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If two primes which end in 3 or 7 and surpass by 3 a multiple of 4 are multiplied, then their product will be composed of a square and the quintuple of another square. In other words, if p, q are of the form 20k + 3 or 20k + 7, then pq = x 2 + 5y 2. Euler later extended this to the conjecture that
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
Thus, the Cayley table of a group is an example of a latin square. An alternative and more succinct proof follows from the cancellation property . This property implies that for each x in the group, the one variable function of y f(x,y)= xy must be a one-to-one map.
It follows that y 1 2 + y 2 2 + y 3 2 + y 4 2 = mr, for some strictly positive integer r less than m. Finally, another appeal to Euler's four-square identity shows that mpmr = z 1 2 + z 2 2 + z 3 2 + z 4 2. But the fact that each x i is congruent to its corresponding y i implies that all of the z i are divisible by m.
Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. [1]
This is because both solutions differ from an integer by the square root of an integer, and so either solution can only be an integer if this square root, and hence the other solution, is also an integer. Every four integers that satisfy the equation in Descartes' theorem form the curvatures of four tangent circles. [33]
However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent. With the bent hypotenuse, the first figure actually occupies a combined 32 units, while the second figure occupies 33, including the "missing" square.
In the 2×2 case, if the coefficient determinant is zero, then the system is inconsistent if the numerator determinants are nonzero, or indeterminate if the numerator determinants are zero. For 3×3 or higher systems, the only thing one can say when the coefficient determinant equals zero is that if any of the numerator determinants are nonzero ...