Diophantine equation in two variables involving floor function

Question

The problem is about finding all $a,b\in\mathbb{N}$ such that $$\left \lfloor{\frac{a^2}{b}}\right \rfloor + \left \lfloor{\frac{b^2}{a}}\right \rfloor +1=\left \lfloor{\frac{a^2+b^2}{ab}}\right \rfloor + ab$$ On the book, there is a hint saying that, based on that we would have $\frac{a^2}{b}+\frac{b^2}{a}+1 > \frac{a^2+b^2}{ab}-1+ab$, and then, that $1 + (a + b − 1)(a^2 + b^2-ab)>a^2b^2$ and then, that $a\geq b\Rightarrow b=1$ but I don't know how to get to the first inequation.

Answer 1

Note that the definition of the floor function gives that

$$x \ge \lfloor x \rfloor \gt x - 1 \tag{1}\label{eq1A}$$

Using the $x \ge \lfloor x \rfloor$ part with the left hand side of your equation, this gives

$$\frac{a^2}{b} + \frac{b^2}{a} + 1 \ge \left\lfloor{\frac{a^2}{b}}\right\rfloor + \left\lfloor{\frac{b^2}{a}}\right\rfloor + 1 \tag{2}\label{eq2A}$$

Using the $\lfloor x \rfloor \gt x - 1$ part with the right hand side of your equation, this gives

$$\left\lfloor{\frac{a^2 + b^2}{ab}}\right\rfloor + ab \gt \frac{a^2 + b^2}{ab} - 1 + ab \tag{3}\label{eq3A}$$

Putting \eqref{eq2A} and \eqref{eq3A} together, and using your equation, gives

$$\frac{a^2}{b} + \frac{b^2}{a} + 1 \gt \frac{a^2+b^2}{ab} - 1 + ab \tag{4}\label{eq4A}$$

Since you only ask about the first inequation, I assume you can deal with the rest of the given hint.