MHB Prove that there is no integer a with P(a)=8.

[anemone]

Let $P(x)$ be a polynomial with integral coefficients. Suppose that there exist four distinct integers $x_1,\,x_2,\,x_3,\,x_4$ with $P(x_1)=P(x_2)=P(x_3)=P(x_4)=5$.

Prove that there is no integer $a$ with $P(a)=8$.

 

[kaliprasad]

Let $P(x)$ be a polynomial with integral coefficients. Suppose that there exist four distinct integers $x_1,\,x_2,\,x_3,\,x_4$ with $P(x_1)=P(x_2)=P(x_3)=P(x_4)=5$.

Prove that there is no integer $a$ with $P(a)=8$.

Spoiler

This is similar to a problem that I had posted months ago. at http://mathhelpboards.com/challenge-questions-puzzles-28/polynomial-11288.html
so solution is similar

Spoiler

let Q(x) = P(x) - 5 is zero for $x_1,x_2,x_3,x_4$So $Q(x) = R(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ where R(x) is a polynomial of degree zero or moreso $Q(x)$ for x an integer is a product of at least 4 different integers as R(x) can be 1but 3 is not a product of at least 4 different integers . it is product of at most 3 different integers (-1) * (-3) * 1so Q(x) cannot be 3 and P(x) cannot be 8 for any integer x

 

[anemone]

Thanks, kaliprasad, for your solution!

And...Ah! I read that polynomial thread before, but didn't remember it at all... sorry for posting a quite similar problem in that thread here, kali! :(