Let $\mathbb{Z}\_p$ be a finite field of order $p$ and $\mathbb{Z}\_p^2$ be a $2$-dimensional vector space over $\mathbb{Z}\_p$. We consider the distance $\lVert \cdot \rVert:\mathbb{Z}\_p^2\to \mathbb{Z}\_p$ defined by $\lVert {x}\rVert:=x\_1^2+x\_2^2$, where ${x}=(x\_1,x\_2)$. Suppose that $E\subset \mathbb{Z}\_p^2$ and consider the function $$\nu\_n(t\_1,\dots,t\_n)=\#\{(x\_1,\dots,x\_{n+1})\in E^{n+1}: \lVert x\_i-x\_{i+1}\rVert=t\_i \ \text{for}\ i=1,\dots n\}.$$


I claim that $$\nu\_n(t\_1,\dots,t\_n)=\sum \limits\_{x\_{n+1}\in E}((\dots((E\*S\_{t\_1})E\*S\_{t\_2})\dots)E\*S\_{t\_n})(x\_{n+1}),$$ where by $\*$ I denote the convolution $f\*g:\mathbb{Z}\_p^2\to \mathbb{C}$ of $f,g:\mathbb{Z}\_p^2\to \mathbb{C}$ defined by $(f\*g)(m)=\sum \limits\_{x\in \mathbb{Z}\_p^2}f(x)g(m-x).$


I checked that this formula is valid for $n=1,2,3,4$. But I cannot prove it via induction, i.e. I have issues with induction step.


I'd be very thankful if someone can show the solution please.


**EDIT:** We use $S\_t$ to denote the sphere of radius $t$ in $\mathbb{Z}\_p^2$: thus $S\_t=\{x\in \mathbb{Z}\_p^2: \lVert x\rVert=t\}.$ $E$ is a subset of $\mathbb{Z}\_p^2$ and $E^{n+1}$ is just a Cartesian product of $E$ with itself $(n+1)$ times. By $E(x)$ and $S\_t(x)$ I denoted indicator functions of $E$ and $S\_t$.



Mathematical induction and the counting function on $\mathbb{Z}_p^2$
For $y\in E$, define
$$
\mu\_n(t\_1,\ldots,t\_n,y)=\#\{(x\_1,\ldots,x\_n)\in E^n\colon \|x\_i-x\_{i+1}\|=t\_i,
$$
where $x\_{n+1}$ is taken to be $y$.


Then your $\nu\_n(t\_1,\ldots,t\_n)$ is just $\sum\_{y\in E}\mu\_n(t\_1,\ldots,t\_n,y)$. It's easy to see that for $z\in E$,
$$
\mu\_{n+1}(t\_1,\ldots,t\_n,t\_{n+1},z)=
\sum\_{y\in E, z-y\in S\_{t\_{n+1}}}\mu\_n(t\_1,\ldots,t\_n,y).
$$
That is,
$$
\mu\_{n+1}(t\_1,\ldots,t\_n,t\_{n+1},z)=
\big([\mu\_n(t\_1,\ldots,t\_n,\cdot)\mathbf 1\_E(\cdot)]\*S\_{t\_{n+1}}\big)(z).
$$
Hence by induction,
$$
\mu\_n(t\_1,\ldots,t\_n,z)=
(\cdots((\mathbf 1\_E\*S\_{t\_1})\mathbf 1\_E)\*S\_{t\_2})\cdots\mathbf 1\_E)\*S\_{t\_n}(z).
$$
Your formula then follows.


