 Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2016-09-20 00:45:35

ElizabethMcArthur
Member
Registered: 2016-09-20
Posts: 1

### Proof: integer solutions of a polynomial

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.

My homework solvers:
math - math helper
algebra - algebra helper

Offline

## #2 2019-05-16 01:31:01

Alg Num Theory
Member Registered: 2017-11-24
Posts: 693
Website

### Re: Proof: integer solutions of a polynomial

We can write

[list=*]
[*]

[/*]
[/list]

where g(x) is a polynomial with integer coefficients. If there were an integer solution, say f(n) = 0, then

[list=*]
[*]

.[/*]
[/list]

since they are all integers. But if a, b, c are distinct, then so are (na), (nb), (nc). This is a contradiction because (na), (nb), (nc) cannot take three distinct values if they can only be ±1.

Thus f(x) = 0 cannot have an integer solution.

Me, or the ugly man, whatever (3,3,6)

Offline

## Board footer

Powered by FluxBB