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I have a question about the below proof, taken from the book by George Andrews, Number Theory, page 72 of Section 5-4. Below is the main part of it.

I am on board until it says near the bottom:

Thus, since the number of solutions of the congruence $g(x)\,\dots\;$ it follows from the induction hypothesis that $g(x)$ is congruent to $0$ mod $p$ for each integer $x$.

I see that $g(x)$ is a polynomial of degree less than $k$. And so $g(x)\equiv 0\bmod p$ can never have $k$ or more incongruent solutions. But I am not following the book's line of reasoning.

Thanks! I understand the part of the proof after this page.

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HFM
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  • Ok, I think I get this now. Basically, we constructed a $g(x)$ of degree $k-1$ which has at least $k$ solutions to $g(x)=0$ and therefore at least $k$ solutions to $g(x) \equiv 0 \bmod p$. For this to NOT yield a contradiction, we must have a situation whereby $g(x)$ is ALWAYS congruent to $0$ mod p for every integer $x$....the author on the next page then uses this to lead to a contradiction of the assumption that a degree $k$ polynomial can have $k+1$ mutually incongruent solutions mod p. – HFM Sep 29 '21 at 22:20
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    I think that there already is a contradiction, that $g(x)$ is a polynomial of degree $< k$ with $k$ mutually incongruent roots; it contradicts the induction hypothesis which is proved for polynomials of degree $<k$.

    Also, how do you reach the conclusion that $g(x)=0$ has $k$ solutions? I think that the fact that $g(x) \equiv 0 \bmod p$ has at least k solutions rather follows from that $g(w_j) = f(w_j) − 0 = f(w_j) \equiv 0 \bmod p$ for $j=1,...,k$.

    I don't really see how $g(x) \equiv 0 \bmod p$ for all integers $x$ avoids any contradiction either.

     – Paradox Sep 30 '21 at 18:46
  • But If you feel sure you've understood the proof, don't mind my comment. I might be missing something. – Paradox Sep 30 '21 at 18:47
  • I agree with you, I also never understood that why $w_{k+1}$ should be solution for $g(x)$. – voyager Jun 25 '23 at 21:44

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