If I have $\lfloor\frac{E}{K}\rfloor =\lfloor \frac{E}{K + m}\rfloor$, what is the upper limit of 'm' in terms of 'E' and 'k'

154 Views Asked by At

Given that E, K, m > 0, then is there a way to find out value of m in terms of E and K for

$$\left\lfloor\frac{E}{K} \right\rfloor =\left\lfloor \frac{E}{K + m} \right\rfloor$$

All I have gotten so far is this:

$$ \frac{E}{K}-1 \lt \left\lfloor\frac{E}{K}\right\rfloor \le \frac{E}{K} $$

and,

$$ \frac{E}{K+m}-1 \lt \left\lfloor\frac{E}{K+m}\right\rfloor \le \frac{E}{K+m} $$

and since $\left\lfloor\frac{E}{K} \right\rfloor =\left\lfloor \frac{E}{K + m} \right\rfloor$,

$$ \implies \frac{E}{K}-1 < \frac{E}{K + m} $$

leading to,

$$ m < \frac{K^2}{E - K} $$

but I am looking for a tighter bound, preferably in terms of equality. Any help will be greatly appreciated.

Edit: I forgot to add that $ E \gt K $