An algebra problem by Anish Roy

Let $\lfloor \sqrt{m} \rfloor=\lfloor \sqrt{m + 125} \rfloor=k$ . Then we know that $k^2 \leq m < m + 125 < (k + 1)^2$ Thus $m + 125 < k^2 + 2k + 1 \leq m + 2k + 1$ This shows that $2k + 1 > 125$ 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 $\lfloor \sqrt{63^2 + 125} \rfloor=\lfloor \sqrt{63^2 + 1 + 125} \rfloor= 63$ but $\lfloor \sqrt{63^2 + 2 + 125} \rfloor= 64$ . Thus we get two values of $m$ such that $\lfloor \sqrt{m} \rfloor=\lfloor \sqrt{m + 125} \rfloor$ for $k = 63$ . Similarly, $65^2 = 64^2 + 129$ so that $\lfloor \sqrt{64^2 + 125}\rfloor=\lfloor \sqrt{64^2 + 1 + 125} \rfloor=\lfloor \sqrt{64^2 + 2 + 125} \rfloor=\lfloor \sqrt{64^2 + 3 + 125} \rfloor= 64$ but $\lfloor \sqrt{642 + 4 + 125}\rfloor= 65$ . Thus we get four values of $m$ such that $\lfloor \sqrt{m} \rfloor=\lfloor \sqrt{m + 125} \rfloor$ 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.