$m^2+n^2=2018$

Given that $m^2 + n^2 = 2018$ , since the RHS is even, this means that the LHS must also be even. Then either $m$ and $n$ are both even or both odd. If both $m$ and $n$ are even then the LHS is a multiple of 4, but the RHS is not a multiple of 4, therefore, both $m$ and $n$ are odd. Let $m=2a+1$ and $n=2b+1$ , where $a,b \in \mathbb N$ . Then we have:

$\begin{aligned} (2a+1)^2 + (2b+1)^2 & = 2018 \\ 4a^2 + 4a+1 + 4b^2 + 4b+1 & = 2018 \\ \implies a^2 + a + b^2 + b & = 504 \\ {\color{#3D99F6}a(a+1)} + {\color{#3D99F6}b(b+1)} & = 504 & \small \color{#3D99F6} \text{Note that }a(a+1) \text{ and }b(b+1) \text{ are even,} \\ {\color{#3D99F6}\frac {a(a+1)}2} + {\color{#3D99F6}\frac {b(b+1)}2} & = 252 & \small \color{#3D99F6} \text{and }\frac {a(a+1)}2 \text{ and }\frac {b(b+1)}2 \text{ are integers,} \\ {\color{#3D99F6}T_a} + {\color{#3D99F6}T_b} & = 252 & \small \color{#3D99F6} \text{and are triangular numbers }T_n. \end{aligned}$

Let $a> b$ . We note that the largest $a=21$ and we find that $b=6$ is a solution, since $T_{21}+T_6 = 231 + 21 = 252$ . From $T_1$ to $T_{21}$ , there is only $T_6$ and $T_{21}$ satisfy the equation. Therefore, $m + n = 2a+1 + 2b+1 = 2(21)+1+2(6)+1 = \boxed{56}$ .

. Here is the same problem

using $Mathematica$

PowersRepresentations[2018,2,2]

returns {{13, 43}}

$43^{2}+13^{2}=2018$

$43+13=56$