x)$ must have a fractional coefficient. Since it doesn't, $r(x)$ contains no terms at all, so $p(x)$ has integer coefficients.

Answer 3

Here's a simpler approach to the proof:

$\rightarrow$Similar to previous approaches, Let $g(x)=f(x)-1991$

Hence, we have: $g(x)=p(x)(x-a)(x-b)(x-c)(x-d)$, where, $p(x)$ is a polynomial with integer coefficients.

Now, assume for the sake of contradiction:

$\exists h \in \mathbb{Z}: f(h)=1993$ $\implies g(h)=f(h)-1991=2$

NOTE: $2$ is a prime number.

From our hypothesis: $$g(h)=p(h)(h-a)(h-b)(h-c)(h-d)=2......(A)$$ Where,$(h-a),(h-b),(h-c),(h-d)$ are all distinct integers [since, $h,a,b,c,d\in\mathbb{Z}$]

METHOD 1: Can the multiplication of $4$ distinct integers be $2$ ?

METHOD 2: For $(A)$ to hold true we need atleast two terms of:$(h-a),(h-b),(h-c),(h-d)$ to be exactly equal to $1$.

Lets say, $(h-a)=(h-b)=1\implies a=b$, which clearly violates the given property of uniqueness for $a,b,c,d$ ; Similar for other casses too.

$\rightarrow\leftarrow$ (CONTRADICTION)

Any corrections for ambiguities is openly accepted. Hope you like my approach to the problem :)