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More generally, an a-by-b rectangle can be covered with square tiles of side length c only if c is a common divisor of a and b. For example, a 24-by-60 rectangular area can be divided into a grid of: 1-by-1 squares, 2-by-2 squares, 3-by-3 squares, 4-by-4 squares, 6-by-6 squares or 12-by-12 squares.
For illustration, a 24×60 rectangular area can be divided into a grid of: 1×1 squares, 2×2 squares, 3×3 squares, 4×4 squares, 6×6 squares or 12×12 squares. Therefore, 12 is the GCD of 24 and 60. A 24×60 rectangular area can be divided into a grid of 12×12 squares, with two squares along one edge (24/12 = 2) and five squares along the ...
The Brahmagupta–Fibonacci identity states that the product of two sums of two squares is a sum of two squares. Euler's method relies on this theorem but it can be viewed as the converse, given n = a 2 + b 2 = c 2 + d 2 {\displaystyle n=a^{2}+b^{2}=c^{2}+d^{2}} we find n {\displaystyle n} as a product of sums of two squares.
There are several ways to find the greatest common divisor of two polynomials. Two of them are: Factorization of polynomials, in which one finds the factors of each expression, then selects the set of common factors held by all from within each set of factors. This method may be useful only in simple cases, as factoring is usually more ...
Most algorithms for finding congruences of squares do not actually guarantee non-triviality; they only make it likely. There is a chance that a congruence found will be trivial, in which case we need to continue searching for another x and y. Congruences of squares are extremely useful in integer factorization algorithms.
It means that 1, 4, and 36 are the only square highly composite numbers. Saying that the sequence of exponents is non-increasing is equivalent to saying that a highly composite number is a product of primorials or, alternatively, the smallest number for its prime signature .
Gauss [10] pointed out that the four squares theorem follows easily from the fact that any positive integer that is 1 or 2 mod 4 is a sum of 3 squares, because any positive integer not divisible by 4 can be reduced to this form by subtracting 0 or 1 from it. However, proving the three-square theorem is considerably more difficult than a direct ...
Another geometric proof proceeds as follows: We start with the figure shown in the first diagram below, a large square with a smaller square removed from it. The side of the entire square is a, and the side of the small removed square is b. The area of the shaded region is . A cut is made, splitting the region into two rectangular pieces, as ...