Is this a true theorem?

Question

I'm trying to prove the existence of the following theorem:

If $n,p \in \mathbb{N}$, then $(p+1)^n = 1 \mod p$

Is this theorem true? I think it is, but I don't know how to prove it!

Thanks!

Answer 1

$$p \equiv 0 \pmod{p} \iff p+1 \equiv 1 \pmod{p} \implies (p+1)^n \equiv 1 \pmod{p}.$$

Answer 2

It is true because $(p + 1)^n - 1 = (p + 1)^n - 1^n = (p + 1 - 1)((p+1)^{n-1} + (p+1)^{n-2} + ... + 1) = kp$ which is divisible by $p$.

Answer 3

Because $x-y$ is a factor of $x^n-y^n$ for any $n\in\mathbb{N}$, we have that $$(p+1)^n-1=(p+1)^n-1^n$$ is divisible by $$((p+1)-1)=p$$

Answer 4

${\rm mod}\ p\!:\,\ p+1\equiv 1\,\Rightarrow\,(p+1)^n\equiv 1^n\equiv 1\ $ by the Congruence Power Rule below.

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Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{blue}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{blue}{AB - ab} $

Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$ By induction $\,g(A)\equiv g(a)\,$ so $\, \color{#0a0}{A g(A)\equiv a g(a)}\,$ by the Product Rule. Hence $\,f(A) = f(0)+\color{#0a0}{Ag(A)}\equiv f(0)+\color{#0a0}{ag(a)} = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv\, a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by applyimg the Polynomial Rule with $\,f(x) = x^{\rm b}).$