Find all $x,y \in \mathbb N^*$ such that $\frac{1}{x} + \frac{1}{y}=5$ My proof:
By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$
So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$
Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5>1 \implies y\mid x$
By the same method we have:
$$y=x(5y-1) \implies x\mid y$$
$x\mid y, y\mid x \implies x=y$, Now we can write the original equation as :
$$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$
Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$
Does this considered a proof?
 A: Your proof is correct. But why make it complicated like that?
Just notice that $x \ge 1$ then $\frac{1}{x} \le 1$. Same for $y$ that $\frac{1}{y} \le 1$.
Hence,
$$\frac{1}{x}+\frac{1}{y}\le 2 <5$$
So, there is no solution.
A: Though here there are obvious direct ad-hoc methods, it's worth emphasis that they are essentially special cases of one case of Lagrange's solution of the general quadratic binary Diophantine equation, namely that completing a square $ $ generalizes to $ $ completing a product as follows
$$\bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\begin{align} axy + bx + cy  &\,=\, d\\[.2em]
\!\!\!\iff (ax+c)(ay+b) &\,=\, ad+bc\end{align}}}\qquad$$
Applied here we get $\, 5xy-x-y = 0\iff (5x-1)(5y-1) = 1\,$ and the rest is easy.
Your proof comes very close to rediscovering this general method.
The same method works in general, where there usually is no such easy direct ad-hoc method.
A: $$\frac{1}{x} \le 1 \text{ and }\frac{1}{y} \le 1.$$
Thus
$$\frac{1}{x}+\frac{1}{y} \le 2.$$
