I would like someone to give me feedback and alternative/more elegant solutions to this problem:
Suppose that you know that $x$ and $y$ are integers and that $xy$ divides $x^2 + y^2$. Prove that $x = \pm y$.
My proof:
Since $xy$ divides $x^2 + y^2$, we know that $x^2 + y^2 = xy \cdot k$ for some integer $k$.
We have $x^2 = xy \cdot \frac{x}{y}$ and $y^2 = xy \cdot \frac{y}{x}$, and therefore the equation can be rewritten as $$xy \cdot(\frac{x}{y} + \frac{y}{x}) = xy \cdot k$$
which is equivalent to $$\frac{x}{y} + \frac{y}{x} = k$$
Now let $\frac{x}{y} = a$ and $\frac{y}{x} = \frac{1}{a}$ then $\frac{x}{y} + \frac{y}{x} = a + \frac{1}{a} = \frac{a^2 + 1}{a}$ and hence $$\frac{\frac{x^2}{y^2} + 1}{\frac{x}{y}} = k$$.
Since the RHS is an integer the LHS also has to be an integer; this can only be true if $\frac{x}{y} \mid 1$, which implies that $x = \pm y$, because only $-1$ or $1$ divide 1. $\blacksquare$
Corrections:
Continuing from $(k-m)(k+m) = 4 $ we have the following cases:
- $k-m = 2$ and $k+m= 2$, which gives us $(k,m) = (2,0)$ as solutions
- $k-m = 4$ and $k+m = 1$, which gives us $(k,m) = (\frac{5}{2}, - \frac{3}{2})$
- (Case omitted since we only care about $k$.)
Feeding either of the solutions back into $a^2 + 1 = ka$, we can conclude that $\frac{x}{y} = \frac{y}{x}$, which implies that $x = \pm y$.
Taking from where you left off: $a^2 + 1 = ka \implies a^2-ka + 1 = 0 \implies \triangle = b^2-4ac = (-k)^2 - 4(1)(1) = k^2-4 = m^2 \implies (k-m)(k+m) = 4, m \in \mathbb{Z}$. Can you take it from here?