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$$x^2+y^2 = 2018^{2019}+2018$$

is expressed as sum of two perfect squares.

Any pair of perfect squares can satisfy?

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closed as off-topic by Did, GNUSupporter 8964民主女神 地下教會, Isaac Browne, JMP, mathematics2x2life Apr 29 '18 at 5:38

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See that your expression equals $2018(2018^{2018}+1).$ If you can do two things, you can solve your problem:

  1. Express $2018$ as a sum of two squares.

  2. Convert the product of two sums of two squares into one sum of two squares.

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Since $1009$ is prime and $1009 \equiv 1 \mod 4,$ $100$9 can be expressed as a sum of two squares. $2=1^2+1^2.$ $2018^{2018}+1=(2018^{1009})^2+1^2.$ From these, you can conclude that $2 \times 1009 \times (2018^{2018}+1)$ is a sum of two squares. (I leave to you as an exercise to prove that the product of a sum of two squares is also a sum of two squares.)

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