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Have problem in understanding logic given elsewhere on an old MSE post's answer.

Let $p(x)$ be a polynomial with integer coefficients satisfying that $p(0)$ and $p(1)$ having odd values. Show that polynomial $p(x)$ has no integer zeros.

Given the polynomial expression $f(x)−f(y)$, it is always divisible by $x−y$ because $x^r−y^r=(x−y)(x^{r−1}+x^{r−2}y+⋯+xy^{r−2}+y^{r−1})$.

Can consider the following cases:

  1. $x=n, y=0 : n \mid (f(n)−f(0))$, &

  2. $x = n, y=1 : (n-1) \mid(f(n)−f(1))$.

Subtracting 1 from 2, get: $−1 \mid(f(0) - f(1))$.

Issue : It is given that $-1$ is divisible by both $n$ and $n−1$.

Can I interpret it as : $-1$ is a linear combination of $n$ and $n-1$, and as they are consecutive numbers, so the $\gcd$ is anyway $1$. Obviously, $-1$ is a multiple of $\gcd=1$, and hence proved.

jitender
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  • Not following. Let $p(x)=x-2$. Then $p(0)=-2,p(1)=-1$ so both are non zero, but $p(2)=0$ so $2$ is an integer root. Please include the link to the older question. – lulu Dec 17 '17 at 12:30
  • @lulu Sorry, I am not too much into roots, and unknowingly changed the wording of the problem - assuming that it would not make a difference. The original question was: "both $p(0)$ and $p(1)$ are having odd values." I do not know it will make a difference or not, but now it is same as the original problem, and the answer given. – jitender Dec 17 '17 at 12:33
  • Please edit your question accordingly. And please include the link to the old question you cited. – lulu Dec 17 '17 at 12:35
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    Note: the statement about odd values is easily proven, as an integer root would have to be a root $\pmod 2$. – lulu Dec 17 '17 at 12:37
  • @lulu Edited. Provided link for the answer as there are many answers to the question. – jitender Dec 17 '17 at 12:38
  • I note that others answer the question the same way I did in my comment. – lulu Dec 17 '17 at 12:41
  • @lulu I am confused about the issue stated in the OP, and now also with your $mod 2$ statement. I feel need to elaborate 2nd issue too, i.e. with your earlier comment:- Why there has to be a root $(mod2)$. Do you mean that it is due to two parameters are being considered that are having values as the residue class of $2$. – jitender Dec 17 '17 at 12:45
  • I don't entirely follow the logic given in the answer you post to...you should post a question to that user directly (just add it as a comment to that answer). Specifically, I don't understand how he concludes that $n,n-1$ divide $-1$. I do however note that he has shown that $n,|,-f(0)$ and $n-1,|,-f(1)$ so, since one of $n,n-1$ is even, we get a contradiction. – lulu Dec 17 '17 at 12:46
  • Please ask the user directly if you are unclear about his post. It is an old post, but that user is still very active on the site. If you read my last comment, I think I have shown how his logic can be used to solve the problem, though I am confused about some of the things he says. – lulu Dec 17 '17 at 12:47
  • @lulu I am thankful for your responses, but if you could explain how the answer has shown: $n \mid -f(0)$. – jitender Dec 17 '17 at 12:51
  • @lulu I would request to tell some source to read all this stuff. I feel that reading from an original source would be far better. – jitender Dec 17 '17 at 12:55
  • No idea what you are saying. I agree with you that going back to the original source is the way to go, so please ask that user your question directly. As to your technical point, we have that $n,|, f(n)-f(0)$ but $n$ is a root so $f(n)-f(0)=-f(0)$. – lulu Dec 17 '17 at 12:58
  • @lulu By the word " original source", I did not mean the book from which the problem is taken, rather I need a definitive and simple source to understand such problems. I hope you could refer such a simple yet comprehensive source. – jitender Dec 17 '17 at 13:01
  • Integer and rational roots of polynomials are discussed in any basic Algebra text. Most will discuss the basics of modular arithmetic as well. – lulu Dec 17 '17 at 13:04
  • @lulu I have read from many texts, but sorry to say such an advanced treatment nobody has given so far. I have read Uspensky's 'Theory of Equations' as the most noticeable one on the topic. Hence, requested you for such an advanced treatment of the polynomials. In absence of your critical help, will only be asking such questions, as have not read such applications from an original source. Or may be, this is the way. But, very doubtful as there is too much gap in my text and such problems. – jitender Dec 17 '17 at 13:10

1 Answers1

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Can consider the following cases:

  1. $x=n, y=0 : n \mid (f(n)−f(0))$, &

  2. $x = n, y=1 : (n-1) \mid(f(n)−f(1))$.

Subtracting 1 from 2, get: $−1 \mid(f(0) - f(1))$.

One can't treat divisibility in this fashion. $$ \begin{gather*} n \mid (f(n)−f(0)) \Longrightarrow k_0 n = f(n) - f(0) \\ (n-1) \mid (f(n)−f(1)) \Longrightarrow k_1(n-1)= f(n) - f(1) \\ \end{gather*} $$ So the author of the answer that you cite can conclude that $$ k_1(n-1)-k_0n = f(0) - f(1) $$ but not that $-1 \mid (f(0) - f(1))$.

Malcolm
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  • I hope you mean that my assessment about the "Issue" having used the linear Diophantine equation, GCD, and linear combinations, is correct. If I am correct, then it is great. – jitender Dec 17 '17 at 13:05
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    Yes, your issue is correct. I misread your question originally. You are correct in raising the issue. I will edit my answer to make that clearer. – Malcolm Dec 17 '17 at 13:09