Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square. I need some help here. I thought starting with $x = n * 10^k$ where $10^k$ represents the number of the zeros at the end of the number and $n$ is the group of digits which end with $2$ could help, but it didn't. Can anyone help?

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    $\begingroup$ It might be more fruitful to decompose $y = n \cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$. $\endgroup$ Jan 18, 2019 at 1:29
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    $\begingroup$ The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9. $\endgroup$
    – TonyK
    Jan 18, 2019 at 1:31

1 Answer 1


Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $\pmod {100}$.


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