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If we remove the condition to be working modulo $p$ prime. Why does the theorem no longer hold? Would I be correct in thinking it is because in the induction we say $f(X)\equiv(X-a)g(X)$ so potentially if $f(X)\equiv0$ then one of the cases to consider is $X\equiv a \mod p$ which has at most one solution?. (I think maybe this is where it would break down if we worked modulo not a prime) But I can't see why explicitly it wouldn't work.

I didn't state the theorem of proof for the sake of brevity as I assume it is well known but if you need any clarification just ask.

Thanks!

Suppose that $p$ is prime and that $f(X)$ is a polynomial with integer coefficients and degree $d ≥ 0$. Then the equation $f(X) ≡ 0 \mod p$ has at most $d$ distinct solutions modulo $p$.

Ryan S
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    Break down where? What Lagrange are you talking about? – Adam Hughes Nov 21 '16 at 22:01
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    Lagrange's theorem talks about the number of distinct solutions modulo a prime. If we remove this and see we can work modulo (any natural number) the theorem no longer holds. I am asking why it stops working when we remove that condition. That is why is that condition necessary in the first place. – Ryan S Nov 21 '16 at 22:05
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    @RyanS What "Lagrange theorem" are you talking about? Solutions to what ? Please do give some link to that. – DonAntonio Nov 21 '16 at 22:08
  • https://en.wikipedia.org/wiki/Lagrange's_theorem_(number_theory) – Ryan S Nov 21 '16 at 22:10
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    @RyanS Never heard of that with that name, and there are several very well known theorems with the name of Lagrange. It may be a good idea to quote exactly the theorem you linked. – DonAntonio Nov 21 '16 at 22:16
  • Thanks for that I apologise for assuming it was well known I have added the theorem as requested. – Ryan S Nov 21 '16 at 22:18
  • See this answer. – Bill Dubuque Jul 27 '17 at 13:34

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You are close. The essence of the issue is a careful analysis of the little theorem:

If $ab \equiv 0 \pmod{m}$, then $a \equiv 0 \pmod{m}$ or $b \equiv 0 \pmod{m}$. If $m$ is prime, this is true, and is just a restatement of Euclid's Lemma: https://en.wikipedia.org/wiki/Euclid's_lemma.

But it never holds modulo a composite $m$. To see this, simply let $m = ab$ with $a$ and $b$ proper factors of $m$.

So back to your question, the problem is with the induction after writing

$$ f(x) \equiv (x-a) g(x) \pmod{m}$$

We would then like to say that the remaining roots of $f$, other than $a$, are all roots of $g$. If $b \not\equiv a \pmod{m}$ is such that $f(b) \equiv 0 \pmod{m}$, then

$$ 0 \equiv (b-a) g(b) \pmod{m}$$.

We assumed that $b - a \not\equiv 0 \pmod{m}$. If $m$ is a prime, then by Euclid's Lemma, we can say $g(b) \equiv 0 \pmod{m}$. One can then finish up the induction.

But if $m$ is composite, then it may be that $g(b) \not\equiv 0 \pmod{m}$ and the proof breaks down.

For example, let $m=65$, $f = x^2 + 1$, and $a=8$. Then

$$ f(x) \equiv (x-8)(x+8) \pmod{65}, $$

so we may set $g(x) = x+8$. But $f(18) \equiv 0 \pmod{65}$, and $g(18) \not\equiv 0 \pmod{65}$.

Notice that we could instead have started with the root $18$ and factored $f(x) \equiv (x-18)(x+18) \pmod{65}$. We can always factor a linear term out from $f$ corresponding a root of $f$ -- but with a composite modulus, if you fix two distinct roots, you may not be able to factor out linear terms corresponding to both roots simultaneously.

Barry Smith
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