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Apologies for pasting a screenshot but it was the fastest way for me to ask the question since it's rather long.

I don't understand why the part where they say $(x^2 \vee 2xy \vee y^2) > 9$

  1. Why do at least one of them have to be greater than $9$?
  2. Why does at least one of them being greater than $9$ make the proof wrong?
Trevor Gunn
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Carpetfizz
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  • Note that for example $13^2 = 169, 31^2 = 961$. Here $x = 1, y = 3$, and none of $x^2, 2xy, y^2$ are greater than $9$, so the digits of $31^2$ are just the digits of $13^2$ reversed. – Michael Lugo Jun 14 '17 at 19:04

2 Answers2

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If any of them are greater than 9 then there is a carry. For instance, take the number $32$ ($x = 3, y = 2$ and $2xy = 12 > 9$). Then $32^2 = 1024$ and $23^2 = 529$. I can also write this as $32^2 = 9(12)4$ where there is a $12$ in the $10$'s place and similarly $23^2 = 4(12)9$. Now carrying, you have $(9 + 1)24$ and $(4 + 1)29$ which are respectively $1024$ and $529$.

Trevor Gunn
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It is wrong because if, e.g., $2xy>9$, this change the first digit of $a^2$ and $b^2$ in different ways so that now they are not $100 x^2$ and $100 y^2$.

Emilio Novati
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