7

I know the proof for $(1+x)^p\equiv 1+x^p\mod p$ using the binomial theorem. Moreover, I know that $x^p \equiv x \mod p$ due to Fermat's theorem.

Hence, is $(1+x)^p\equiv(1+x)\equiv1+x^p \mod p$ a correct proof of this relation?

After thinking about it for a bit, one of the proof of Fermat's theorem uses the binomial theorem so my comments might have been redundant (although Fermat's theorem can be deduced from Lagrange's theorem).

I guess you can try to prove Fermat's by Pigeon hole:

Assuming $a\not\equiv 0$

$$a^i,\;\;1\le i \le p$$ takes $p$ values but $\mod p$ can only take $p-1$ distinct values ($a^p \not \equiv 0$ unless $a=0$) and so $a^i\equiv a^j$ and take inverses and the rest follows. If $a\equiv 0$, then the statement follows.

This can be extended to $(1+x)^{p^n}\equiv (1+x^{p^n})$ without using induction since $p |p^n$, then $$(1+x)^{p^n}\equiv (1+x)^{p\cdot{p^{n-1}}}\equiv 1+x \equiv 1+x^{p^n}$$

I have a gut feeling that says I overlooked something important.

If this is correct, what additional instructive value does the proof using the binomial theorem has that the Fermat's proof doesn't?

PS: I don't know what to say on the title. Feel free to edit it.

  • 3
    Yep, it is correct (of course, $;p;$ is a prime) . – DonAntonio Jun 14 '20 at 19:47
  • 4
    I'd advise always using the congruence sign ($\equiv$) with congruences; $a\equiv b\pmod n$ is fine, $a=b\pmod n$ is suspicious. – Angina Seng Jun 14 '20 at 20:03
  • 1
    @AnginaSeng: $a\equiv b\bmod n$ means that $b-a$ is divisible by $n$. But $a=b\bmod n$ means that $a$ is the canonical representative of the residue class of $b$, that is to say, $a\equiv b\bmod n$ and $0\le a<n$. (In this case, you are quite right to call out the OP.) – TonyK Jun 14 '20 at 20:24
  • 3
    You may be missing the fact that $(1+x)^p\equiv1+x^p$ may be meant to be a congruence of formal polynomials (in my opinion that is the most useful way to interpret it). See here for a discussion as to why we want our polynomials to be formal (as opposed to functions). – Jyrki Lahtonen Jun 14 '20 at 20:27
  • @TonyK Was that what was intended in the proposer's original question (since edited) when they wrote "$(1+x)^p=(1+x)=1+x^p \mod p$"? – Angina Seng Jun 14 '20 at 20:27
  • @AnginaSeng: No indeed! That is why I added that last sentence in parentheses. – TonyK Jun 14 '20 at 20:28

3 Answers3

3

That's not only valid (albeit only for $\Bbb F_p$) but the neatest proof of it I've ever seen, much more cunning than using the binomial theorem (which, as @reuns & @JyrkiLahtonen noted, covers $\Bbb F_p(t)$ on a commutative ring of characteristic $p$).

J.G.
  • 115,835
3

$$(1+x)^p = 1+x^p$$ is true not only for $x\in \Bbb{F}_p$ but also for $x\in \Bbb{F}_p[t]$ or any commutative ring of charactistic $p$.

In contrary $a^p = a$ is true only for $a\in \Bbb{F}_p$ (as $t\ne t^p$ in $\Bbb{F}_p[t]$)

reuns
  • 77,999
2

I would read the congruence $$(1+x)^p\equiv1+x^p\pmod p$$ to involve polynomials in the ring $\Bbb{Z}[x]$.

Two polynomials $a(x)=a_0+a_1x+\cdots+a_nx^n$ and $b(x)=b_0+b_1x+\cdots+b_mx^m$ with integer coefficients are said to be congruent modulo $p$, if and only if $a_i\equiv b_i$ for all $i$ in the range $[0,\max\{m,n\}]$.

You must not think of $x$ as an integer. Instead, it is an indeterminate. Meaning that $x^p$ and $x$ are not congruent modulo $p$. For example their linear terms have coefficients $0$ and $1$ respectively. And $0\not\equiv1\pmod p$.

Observe that, by the binomial theorem, $$ (1+x)^p\equiv 1+x^p\pmod p $$ even though $x$ and $x^p$ are not congruent modulo $p$.

Another way to look at this is to say that by the binomial theorem we have the identity (not a congruence!) in the polynomial ring $\Bbb{Z}_p[x]$: $$ (1+x)^p=1+x^p. $$ Observe that, for example, in the polynomial ring $\Bbb{Z}_2[x]$ we don't have the identity $x=x^2$. The polynomial on the left has degree one whereas the polynomial on the right hand side has degree two.

Yet another way to see the difference is to let $x$ take values from e.g. the ring of Gaussian integers $\Bbb{Z}[i]$. For example, with $p=3, x=1+i$ we see that $$(1+i)^3=-2+2i.$$ This is not congruent to $(1+i)$ modulo $3$, so $x\not\equiv x^3\pmod 3$ for this choice of $x$. But, we have $$ (1+x)^3=(2+i)^3=2+11i $$ and $$ 1+x^3=1+(1+i)^3=-1+2i. $$ Their difference is $$ (1+x)^3-(1+x^3)=3+9i=3(1+3i) $$ a multiple of three. All in line with the congruence $$(1+x)^3\equiv1+x^3\pmod3.$$

Jyrki Lahtonen
  • 133,153
  • Hmm, if $a(x)$ and $b(x)$ had all their roots in $\mathbb{Z}$ and they have the same roots with the same multiplicity wouldn't that imply $a=b$? or is it the fact that their coefficients are the same that establishes their identity? –  Jun 14 '20 at 20:33
  • @crystal_math By definition the polynomials $a(x)$ and $b(x)$ are the same if and only if all the corresponding coefficents are equal. In other words $a_i=b_i$ for all $i=0,1,\ldots,\max{m,n}$. That's why they are called formal polynomials. – Jyrki Lahtonen Jun 14 '20 at 20:35
  • Brilliant example. Thank you! –  Jun 14 '20 at 20:52