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I need to prove the following result:

Let $a\in(\mathbb{Z}/q\mathbb{Z})[X]$ with $q$ prime. Then $$a(1)\equiv 0 \pmod{q} \;\Rightarrow\; (X-1)\,|\,a(X) \;\text{ in }\; (\mathbb{Z}/q\mathbb{Z})[X]$$

So far I have only managed to prove the converse. Any hint toward the right direction is therefore appreciated.

Kristian S. Jensen
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  • I have attempted to use polynomial long division to compute $a(X)$ divided by $X-1$. Is the point that the remainder should be $a_n+a_{n-1}+\cdots+a_0=a(1)$, and then we apply $a(1)\equiv 0 \pmod{q}$ to conclude that the remainder is zero, thus implying $(X-1),|,a(X)$? – Kristian S. Jensen Mar 26 '21 at 12:36
  • That would work. – robjohn Mar 26 '21 at 13:12

1 Answers1

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Let $a(X)= \sum a_k X^k$ then, since $a(1)=\sum a_k = 0 $ :

$a(X)= a(X)-a(1)=\sum a_k(X^k-1)=\sum a_k(X-1)(X^{k-1}+...+1)=\\(X-1)\times(\sum a_k(X^{k-1}+...+1) )$

Therfore $X-1 | a(X)$

Mohammed M. Zerrak
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