its obvious that a=1 , 2, 3 doesn't work i show that a=4 doesn't work and we have an obvious example for a=5 but i cant show the sketch of my proof because i cant work with geogebra but i draw my picture in geometer' sketchpad and i cant upload it here can anyone help me?

Is it clear we should find the least such integer $a$ ? There is no indication we are not in fact asked to find the least real number $a$ with that property.

sorry.i made a mistake i don't pay attention to this fact but i believe that it must be natural because i think that if $a$ works then there exist a real positive $e$ such that e is very small and $a-e$ works too and it shows that we dont have minimum in the reals and another reason is that its balticway competition and they wanna tricky idea in their problems not troublesome problems

No, this argument invoked by you seems to imply "least integer" (which always exists), since "least real" may not exist; but your argument is flawed. If $a$ works, then there does not necessarily exist a small $e$, such that $a-e$ also works - since for the minimal $a$, some disks will touch the border of the square, so we cannot shrink it. This can be formally argued by Weierstrass' theorem, the case be. Second-guessing the composers is also not that good an idea

Quote: This can be formally argued by Weierstrass' theorem can you explain more about this theorem and its statement ? meanwhile if the problem rearrangement such that any two disks dont even touch each other then my argument works and my example for a=5 works too

I think the idea is to reason something like this: Consider the smalles square that can contain 2 disks of radius 1. It is intuitive and not hard to prove that the square would contain the 2 disks in a diagonal arrangement. The length of the diagonal of the square would be $2 + 2\sqrt{2}$, so its side would be $2 + \sqrt{2}$. Now let s be the length of some square that contains 5 disks of radius 1. Divide the square into 4 squares by adding a vertical and horizontal line. By the pigeon hole principal, at least 1 of these 4 smaller squares contains the centers of 2 of the disks. Let's say it is the square in the lower left. If we enlarge this smaller square by 1 unit upward and to the right, we now have a square of side $\frac{s}{2} + 1$ that completely contains 2 unit disks. So we have the formula $\frac{s}{2} + 1 \geq 2 + \sqrt{2}$, or $s \geq 2 + 2\sqrt{2}$. But we can achieve equality by placing 1 disk in the exact center of a square and placing the other 4 disks each tangent to the center disk and 2 of the square's sides. So we conclude a = $2 + 2\sqrt{2}$.