Additive energy of unit circle in $\mathbb{R}^2$ and $\mathbb{Z}_p^2$

Question

1. Suppose that $u,v,u',v'\in S^1$ with $u+v=u'+v'$. Prove that $u=u',v=v'$ or $u=v',v=u'$.

Suppose $\langle \cdot , \cdot\rangle$ is usual dot product in $\mathbb{R}^2$. I was able to show $\langle u , v\rangle=\langle u' , v'\rangle$. Then I do not know how to obtain the desired result.

I'd be very thankful if someone can show how to finish this problem.

2. Suppose that $\mathbb{Z}_p$ be a finite field and $\mathbb{Z}_p^2$ be a $2$-dimensional vector space over $\mathbb{Z}_p$. Define the function $\lVert \cdot\rVert:\mathbb{Z}_p^2\to \mathbb{Z}_p$ in the following way: $\lVert x\rVert=x_1^2+x_2^2,$ where $x=(x_1,x_2)$. Denote by $S_t$ a sphere of radius $t\in \mathbb{Z}_p$, i.e. $S_t=\{x\in \mathbb{Z}_p^2: \lVert x\rVert=t\}$. Suppose that $t\neq 0$ and $u,v,u',v'\in S_t$ such that $u\neq v$. Then $u+v=u'+v'$ if and only if $u=u',v=v'$ or $u=v',v=u'$.

I was wondering how to prove this result in the case of vector space over finite field.

Answer 1

Suppose that $u,v\in S^1$ and we know the coordinates of the vector $u+v$ (also suppose that $u+v=(z_1,z_2)\neq (0,0))$. Our goal to find vectors $u$ and $v$.

Possible solution: We see that $|u+v|=\sqrt{z_1^2+z_2^2}=:r>0$. Since $|u+v|^2=|u|^2+|v|^2+2u\cdot v$ then $u\cdot v=\dfrac{r^2-2}{2}.$ Hence the angle between vectors $u$ and $v$ is $\beta:=\arccos \left(\frac{r^2-2}{2}\right)\in [0,\pi).$

Consider the line which passes through the origin and the vector $u+v$ then this line intersects unit circle at two points. Let's take the point which lies on the smaller arc (this is possible since angle between vectors is $[0,\pi)$) and denote this point by $P$. Then $P=(\cos \varphi,\sin \varphi)$, where $\cos \varphi=\frac{z_1}{r},\sin \varphi=\frac{z_2}{r}$.

Let's rotate this point $P$ clockwise by angle $\frac{\beta}{2}$ and counter-clockwise by angle $\frac{\beta}{2}$. Thus we obtain two points $$a=(\cos \varphi, \sin \varphi)\begin{pmatrix} \cos \dfrac{\beta}{2} & -\sin \dfrac{\beta}{2} \\ \sin \dfrac{\beta}{2} & \cos \dfrac{\beta}{2} \end{pmatrix}$$ and $$b=(\cos \varphi, \sin \varphi)\begin{pmatrix} \cos \dfrac{\beta}{2} & \sin \dfrac{\beta}{2} \\ -\sin \dfrac{\beta}{2} & \cos \dfrac{\beta}{2} \end{pmatrix}.$$

Easy to check that $a,b\in S^1$ and $a+b=u+v$.

I was wondering how to show that $(u,v)=(a,b)$ or $(u,v)=(b,a)$.

Answer 2

We have $u+v=u'+v' =x \neq 0$. Since $4 \langle u-u', u+v \rangle = \|u+v+u-u'\|^2 - \|v-u'\|^2$, and $u-v-u'=v'$ we see that $\langle u-u', x \rangle = \langle u-u', u+v \rangle = 0$.

If $u=u'$ then we must have $v=v'$, so assume $u \neq u'$.

It is straightforward to check that $\langle u-u', u+u' \rangle = 0$.

We have $u = \frac12 (u+u')+\frac12 (u-u')$, $u' = \frac12 (u+u')-\frac12 (u-u')$ and so $1 = {1 \over 2^2} \|u+u'\|^2 + {1 \over 2^2} \|u-u'\|^2 $.

Since $x \bot u-u'$ we must have $x = \lambda (u+u')$ and $1=\|x-u\|^2 = \|\lambda(u+u')-u\|^2 = (\lambda -{1\over 2})^2 \|u+u'\|^2 + {1 \over 2^2} \|u-u'\|^2 $ from which we get $\lambda = 1$.

Hence $x = u +u'$. Since $x=u+v$ we have $v=u'$ and similarly $u=v'$.