Question

Write 2018 to the power of 2019 + 2018 as the sum of two perfect squares

Answer 1

$2018^{2019} + 2018=2018( 2018^{2018} +1)=( 13^{2}+ 43^{2})(( 2018^{1009})^{2} + 1^{2})$
According to Brahmagupta-Fibonnaci identity 
$( a^{2} + b^{2}) (c^{2}+ d^{2}) = (ac+bd)^{2} + (ad-bc )^{2}$
Then, we can write that
$(13^{2}+ 43^{2})(( 2018^{1009})^{2} + 1^{2})$
$[(13*2008^{1009}+43)^{2} +(13-2008^{1009}*43)^{2}]$ 
Q.E.D