# The number of solutions of the equation $x^n+y^n=1$ over $\mathbb{Z}_p$ with $\mathbb{Z}_p=\{\alpha\in \mathbb{Q}_p:|\alpha|_p\le 1\}$.

The number of solutions of the equation $x^n+y^n=1$ over $\mathbb{Z}_p$ with $\mathbb{Z}_p=\{\alpha\in \mathbb{Q}_p:|\alpha|_p\le 1\}=\{\sum_{i=0}^\infty a_ip^i:0\le a_i \le p-1\}$.

I was trying to show that this equation actually has infinitely many solutions. I felt that Hensel's lemma will be helpful here, but I cannot figure out how to apply it.

-

Edit Correct a typo and make things more precise.

We want to show that for any $N$ big enough and any $a\in \mathbb Z_p$, there exists a solution with $y=p^Na$ and $x\equiv 1 \mod p$.

1. If $n$ is prime to $p$, then for any $b\in\mathbb Z_p$, the equation $(1+pw)^n=1+pb$ has a solution with $w\in \mathbb Z_p$. Or, if you prefer, $1+pb$ has a $n$th root in $\mathbb Z_p$, congruent to $1$ mod $p$.

2. If $n=p$, then for any $b\in\mathbb Z_p$, the equation $(1+pw)^n=1+p^2b$ has a solution with $w\in \mathbb Z_p$.

Both statements are proved by developping the left-hand side, simplifying by $1$ and dividing by $p$ (resp. $p^2$) to get an equation $$F(w)=0$$ with $F(w)\in \mathbb Z_p[w]$ (not monic) and $F(w)\equiv w+b \mod p$· Then apply Hensel Lemma which states that there is a solution. Moreover, inspecting the form of $F(w)$, we see that any solution satisfies $v_p(w)=v_p(b)$.

For a general $n$, write $n=mp^r$ with $m$ prime to $p$. Using repeatly (1) and (2) above, we see that for any fixed $a\in\mathbb Z_p$, the equation $$(1+pw)^n=1+p^{2r+1}a$$ has a solution with $w\in\mathbb Z_p$.

Now back to your equation. Fix any $a\in \mathbb Z_p$ and any $N\ge 2$. We are looking for solutions of the form $y=p^Na$ and $x=1+pw$. Then your equation becomes: $$(1+pw)^n=1-p^{nN}a^n$$ (I forgot the $n$th power in the RHS) with $w\in \mathbb Z_p$. As $nN\ge 2r+1$, this equation has a solution. By varying $N$ or $a$, we get infinitely many solutions for $x^n+y^n=1$ with $x=1+pw$ and $y=p^Na$.

-
Thanks for answering! Could you elaborate your solution a little bit? – Qiang Zhang Apr 18 '12 at 16:31
Can you first try the case $v_p(n)=0$ ? – user18119 Apr 18 '12 at 17:02
I don't get why do we write $y$ as $-p^N z$ and why does the induction really talk about. Grateful for your explanation. – Qiang Zhang Apr 18 '12 at 18:37
Sorry. The idea is to find solutions with $y\equiv 0$ and $x\equiv 1$. Then one has to show that when $y$ is small enough (in the $p$-adic sense), $1+y$ has a $n$th root close to $1$. If $n=pm$, then taking an $n$th root is same as taking first an $m$th root, then a $p$th root of the $m$th root. Hope this helps. – user18119 Apr 19 '12 at 0:07
@QiL, thanks for the expansion! It helps a lot! Btw, 'Shwo' is mistyped. – Qiang Zhang Apr 21 '12 at 15:44