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The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.
In this vein, the discriminant is a symmetric function in the roots that reflects properties of the roots – it is zero if and only if the polynomial has a multiple root, and for quadratic and cubic polynomials it is positive if and only if all roots are real and distinct, and negative if and only if there is a pair of distinct complex ...
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
Fermat's method works best when there is a factor near the square-root of N. If the approximate ratio of two factors ( d / c {\displaystyle d/c} ) is known, then a rational number v / u {\displaystyle v/u} can be picked near that value.
The square root of 2 is an algebraic number equal to the length of the hypotenuse of a right triangle with legs of length 1.. An algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients.
A number where some but not all prime factors have multiplicity above 1 is neither square-free nor squareful. The Liouville function λ(n) is 1 if Ω(n) is even, and is -1 if Ω(n) is odd. The Möbius function μ(n) is 0 if n is not square-free. Otherwise μ(n) is 1 if Ω(n) is even, and is −1 if Ω(n) is odd.
We are not taking the square root of any negative values here, since both and are necessarily positive. But we have lost the solution x = − 2. {\displaystyle x=-2.} The reason is that x {\displaystyle x} is actually not in general the positive square root of x 2 . {\displaystyle x^{2}.}
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 ...