
Solution: Let $[\sqrt{m}]=[\sqrt{m+125}]=k$. Then we know that

$$
k^{2} \leq m125$ or $k>62$. Using $k^{2} \leq 5000$, we get $k \leq 70$. Thus $k \in$ $\{63,64,65,66,67,68,69,70\}$. We observe that $63^{2}=3969$ and $64^{2}=63^{2}+127$. Hence

$$
\left[\sqrt{63^{2}+125}\right]=\left[\sqrt{63^{2}+1+125}\right]=63
$$

but $\left[\sqrt{63^{2}+2+125}\right]=64$. Thus we get two values of $m$ such that $[\sqrt{m}]=[\sqrt{m+125}]$ for $k=63$. Similarly, $65^{2}=64^{2}+129$ so that

$$
\left[\sqrt{64^{2}+125}\right]=\left[\sqrt{64^{2}+1+125}\right]=\left[\sqrt{64^{2}+2+125}\right]=\left[\sqrt{64^{2}+3+125}\right]=64
$$

but $\left[\sqrt{64^{2}+4+125}\right]=65$. Thus we get four values of $m$ such that $[\sqrt{m}]=[\sqrt{m+125}]$ for $k=64$. Continuing, we see that there are $6,8,10,12,14,16$ values of $m$ respectively for $k=65,66,67,68,69,70$. Together we get

$$
2+4+6+8+10+12+14+16=2 \times \frac{8 \times 9}{2}=72
$$

values of $m$ satisfying the given requirement. $\qquad$



6. How many integers $m$ satisfy both the following properties:

(i) $1 \leq m \leq 5000$; (ii) $[\sqrt{m}]=[\sqrt{m+125}]$ ?

(Here $[x]$ denotes the largest integer not exceeding $x$, for any real number $x$.)
