Question

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

Answer 1

Answer: I, III, and IV only

Explanation:

Answer 2

`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