Squares Within Squares

For positive $n$ , you can construct a solution for the following two cases:

• $2n+2$
• $n+3$ (If there is a solution for $n$ )

The first one is done by drawing a square, then drawing $n$ squares along each of two adjacent edges and $1$ at the corner.

The second one can be done by taking any arrangement of $n$ squares that forms a square and subdividing one square into $4$ parts which will add $3$ to the total square count.

This means that you are good to go for all even $N > 2$ :

• $4,6,8,10,...$

And, you can add three to any of those, giving you:

• $7,9,11,13,...$

Now that you have a sequence of three (namely $6,7$ and $8$ ) you can make any higher number by adding $3$ to the lowest of the three, in this case $6$ , giving you $9$ , producing a new sequence of three ( $7,8$ and $9$ ). Rinse wash and repeat! :)

The only numbers that can't be constructed in this way, then, are $2, 3$ , and $5$ , the largest of which is $\boxed{5}$

And after playing around with these three numbers I convinced myself (without a rigorous proof... :-/ ) that there was no way to construct a square with any of these numbers of squares.

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For more examples of squares in squares, see http://www.squaring.net .