Find all $x,y \in \mathbb N^*$ such that $\frac{1}{x} + \frac{1}{y}=5$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

$$\frac{1}{x} \le 1 \text{ and }\frac{1}{y} \le 1.$$
Thus $$\frac{1}{x}+\frac{1}{y} \le 2.$$