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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.
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 ...
A powerful number is a positive integer m such that for every prime number p dividing m, p 2 also divides m.Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a 2 b 3, where a and b are positive integers.
A square has even multiplicity for all prime factors (it is of the form a 2 for some a). The first: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 (sequence A000290 in the OEIS ). A cube has all multiplicities divisible by 3 (it is of the form a 3 for some a ).
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.
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.
All square triangular numbers have the form , where is a convergent to the continued fraction expansion of , the square root of 2. [ 4 ] A. V. Sylwester gave a short proof that there are infinitely many square triangular numbers: If the n {\displaystyle n} th triangular number n ( n + 1 ) 2 {\displaystyle {\tfrac {n(n+1)}{2}}} is square, then ...
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}.}