Let $a>0$. Prove $\nexists x\in\mathbb R^+$ s.t.
$$\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$$
I know that $$k\in\mathbb Z\implies\left(\forall x\in\mathbb R\right) \lfloor x+k\rfloor=\lfloor x\rfloor + k.$$
I can say that $\left\lfloor\frac{144}{x+a}-1\right\rfloor=\left\lfloor\frac{144}{x+a}\right\rfloor-1$.
How should I proceed after this? Does this help in any way?
Hint: The floor function, $\lfloor x\rfloor$, is bounded by $x-1< \lfloor x\rfloor\le x$ so you have $\text{RHS}\le \left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1<\text{LHS}$. By considering their (single) intersection, consider how you can form an inequality with $\left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1$, then check whether this intersection is a solution and find an inequality relating all four terms.