All n > 0 except 2, 3, and 5.
One and four were given in the problem statement,
six, seven, and eight will be diagrammed below, and
the rest will follow by induction: If a square can
be partitioned into k squares k≥6, then it can be partitioned
into k+3 squares.
Let a square be partitioned into k squares where k≥6,
then one of those k squares can be partitioned into four squares.
Thus, k-1+4 = k+3.
+--+--+--+ +-----+-----+ +--+--+--+--+
| | | | | | | | | | | |
+--+--+--+ | | | +--+--+--+--+
| | | | | | | | |
+--+ | +-----+--+--+ +--+ |
| | | | | | | | | |
+--+-----+ | +--+--+ +--+ |
| | | | | | |
n=6 +-----+--+--+ +--+--------+
n=7 n=8
Note: See Jer's post for why a square cannot be partitioned into 2, 3, or 5 squares.
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