$(n-m)^2(n+m)=2023=17^2\cdot7$. Thus $n-m|17$. If $n-m=1$ then $n+m=2023$ and $(n,m)=(1012,1011)$. If $n-m=17$ then $n+m=7$ and $(n,m)=(12,-5)$.

qwedsazxc wrote: $(n-m)^2(n+m)=2023=17^2\cdot7$. Thus $n-m|17$. If $n-m=1$ then $n+m=2023$ and $(n,m)=(1012,1011)$. If $n-m=17$ then $n+m=7$ and $(n,m)=(12,-5)$. Your solution is well however you should've mentioned that $n-m$≠$-17$ in fifth line because of $n$≥$m$ for a complete solution.

$ n^3 + m^3 -nm(n + m) = (n + m)(n^2 -2nm + m^2) = (n + m)(n - m)^2 = 2023 $ Case 1: $ n - m = 1 , n + m = 2023 \rightarrow\ n = 1012 , m = 1011 $ Case 2: $ n - m = 17 , n + m = 7 \rightarrow\ n = 12 , m = -5 $

I missed a silver medal because I overlooked $n\ge m$.

Notice that $(n+m)(m^2 -2mn +n^2) = 2023$. You can write this as $(m+n)(m - n)^2 = 2023$. The positive integer divisors of 2023 are $ 1, 7, 17, 119, 289, 2023$. There are two possibilities : $ 1) : (m-n)^2 = 1$ and $(n+m) = 2023$. In this case, due to $ n \ge m $, $ (m-n)= -1$ and $ (m+n)= 2023$. If you solve this equation, $ (m,n) = (1011,1012) $. $ 2) : (m-n)^2 = 289 $ and $ (m+n) = 7$. In this case, due to $ n \ge m $, $ (m-n)= -17 $ and $ (m+n) =7 $. If you solve this equation, $ (m,n) = (-5, 12)$. There is none of another solution. $\square$

It is clear that \[2023=n^3+m^3-nm(n+m) = (n+m)(n^2-nm+m^2-nm) = (n+m)(n-m)^2\]But, if any prime $p \neq 17 \mid n-m$ we have that $p^2 \mid 2023$ which is clearly impossible. Further, if $\nu_{17} (n-m)\geq 2$, \[\nu_{17}(n+m)(n-m)^2 \geq 4 > 2\]which is also a clear contradiction. Thus, we conclude that $n-m=1$ or $n-m=17$. Solving the resulting systems of equations gives us the solution pairs, \[(m,n)=(1011,1012);(-5,12)\]