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This question relates to the one I just asked several hours ago: Find $x+y$ if $x,y \in \mathbb{N}$ and $x^2+y^2 = 2019^2$. There is a short cut to this equation but I haven’t found it yet. I tried estimate $x+y$ but it has lots of possible outcomes. Any one sees the trick here? Thank you in advance.

DeepSea
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1 Answers1

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We have $2019=3\cdot 673$ and $0^2+3^2=3^2$, $$ 385^2+552^2=673^2 $$ by using the formulas here. Now we can use two square theorem by Brahmagupta to solve $2019^2=x^2+y^2$.

Dietrich Burde
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  • Very nice answer. I wonder how you express the $673^2$ as the sum of the squares of $385$ and $552$. – DeepSea Feb 28 '20 at 11:00
  • I used the wikipedia link with the formula $(m^2-n^2,2mn,m^2+n^2)$ and $(m,n)=(23,12)$. Then $m^2-n^2=(m+n)(m-n)=35\cdot 11=385$ and so on. – Dietrich Burde Feb 28 '20 at 11:47