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The case (x, y, z) = (3, 5, 5) and all its permutations were proven by Bjorn Poonen in 1998. [26] The case (x, y, z) = (3, 6, n) and all its permutations were proven for n ≥ 3 by Bennett, Chen, Dahmen and Yazdani in 2014. [6] The case (x, y, z) = (2n, 3, 4) and all its permutations were proven for n ≥ 2 by Bennett, Chen, Dahmen and Yazdani ...
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...
The polynomial 3x 2 − 5x + 4 is written in descending powers of x. The first term has coefficient 3 , indeterminate x , and exponent 2 . In the second term, the coefficient is −5 .
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
That 641 is a factor of F 5 can be deduced from the equalities 641 = 2 7 × 5 + 1 and 641 = 2 4 + 5 4. It follows from the first equality that 2 7 × 5 ≡ −1 (mod 641) and therefore (raising to the fourth power) that 2 28 × 5 4 ≡ 1 (mod 641). On the other hand, the second equality implies that 5 4 ≡ −2 4 (mod 641
(x 2 + y 2)(x 2 − y 2) = z 2. Since x and y are coprime (this can be assumed because otherwise the factors could be cancelled), the greatest common divisor of x 2 + y 2 and x 2 − y 2 is either 2 (case A) or 1 (case B). The theorem is proven separately for these two cases.
Similarly, Baudhayana discovered that x = 17, y = 12 and x = 577, y = 408 are two solutions to the Pell equation, and that 17/12 and 577/408 are very close approximations to the square root of 2. [6] Later, Archimedes approximated the square root of 3 by the rational number 1351/780.
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.