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When a triple of numbers a, b and c forms a primitive Pythagorean triple, then (c minus the even leg) and one-half of (c minus the odd leg) are both perfect squares; however this is not a sufficient condition, as the numbers {1, 8, 9} pass the perfect squares test but are not a Pythagorean triple since 1 2 + 8 2 ≠ 9 2. At most one of a, b, c ...
Square packing in a square is the problem of determining the maximum number of unit squares (squares of side length one) that can be packed inside a larger square of side length . If a {\displaystyle a} is an integer , the answer is a 2 , {\displaystyle a^{2},} but the precise – or even asymptotic – amount of unfilled space for an arbitrary ...
Perfect: All 3m planar arrays must be pandiagonal magic squares. In addition, all pantriagonals must sum correctly. These two conditions combine to provide a total of 9m pandiagonal magic squares. The smallest normal perfect magic cube is order 8; see Perfect magic cube. Nasik; A. H. Frost (1866) referred to all but the simple magic cube as Nasik!
Packing circles in a square - closely related to spreading points in a unit square with the objective of finding the greatest minimal separation, d n, between points. To convert between these two formulations of the problem, the square side for unit circles will be L = 2 + 2 / d n {\displaystyle L=2+2/d_{n}} .
Since each 2 × 2 subsquare sums to the magic constant, 4 × 4 pandiagonal magic squares are most-perfect magic squares. In addition, the two numbers at the opposite corners of any 3 × 3 square add up to half the magic constant. Consequently, all 4 × 4 pandiagonal magic squares that are associative must have duplicate cells.
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A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. From Gulley (2010). The n th coloured region shows n squares of dimension n by n (the rectangle is 1 evenly divided square), hence the area of the n th region is n times n × n.
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