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For example, when d=4, the hash table for two occurrences of d would contain the key-value pair 8 and 4+4, and the one for three occurrences, the key-value pair 2 and (4+4)/4 (strings shown in bold). The task is then reduced to recursively computing these hash tables for increasing n , starting from n=1 and continuing up to e.g. n=4.
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.
If the polynomial has rational roots, for example x 2 − 4x + 4 = (x − 2) 2, or x 2 − 3x + 2 = (x − 2)(x − 1), then the Galois group is trivial; that is, it contains only the identity permutation. In this example, if A = 2 and B = 1 then A − B = 1 is no longer true when A and B are swapped.
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 ...
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.
This counterintuitive result occurs because in the case where =, multiplying both sides by multiplies both sides by zero, and so necessarily produces a true equation just as in the first example. In general, whenever we multiply both sides of an equation by an expression involving variables, we introduce extraneous solutions wherever that ...
144 (one hundred [and] forty-four) is the natural number following 143 and preceding 145. It is coincidentally both the square of twelve (a dozen dozens , or one gross .) and the twelfth Fibonacci number , and the only nontrivial number in the sequence that is square.
The modular square root of can be taken this way. Having solved the associated quadratic equation we now have the variables w {\displaystyle w} and set v {\displaystyle v} = r {\displaystyle r} (if C {\displaystyle C} in the quadratic is a natural square).