Let $f(x)$ be a polynomial with complex coefficients such that $\exists n_0 \in \mathbb Z^+$ such that $f(n) \in \mathbb Z , \forall n \ge n_0$, then is it true that $f(n) \in \mathbb Z , \forall n \in \mathbb Z$ ?
Asked
Active
Viewed 152 times
2 Answers
1
No. The ring of integer-valued polynomials (polynomials in $\mathbf Q[x]$ with integer values in $\mathbf Z$) is a counter-example. A basis, as a $\mathbf Q$-vector space made up of the generalised binomial coefficients: $$\binom{x}{n}=\frac{x(x-1)\dots(x-n+1)}{n!}.$$
Bernard
- 175,478
-
2I asked whether $f$ is integer valued for all sufficiently large integers imply $f$ is integer valued for all integers , then how come "integer valued polynomials" is a counter example ?!? – Apr 02 '15 at 08:41
-
1He who can do more can do less. :o) – Bernard Apr 02 '15 at 08:44
1
Proof by induction on the degree of the polynomial. Trivially true for degree zero. Assume true for degree $n$. Let $f$ have degree $n+1$ and be integral for all sufficiently large integer arguments. Then same is true for $g(x)=f(x+1)-f(x)$, a polynomial of degree $n$. By the induction hypothesis, $g$ is integer-valued for all integer arguments. Then it follows that $f$ is.
Gerry Myerson
- 179,216
-
How does it follow that $f$ is? We just know that $f(x+1)-f(x)\in \mathbb{Z}$. However, this might also mean that $f(x)=n+\frac{1}{2}$, $n\in\mathbb{Z}$? – user2520938 Apr 02 '15 at 09:27
-
If $f(17)$ is an integer, then so is $f(16)=f(17)-g(16)$, and then so is $f(15)=f(16)-g(15)$, and then...and then.... – Gerry Myerson Apr 02 '15 at 11:42
-
AH yes of course, i forgot that fx is integer for large enough x. Thanks – user2520938 Apr 02 '15 at 11:43