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Let ABC be a triangle with side lengths a, b, and c, with a 2 + b 2 = c 2. Construct a second triangle with sides of length a and b containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length c = √ a 2 + b 2, the same as the hypotenuse of the first triangle.
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 shown in the second diagram. The larger piece, at the top, has width a and height a-b.
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
The prime decomposition of the number 2450 is given by 2450 = 2 · 5 2 · 7 2. Of the primes occurring in this decomposition, 2, 5, and 7, only 7 is congruent to 3 modulo 4. Its exponent in the decomposition, 2, is even. Therefore, the theorem states that it is expressible as the sum of two squares. Indeed, 2450 = 7 2 + 49 2.
If a and b have different parity, let p be any factor of a 2 + b 2 such that p 2 < a 2 + b 2. Then c = a 2 + b 2 − p 2 / 2p and d = a 2 + b 2 + p 2 / 2p . Note that p = d − c. A similar method exists [5] for generating all Pythagorean quadruples for which a and b are both even. Let l = a / 2 and m = b / 2 and ...
Since x 2 represents the area of a square with side of length x, and bx represents the area of a rectangle with sides b and x, the process of completing the square can be viewed as visual manipulation of rectangles. Simple attempts to combine the x 2 and the bx rectangles into a larger square
In algebra, the Brahmagupta–Fibonacci identity [1] [2] expresses the product of two sums of two squares as a sum of two squares in two different ways. Hence the set of all sums of two squares is closed under multiplication.
( ( a / b ) n = a n / b n ) ( a 2 and b 2 are integers) Therefore, a 2 is even because it is equal to 2b 2. (2b 2 is necessarily even because it is 2 times another whole number.) It follows that a must be even (as squares of odd integers are never even). Because a is even, there exists an integer k that fulfills =.