Perfect squares that ends in $4$s

Question

1. Prove that there is an infinity of perfect squares which have $4$ as the last 3 digits.
2. Prove that there are no perfect squares which have $4$ as the last 4 digits.

This is a problem form a math contest (Romanian Math Olympiad, county level) for fifth graders. No calculators allowed. I couldn't even come quickly with a solution at highschool level, so any help is highly appreciated.

Answer 1

If you observe that $1444=38^2$, the perfect squares

$$(1000k+38)^2=1000000k^2+76000k+1444$$ end in $444$.

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For the second question, notice that the last four digits of a perfect square only depend on the last two digits of the root, and we have

$$66^2<4444<67^2.$$

Answer 2

Here's a quick proof for the second part:

Every natural number which ends in $4444$ is of the form $$10000k+4444=4\times (2500k+1111)$$ for some $k\in \mathbb N$.

Of course, in order for this to be a square, $2500k+1111$ would also have to be a square.

$2500$ is divisible by $4$ and $1111=4\times 277+3$ so $2500k+1111$ is of the form $4l+3$ but it is easy to see that any square is of the form $4l$ or $4l+1$ so we are done.