2015 and square root madness

This question doesn't have any trick with equation system, you can just solve it one-by-one.

Since $b$ is an integer, $2015-a$ must be a perfect square. The condition that $a$ is the least possible shows that $2015-a$ is the maximum perfect square possible that is less than $2015$ . Since $44^{2} = 1936$ and $45^{2} = 2025$ , we now know that $2015-a$ is the square of $44.$ $2015 - a = 44^{2}$ $\sqrt{2015 - a} = \sqrt{44^{2}}$ $\sqrt{2015 - a} = 44 \cdots \star$ $2015 - a = 1936$ $a = 2015 - 1936$ From above, we get $a = 79$ and from $\star$ , $b = 44$

Using the almost similar trick to the second equation.

Since $d$ is an integer, $a+c=79+c$ must be a perfect square. The condition that $c$ is the least possible shows that $79+c$ is the minimum perfect square possible that is more than $79$ . Since $8^{2} = 64$ and $9^{2} = 81$ , we now know that $79+c$ is the square of $9.$ $79+c = 9^{2}$ $\sqrt{79+c} = \sqrt{9^{2}}$ $\sqrt{79+c} = 9 \cdots \star$ $79+c = 81$ $c = 81-79$ From above, we get $c = 2$ and from $\star$ , $d = 9$

Now, to the last equation, $\sqrt{a+b+c+d+1} = m\sqrt{n}$ Putting everything together we will get $\sqrt{79 + 44 +2 + 9 + 1} = \sqrt{135} = m\sqrt{n}$ If $n$ is the least positive integer possible, then $n$ cannot have common primes. Using prime factorization to get $135 = 3^{3} \times 5 = 3^{2} \times 3 \times 5 = 3^{2} \times 15$ $\sqrt{135} = \sqrt{3^{2} \times 15} = 3\sqrt{15}$

At last, $m + n = 3 + 15 = \boxed{18}$