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Let $p$ be a prime number. Then if $ f(x) = (1+x)^p$ and $g(x) = (1+x)$, then is $f \equiv g \mod p$?

I'm trying to prove that for integers $a > b > 0$ and a prime integer $p$, ${pa\choose b} \equiv {a \choose b}.$ To do this I use FLT to show that $(1+x)^{pa} \equiv (1+x)^a \mod p$ and compare the coefficients of $x^b$ to complete the proof. Am I applying FLT correctly?

In general, do most theorems regarding integers/reals generalize to polynomials over the integers/reals? Are there some common pitfalls that I could make when trying to generalize such theorems?

user1299784
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2 Answers2

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Possibly you have encountered a common stumbling point: confusing polynomial functions with formal polynomials. Over $\,\Bbb F_p,\,$ by $\,\mu$Fermat, $\,f-g\, =\, x^p -x \,$ equals the constant function $\,0,\,$ but it is not equal to $\,0\,$ as a formal polynomial since, by definition, formal polyomials are equal iff their corresponding coefficients are equal. See here for elaboration.

Bill Dubuque
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  • I think I follow you. I'm trying to prove that for integers $a > b > 0$ and a prime integer $p$, ${pa\choose b} \equiv {a \choose b}.$ To do this I use FLT to show that $(1+x)^{pa} \equiv (1+x)^a \mod p$ and compare the coefficients of $x^b$ to complete the proof. You are saying that this proof is incorrect, since even though$(1+x)^{pa} \equiv (1+x)^a \mod p$, they are not formally equivalent, so therefore their coefficients are not necessarily equivalent. Can you help me prove it correctly? – user1299784 Jan 27 '14 at 04:02
  • @user1299784 $\ {pa\choose b} \equiv {a \choose b}\ $ is $\ 2 p\equiv 2\ $ for $\ a,b = 2,1,, $ How is that true? – Bill Dubuque Jan 27 '14 at 04:21
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    So what I'm trying to prove is wrong in the first place. I was really trying to prove ${pa\choose pb} \equiv {a \choose b} \mod p.$ which I'm certain is true...I guess I have to restart since I'm going in the wrong direction. Thanks for the help. – user1299784 Jan 27 '14 at 04:28
  • Yes, I remember puzzling over this before – suckling pig Oct 28 '24 at 01:38
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The answer to the first question is NO. When you expand $g(x)$ by the binomial theorem, you will see that all of the coefficients, except for the first and last, are zero, so $g(x)$ is $1+x^p$ modulo $p.$ The difference between $f(x)$ and $g(x)$ has $p$ roots in the algebraic closure of $\mathbb{F}_p,$ which is not the same as the number of roots of the zero polynomial.

I think that gives an indication of the problems for your second question.

Igor Rivin
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  • Thanks for the reply -- I'm guessing you meant $f(x) = (1+x)^p \equiv 1+x^p$. The difference between $f$ and $g$ is therefore $x^p - x$. You say that this expression has $p$ roots, not infinite roots like the zero polynomial: but is that necessary? The difference $x^p - x$ is divisible by $p$ for all integers by FLT. It follows that since the difference between $f$ and $g$ is divisible by $p$ for all integers, $f \equiv g$ and FLT applies to polynomials, right? – user1299784 Jan 27 '14 at 03:42
  • Yes, I did mean $f(x)$ sorry. Otherwise, the point is that there is more to life than the $\mathbb{F}_p$ (or integers). – Igor Rivin Jan 27 '14 at 03:47
  • I'm not sure I follow. You're saying that FLT is fine for polynomials over integers modulo p - but what more is there? I thought FLT only applied to integers modulo a prime number anyway. – user1299784 Jan 27 '14 at 03:49
  • +70k............ – suckling pig Oct 28 '24 at 03:04