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Can someone please explain it to me like you would explain to an idiot?

tried to read about it in Burton's, watch videos and read answer's from here on questions about the subject and I don't get it yet. Here's a document presenting the proof I found on the internet: https://people.maths.bris.ac.uk/~mazag/nt/lecture6.pdf

we try to prove by induction that any function $f(x)$ in some degree n has n solutions or less mod p, when p is a prime. its easy to prove it for the 1st degree.

then we assume solution a for the polynomial $f(x)$ and we define $f(x)-f(a)=(x-a)g(x)$ , where starts the part that really confuses me. how is $-ag(x)=-f(a)\,$? shouldn't it be $g(a)$ instead?... the reasoning behind our choice to proceed through this idea is really unintuitive for me.

Thanks a lot!

user670028
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    Welcome to stackexchange. We can't help you with just this fragment. Please [edit] the question to include the whole algebraic calculation that gives you trouble, along with enough context so we know what the variables mean. Point to the particular place you are stuck. – Ethan Bolker Apr 30 '19 at 22:09
  • If $f(x)$ is a polynomial, then $f_1(x)=f(x)-f(a)$ is a polynomial as well and $f_1(a)=f(a)-f(a)=0$. Thus $f_1(x)$ is divisible by $x-a$. Hence $f_1(x)=(x-a)g(x)$, for some $g$. There's no reason for $-ag(x)=-f(a)$ nor for $-ag(a)=-f(a)$. – egreg Apr 30 '19 at 22:50
  • This does make it much clearer. just not completely sure why does this consists a proof, is it because $f1(x)$ is a function of n degree and g(x) of n-1 and we prove it for every function? how is showing this for this function is viewed as a proof about all other functions of n degree? – user670028 Apr 30 '19 at 22:58
  • There are already many proofs here of the Factor Theorem e.g. here But that doesn't help us debug your problem. Where did you get $\ - a g(x) = -f(a)?\ $ – Bill Dubuque Apr 30 '19 at 23:19
  • they probably got it from thinking $f(x)=x$ @BillDubuque –  Apr 30 '19 at 23:36
  • I thought about $g(x)(x-a)=f(x)-f(a)$ when $g(x)x=f(x)$ which made sense because it is a polynomial in degree $n-1$. but I understand now that if $f'(x)=f(x)-f(a)$ then $f'(x)=f(x) mod(p)$, so we derive from that that $(x-a)|f(x)$, and we can define a polynomial $g(x)=f(x)-f(a)$. I don't get why $g(x)$ is degree $(n-1)$ and why is it a function of $x$. I guess that I should read about the Factor theorem to understand these ? – user670028 May 01 '19 at 18:38

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A few things:

  • not all functions are polynomials, it applies to polynomials.

  • by division by $x-a$ we have $g(x)=\frac{f(x)-f(a)}{x-a}$

  • $f(x)=a_nx^ n + a_{n−1}x ^{n−1} +\cdots + a_0$ which evaluated at a is $f(a)=a_na^ n + a_{n−1}a ^{n−1} +\cdots + a_0$

  • This makes their difference, the coefficients times a difference of relevant powers termwise.
  • those power differences have $x-a$ as a factor. factoring it out, leaves a polynomial with smaller degree.
  • We've assumed that all lower degrees work.
  • Therefore, our next higher degree is proven.
  • Thank you, i'm starting to understand this now. However i'm not sure about how factoring $x-a$ out ensures function $g(x)$ which is in degree $n-1$, as this is essential for the proof? – user670028 May 01 '19 at 18:27
  • $x^m=((x-a)+a)^m $ which has every term except the two $(x-a)^m, a^m$ of lower exponent you are subtracting the latter, and dividing out the $x-a$ ,decreases the degree of the former. –  May 01 '19 at 18:35
  • Sorry, could you rephrase that or link me to a page explaining this? – user670028 May 02 '19 at 06:26
  • https://en.m.wikipedia.org/wiki/Polynomial_remainder_theorem –  May 02 '19 at 08:47