Inspired by Muhammad Rasel Parvej

[This solution is incomplete.]

If the unit squares had to be parallel to the sides of the larger square, then the answer would be 3. See this solution for why that is the case.

The difficulty of this problem comes about from being able to tilt the squares . While it is clear that the large square has area at least 5 hence side length at least 2.23, and that a large square of side length 3 can fit 9 unit squares hence gives us an upper bound, it's not immediately apparent what the actual minimum is.

The generalized problem is still open, though specific cases have been worked on. Proving the minimum is extremely hard, even for small values. See this survey of known results .

Here is how we can pack 5 unit squares into a square of length $2 + \frac{ \sqrt{2}}{2}$ .

There is a relatively simple proof. Can you find it?

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Hint: Here is a proof that the minimum side length of a square which can fit 3 squares is 2.

We will show that a square of side length $t < 2$ allows only 1 unit square.
Orientate the square $T$ with the lower left corner at the origin, meaning that the square has vertices $(0,0), (t, 0), (t, t), (0, t)$ .
Consider the point $(t/2, t/2)$ . Since it is distance $t/2 < 1$ away from any sides of T, thus any unit square in $T$ will have to contain this point.
Since all unit squares must contain this point, hence we can place at most 1 non-overlapping unit square in T.
Conversely, it is clear how we can fit 3 unit squares into a square of slide length 2.

How can we extend this argument to the case of 5 squares?

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Here is an outline of the proof that no square of side length $t < 2 + \frac{ \sqrt{2} } { 2 }$ will allow 5 squares.

Set $L = 2 + \frac{ \sqrt{2} } { 2 }$ .
Orientate the square $T$ with the lower left corner at the origin, meaning that the square has vertices $(0,0), (t, 0), (t, t), (0, t)$ .
Consider the 4 points $( \frac{t}{L}, \frac{t}{L} ), ( \frac{ t}{L}, \frac{ (L-1)t} { L } ), ( \frac{ (L-1)t} { L } , \frac{ t}{L}), ( \frac{ (L-1)t} { L } , \frac{ (L-1)t} { L } )$ .

(Claim, which is harder than I initially expected) Any unit square in $T$ must contain at least one of these 4 points in the interior.
Hence, we can place at most 4 non-overlapping unit squares in T.

Note: For $t = L$ , these 4 points correspond to the touching points of the 5 square setup above, which is why it makes sense to consider these 4 points.