Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $n>1$, let $a_1, a_2, \dots, a_n$ be $n$ distinct integers. Prove that the polynomial $$f(x)=(x-a_1)(x-a_2)...(x-a_n) - 1$$ cannot be written as the product of two nonconstant polynomials with integer coefficients.

My Proof (or attempt)

Assume that $f(x)$ can be written as $h(x)\cdot g(x)$. Note for any $x$, $f(x)$ must be prime. This means either $h(x)$ or $g(x)$ must equal $1$, but since these polynomials must be non-constant, we have a contradiction, and we are done.

Source: Art of Problem Solving Vol. 2

share|improve this question
4  
In short, "yes". What leads you to conclude that f(x) must be prime for any x? (surely you mean any integer x, but even that isn't possible). –  NKS Dec 31 '11 at 18:30
1  
To expand on the above comment, was the OP's intuition that if you take a product of "many" integers, then subtract 1, you get a prime? This is false. –  idmercer Dec 31 '11 at 22:58
    
This would just give us a fantastic generator for prime numbers ! :D –  user20010 Dec 31 '11 at 23:18
    
So yes, you proof is wrong. –  user20010 Dec 31 '11 at 23:24
    
@idmercer Haha yes, it was. I don't know why I believed so, but it just felt "right" but Ismail brings up a nice point ;) –  MathMathCookie Jan 1 '12 at 16:59

1 Answer 1

up vote 8 down vote accepted

Suppose $f=gh$. Then $-1=f(a_i)=g(a_i)h(a_i)$. Since $g(a_i)$ and $h(a_i)$ are integers, we have $g(a_i)=-h(a_i)$ for all $i$. If both $g$ and $h$ have degree less than $n$, this implies that $g=-h$. But then $f=-g^2$, which cannot happen because the leading coefficients cannot match.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.