Prove that every square?

Two constructive methods.

The simpler one I thought of later is shown below.

UPDATE:

At the end I address the problem from a number theory perspective.

METHOD #1:

Draw the original square with opposite corners at (0, 0) and (a, a).

Then divide the line segment (0,0) -- (a,0) into

k equal parts, k >= 2.

Draw a horizontal line at y = a/k, and a vertical line at x = a/k.

Now you have divided the square into 4 regions:

lower left = square (a/k) x (a/k)

upper right = square (a - a/k) x (a - a/k)

and two rectangles which are

(a/k) x (k-1) a / k

Each of those two rectangles can be divided into (k-1) squares, of size (a/k) x (a/k)

yielding a total of

2 + 2 (k-1) squares = 2k.

So for any k >= 2, you can do it for an even number of squares, 2k, starting with 4.

By the same reasoning, you divide either of the squares inside the original

square into an even number of squares, starting with 4.

Adding those to the 2 (k - 1) + 1 other squares from the construction,

we get any odd number

2k - 1 + 2j, for k >= 2, j >= 2

which is at minimum 4 - 1 + 4 = 7.

And that satisfies the demands of the problem.

METHOD #2: simpler than the above

You can do 6 as described above like this:

A b b

A b b

A A A

where each A represents a complete square

and the 4 b's together represent 1 square.

And you can do 7 as follows:

Divide the original square into 4 quadrants

A | B

--+--

C | D

Then divide one of the quadrants simlarly, adding 3 more to give 7.

8 can be done as in the first method above:

A b b b

A b b b

A b b b

A A A A = 7 small A squares and 1 large b square.

Then you can take any existing arrangement,

and subdivide one of the squares into 4 quadrants,

thereby adding 3 more squares.

Then since we have above a way to construct

6, 7, or 8 (3 consecutive numbers),

we can add any multiple of 3 to any of them,

and thereby reach any number >= 6.

ADDED:

Number Theory Argument.

In 1770, Lagrange proved that every integer can be expressed as the sum of 4 squares.

This problem is a generalization (maybe reverse generalization)

of that since it boils down to "4 or 6 or more squares" instead.

So if the number can be broken down into 4 non-zero

squares, then we can divide the square into quadrants,

and then put an n × n square grid in each quadrant,

corresponding to the 4 (smaller) squares.

Quoting from Wikipedia:

The sequence of positive integers which cannot be represented as a sum of four non-zero squares is:

1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41, 56, 96, 128, 224, 384, 512, 896 ... (sequence A000534 in OEIS).

These integers consist of the eight odd numbers 1, 3, 5, 9, 11, 17, 29, 41 and all numbers of the form 2 × 4n, 6 × 4n or 14 × 4n.

Disregarding the few small odd numbers which we can

handle as described above, the others are all multiples of

4: 4 × (2n, 6n, or 14n).

For those, we simply divide the square into quadrants,

and then subdivide each quadrant by whatever means.

In general, there will be lots of ways to do it, such as

using a "divide and conquer" approach.

For example, take 300.

First divide it into 4 × 75.

Each of the 75 can be divided into 25 + 25 + 16 + 9.

Or 16, 16, 16, 27.

Then 27 can be done as 8+8+7+6 or 9+8+6+4.

Or 273:

First do 8 × 8, giving a grid of 64 squares.

Do 60 of them as 2 × 2, for a total of 240.

That leaves 33, which can be done in the other 4

as 16 + 9 + 4 + 4.

Or numerically: 273 = (2 × 2) * 60 + 16 + 9 + 4 + 4.

n is the two strange or regardless of if n is even n^7 mod 4 = 0 so n^7+7 = 3 mod 4 yet for an strange sq. (2k+a million)^2 mod 4 = a million so n^7+7 isn't applicable sq. whilst n is even now for n = -a million mod 4 n^7+7 = 2 mod 4 which isn't a applicable sq. so the case continues to be for n =a million mod 4 I would be lower back if i discover answer for the comparable/.