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Let $$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$$ be a polynomial with integer coefficients and whose degree is at least 2. Suppose each $a_i$, ($0\leq i\leq {n-1})$ is of the form $\displaystyle a_i=\pm \frac{17!}{r!(17-r)!}$. Show that $f(x)=0$ has no integral solution.

My work :

Choose $p=17$ . We see that each $a_i$ ,is divisible by $17$ but $p^2$ doesn't divide $a_0$ due to this Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$ .

So $f(x)$ is irreducible over integers. Hence has no zero. Is my proof correct?

user26857
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Suman Kundu
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    your proof is correct but you should add that the result you are using, is Eisenstein's criterion and how all it's requirements are met.. – Anurag A Jun 02 '17 at 20:50
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    If you don't suppose $0<r<p$ for all $i$ (is it a possibly different $r$ for each $a_i$ ? You don't precise this either), then you proof is not correct.

    For example, $a_i=(-1)^{i+1}$ for all $i$ (case $r_i=0$ or $p$ for all $i$), and $n$ is odd, then $1$ is root of $f(x)=x^n-x^{n-1}+x^{n-2}-\dots+x-1$.

     – Nicolas FRANCOIS Jun 02 '17 at 21:39
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    The question in the title is different from the actual question. It might be worth phrasing the title question a little more generically ('showing a certain polynomial has no integer roots' or the like) to avoid confusion, because the title's question as it stands is false. – Steven Stadnicki Jun 02 '17 at 22:01

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