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

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.

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

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

Re: Proof: integer solutions of a polynomial

We can write



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



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)


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