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Find all polynomials $f(x)$ such that $\deg(f)=4$ and such for all $x\in Z$ then $f(x)\in Z$.

My try: since $f(x)=x^4$ is such it

because $$x\in Z\Longrightarrow x^4\in Z$$ and $$f(x)=a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0},~~~~~~~~~~~~~~a_{0},a_{1},a_{2},a_{3},a_{4}\in Z$$ is such it and have other form? Thank you.

Adriano
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math110
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    Consider the polynomials $\frac{x(x-1)(x-2)\cdots (x-k+1)}{k!}$. All integer linear combinations of these have the desired property, and that's it. In the case degree $4$, we have a positive coefficient term with $k=4$, and lower degree terms. – André Nicolas Dec 19 '13 at 04:10
  • oh,Thank you,But this have other form? Thank you – math110 Dec 19 '13 at 04:16
  • What is $Z$? Do you mean $\mathbb{Z}$? – Lost Dec 19 '13 at 04:25
  • The result mentioned in the above comment is the only general characterization of the the general polynomials that are integers at integers that I know. Probably a proof has appeared on MSE. – André Nicolas Dec 19 '13 at 04:25

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Hint $\ $ Write $\displaystyle\ f(x) = c_0 + c_1 {x \choose 1} + \cdots + c_4 {x \choose 4},\ $ possible since $\displaystyle {x\choose k}\,$ has degree $k$. Then

$$\begin{eqnarray} f(0) &=& c_0 \\ f(1) &=& c_0 + {1\choose 1} c_1 \\ f(2) &=& c_0 + {2\choose 1} c_1 + {2\choose 2} c_2 \\ f(3) &=& c_0 + {3\choose 1} c_1 + {3\choose 2} c_2 + {3\choose 3} c_3 \\ f(4) &=& c_0 + {4\choose 1}c_1 + {4\choose 2}c_2 + {4\choose 3} c_3 + {4\choose 4} c_4 \end{eqnarray}$$

Successively solving the above triangular system for $\,c_0,\ c_1,\ c_2,\ldots$ shows that all $\,c_i\in \Bbb Z$, since all $f(i), {j\choose k}\in \Bbb Z\,$ and each linear equation in $\,c_i\,$ is monic, i.e. has lead coef $= {i\choose i} = 1.\,$ Conversely, if all $c_i\in \Bbb Z$ then $f(n)\in \Bbb Z$ since all ${x \choose i}$ are integer valued.

Thus $f(x)$ is integer-valued $\iff$ its coefficients in a binomial basis are all integers.

Clearly the same proof works for polynomial of any degree. This is a well-known result of Polya and Ostrowski.

Bill Dubuque
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  • oh,Nice,so we can have this relsut: if $f(x)$ is real polynomial of the degerr is $n$,such as $f(0),f(1),\cdots,f(n)$ is integer,then $f(x)\in Z,\forall x\in Z$ – math110 Dec 19 '13 at 04:37
  • This is essentially a proof by Newton interpolation, for the sample points $0,1,2,3,4$, if one want an official name for the method. – Lutz Lehmann Dec 19 '13 at 11:57