Proof: polynomial with integer solutions

[Johulus]

I would appreciate it if someone could help me with this proof that I should perform. The task says:


If f(x) is a polynomial with integer coefficients, and if f(a)=f(b)=f(c)=-1, where a,b,c are three unequal integers, the equation f(x)=0 does not have integer solutions. Prove!


I know that if polynomial f(x) has integer coefficients and if it has integer solutions then that same solutions are divisors of coefficient that does not have x next to it. Now, I do not know how to include this with all the information that I got.

 

[Ishuda]

I would appreciate it if someone could help me with this proof that I should perform. The task says:


If f(x) is a polynomial with integer coefficients, and if f(a)=f(b)=f(c)=-1, where a,b,c are three unequal integers, the equation f(x)=0 does not have integer solutions. Prove!


I know that if polynomial f(x) has integer coefficients and if it has integer solutions then that same solutions are divisors of coefficient that does not have x next to it. Now, I do not know how to include this with all the information that I got.

Write f(x) as
f(x) = (x-a) (x-b) (x-c) g(x) - 1
where g(x) is any polynomial with integer coefficients. What would g(x₀) have to be if f(x₀) were zero?

Why can you write f(x) that way?

 

[Johulus]

Write f(x) as
f(x) = (x-a) (x-b) (x-c) g(x) - 1
where g(x) is any polynomial with integer coefficients. What would g(x₀) have to be if f(x₀) were zero?

Why can you write f(x) that way?

\(\displaystyle f(a)=f(b)=f(c)=-1 \implies f(a)+1=f(b)+1=f(c)+1=0\)

If we imagine that f'(x) is a polynomial and f'(x)=f(x)+1 then:

\(\displaystyle f'(x)=0 \implies f(x)=-1 \)

But we know that \(\displaystyle f(a)=f(b)=f(c)=-1 \), so then we have:

\(\displaystyle f'(a)=f'(b)=f'(c)=0 \).

And we know that every polynomial can be writen as:

\(\displaystyle f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=a_n(x-x_1)(x-x_2) \cdot \cdot \cdot (x-x_n) \), where \(\displaystyle x_1,...,x_n \) are zeros of the polynomial.

If we assume that the leading coefficient of f(x) \(\displaystyle a_n=1\) then the leading coefficient of f'(x) is the same.

Since a,b,c are zeros of the polynomial f'(x) then we have:

\(\displaystyle f'(x)=(x-a)(x-b)(x-c) \cdot \cdot \cdot (x-x_n) \).

Now if we note that all other remaining factors up to \(\displaystyle (x-x_n) \) form another polynomial g(x), then we have:

\(\displaystyle f'(x)=(x-a)(x-b)(x-c)g(x) \).

We used \(\displaystyle f'(x)=f(x)+1 \), so now we have:

\(\displaystyle f(x)+1=(x-a)(x-b)(x-c)g(x) \implies f(x)=(x-a)(x-b)(x-c)g(x)-1 \).

\(\displaystyle f(x)=0 \implies (x-a)(x-b)(x-c)g(x)-1=0 \implies (x-a)(x-b)(x-c)g(x)=1 \).

Task states that \(\displaystyle \nexists x \in\mathbb{Z} : f(x)=0 \), but let's assume \(\displaystyle \exists x \in\mathbb{Z} : f(x)=0 \).

And let's say that \(\displaystyle d \in\mathbb{Z}: f(d)=0 \), then we have:

\(\displaystyle f(d)=(d-a)(d-b)(d-c)g(d)-1=0 \implies (d-a)(d-b)(d-c)g(d)=1 \).

Since g(x) is a polynomial with integer coefficients \(\displaystyle g(x) \in\mathbb{Z}, \forall x \in\mathbb{Z} \), and \(\displaystyle a,b,c,d \in\mathbb{Z} \implies (d-a), (d-b),(d-c) \in\mathbb{Z} \), then \(\displaystyle (d-a)(d-b)(d-c)g(d)=1 \) is possible when g(d)=1, two other factors are -1 and one is 1, when g(d)=-1, one factor is -1 and two others are 1, g(d)=1 and all other factors are 1, or g(d)=-1 and all other factors are -1. So, at least two of the factors \(\displaystyle (d-a),(d-b),(d-c) \) are the same. So, let's assume that:

\(\displaystyle (d-a)=(d-b)=1, (d-c)=-1, g(d)=-1 \, \implies \, d-a=d-b \qquad a=b \), but task states that \(\displaystyle a \neq b \neq c \). We have contradiction, so our assumption that \(\displaystyle \exists x \in\mathbb{Z} : f(x)=0 \) is wrong because for \(\displaystyle x=d, d \in\mathbb{Z}, f(d)=0 \) we got contradiction \(\displaystyle a=b \qquad vs. \qquad a \neq b \neq c \). Therefore \(\displaystyle \nexists x \in\mathbb{Z} : f(x)=0 \).