Prove that $\{m,n\} =\{k,l\}$ 
Let $n,m,k,l$ be positive integers and $p$ and odd prime with $0 \leq n,m,k,l \leq p-1$ and $$n+m \equiv k+l \pmod{p}, \quad n^2+m^2 \equiv k^2+l^2 \pmod{p}.$$ Prove that $\{m,n\} =\{k,l\}$.

We can rearrange the third condition to get $(n-l)(n+l) \equiv (k-m)(k+m) \pmod{p}$. How do we continue from here?
 A: This holds true in any field in which $2$ is invertible. Suppose 
$$a+b=c+d \tag{1} $$
$$a^2+b^2=c^2+d^2 \tag{2} $$
Square $(1)$, subtract $(2)$, then divide by $2$ to get
$$ ab=cd \tag{3} $$
This means we have an equality of polynomials
$$(x-a)(x-b)=(x-c)(x-d) \tag{4}$$
So both $\{a,b\}$ and $\{c,d\}$ are the roots of this polynomial.
There are two key ideas here:


*

*To know how to get from $(1)$ and $(2)$ to $(3)$, be familiar with the relationship between elementary symmetric polynomials and power symmetric polynomials (at least the first couple of each type).

*To know to combine $(1)$ and $(3)$ to $(4)$, know Vieta's formulas.
(A field is a number system with all the usual operations and properties - addition, multiplication, associativity, commutativity, distributivity, etc. - with the property that all nonzero elements have multiplicative inverses.)
A: For a more elementary solution: 
From the relations it follows that $2mn\equiv 2kl \pmod{p}$. Subtracting from the second gives $(m-n)^2\equiv(k-l)^2\pmod{p}$. Because $|m-n|<p$ and $|k-l|<p$ and $p$ is prime, it follows that either $m-n\equiv k-l\pmod{p}$ or $m-n\equiv l-k\pmod{p}$. Either way, adding to the first relation gives $m\equiv k\pmod{p}$ or $m\equiv l\pmod{p}$. This gives immediately what you want, considering that all $m,n,k,l$ are smaller than $p-1$, and $p$ is prime. 
A: First, observe that $n^2 + 2mn + m^2 \equiv (n+m)^2 \equiv (k+l)^2 \equiv k^2 + 2kl + l^2$, so $kl \equiv mn$ (since $p$ is odd, 2 is a unit).
If $k \equiv 0$, then $0 \equiv kl \equiv mn$. Without loss of generality we have $m \equiv 0$, and thus $n \equiv l$. The claim follows then from $0 \le m,n,k,l \le p-1$.
Otherwise, let $k \not\equiv 0$. This means $k$ is a unit, and we can find $a$ in $\mathbb{F}_p$ with $m \equiv ak$ (choose e.g. $a = k^{-1} m$). It follows that $kl \equiv akn$, and thus $l \equiv an$ (by multiplying both sides with $k^{-1}$). Furthermore, $k + an \equiv k + l \equiv n + m \equiv n + ak$.
If $a \equiv 1$, then we immediately have $l \equiv n$ and $m \equiv k$, and the claim follows as before from $0 \le m,n,k,l \le p-1$.
Else, we get $k \equiv n$ and thus $l \equiv an \equiv ak \equiv m$. As before, $\{ k,l \} = \{ m,n \}$.
