Thai Mathematical Olympiad

Let $5n+1=m^2\equiv 1(\mathrm{mod}5)$. Thus $m=5k\pm 1$ for some integer $k$. We have $$n+1=\frac{(5k\pm 1{)}^2+4}{5}=5k^2\pm 2k+1=4k^2+(k\pm 1{)}^2$$ which can be written as a sum of 5 perfect squares as desired.