Respuesta :

sarkhan2018
[tex] 2018^{2019} + 2018=2018( 2018^{2018} +1)=( 13^{2}+ 43^{2})(( 2018^{1009})^{2} + 1^{2}) [/tex]
According to Brahmagupta-Fibonnaci identity 
[tex]( a^{2} + b^{2}) (c^{2}+ d^{2}) = (ac+bd)^{2} + (ad-bc )^{2} [/tex]
Then, we can write that
[tex](13^{2}+ 43^{2})(( 2018^{1009})^{2} + 1^{2})[/tex]
[tex][(13*2008^{1009}+43)^{2} +(13-2008^{1009}*43)^{2}] [/tex] 
Q.E.D
ughiwishlolo

Answer: I, III, and IV only

Step-by-step explanation:

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