Answer without using mod Let $\frac{a}{b}=1+\frac{1}{2}+... +\frac{1}{p-2}$. Show that $a-b$ is divisible by $p$.

Question

PROBLEM

Let the natural numbers $a,b,p$ be non-zero, and $p$ be an odd prime number, so that the relation $\frac{a}{b}=1+\frac{1}{2}+... +\frac{1}{p-2}$. Show that $a-b$ is divisible by $p$.

WHAT I THOUGHT OF

I think that the formula $\frac{y-x}{x*y}=\frac{1}{x}+\frac{1}{x}$ as we can rewrite the second part to fit. I don't know what else to do or how to start solving it! Hope one of you can help me!

Answer 1

Interpret everything modulo $p$. First of all, we have $$ 1^{-1} + 2^{-1} + \ldots + (p-1)^{-1} = 1 + 2 + \ldots + (p-1) = \frac{p-1}{2} \cdot p \equiv 0 \mod p. $$ Because $(p-1)^{-1} = -1$, it follows that $$ 1^{-1} + 2^{-1} + \ldots + (p-2)^{-1} \equiv 1 \mod p, $$ and we obtain $a \equiv b \mod p$.

Answer 2

The cyclic group $\mathbb{Z}/p\mathbb{Z}-\{ 0 \}$ is generated by element $a$. Now $1+\frac{1}{2}+…+\frac{1}{p-2}=\frac{N}{(p-2)!}$ where $N=(p-2)!+\frac{(p-2)!}{2}+…+\frac{(p-2)!}{p-2}$. From $a(p-2)!=bN$ we have $a=bN$ mod $p$. We only need to prove that $N=1$ mod $p$. Now $N-1=\sum_{0 \leq j \leq p-2, j \neq \frac{p-1}{2}}a^{-j}-1=\sum_{0 \leq j \leq p-2, j \neq \frac{p-1}{2}}a^{j}+a^{\frac{p-1}{2}}=\sum_{0 \leq j \leq p-2}a^{j}=0$ mod $p$. Therefore we have proven the statement.

Answer 3

Because of the unicity of inverses there is bijection between the inverses and the elements in $\mathbb F_p$ therefore $$1^{-1} + 2^{-1} + \ldots + (p-2)^{-1}+(p-1)^{-1}=1+2+\cdots+(p-2)+(p-1)=0$$ which implies $$\frac ab=-(p-1)\equiv 1\pmod p\iff a-b\equiv 0\pmod p$$ We are done.