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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 ...
The 2PL is equivalent to the 3PL model with =, and is appropriate for testing items where guessing the correct answer is highly unlikely, such as fill-in-the-blank items ("What is the square root of 121?"), or where the concept of guessing does not apply, such as personality, attitude, or interest items (e.g., "I like Broadway musicals. Agree ...
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.
For example, to factor =, the first try for a is the square root of 5959 rounded up to the next integer, which is 78. Then b 2 = 78 2 − 5959 = 125 {\displaystyle b^{2}=78^{2}-5959=125} . Since 125 is not a square, a second try is made by increasing the value of a by 1.
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 ...
The same technique can be applied to the square roots of any other value, particularly the square roots of −1 mentioned in § Combining multiple tests. If two (successful) strong probable prime tests find x 2 ≡ −1 (mod n) and y 2 ≡ −1 (mod n), but x ≢ ±y (mod n), then gcd(x − y, n) and gcd(x + y, n) are nontrivial factors of n. [10]
Such an n is easy to factor, because in this case, n+1 = (p+1) 2 is a perfect square. One can quickly detect perfect squares using Newton's method for square roots. By combining a Lucas pseudoprime test with a Fermat primality test, say, to base 2, one can obtain very powerful probabilistic tests for primality, such as the Baillie–PSW ...
The first couple of times do not yield anything interesting (the result was still 1 modulo 31697), but at exponent 3962 we see a result that is neither 1 nor minus 1 (i.e. 31696) modulo 31697. This proves 31697 is in fact composite (it equals 29×1093). Modulo a prime, the residue 1 can have no other square roots than 1 and minus 1.