Fake proof that the digits of $a$ squared then reversed equal the digits of $a$ reversed then squared

Question

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?

Answer 1

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 $529$.

Answer 2

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 $100 y^2$.