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  • ...} [http://www.garlic.com/~wedgingt/MMPstats.txt], a [[bit level]] over 169.4. The current version of [[Prime95]] cannot handle numbers this large, nor c
    2 KB (354 words) - 14:52, 19 September 2021
  • Prove that N = 811 is prime knowing that N-1 = 2 &times; 3<sup>4</sup> &times; 5
    1 KB (177 words) - 14:31, 17 February 2019
  • ...= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 + ...</math> <u> 1 2. 3 4 </u>
    13 KB (1,873 words) - 16:52, 24 October 2020
  • ...of other numbers, like the [[Generalized Fermat number]]s <math>F_{n,2} = 4^{3^n}+2^{3^n}+1</math> with k = 5 instead of k = 3.
    2 KB (401 words) - 14:40, 6 March 2019
  • ...ally taken to be 1, but that is not essential. In some proofs (see example 4 below) we have to show that the statement is true for several values of n. :<math>\sum_{k=1}^{n}k^{3}\,=\,\frac{n^{2}(n+1)^{2}}{4}</math>
    4 KB (679 words) - 13:57, 20 February 2019
  • ...of the Miller-Rabin test, for example, has a probability of only <math>{1/4}^{100}</math> of being composite, which is less than <math>10^{-60}</math>.
    1 KB (155 words) - 20:32, 25 July 2020
  • ===Modulus congruent to 3 modulo 4=== :<math>r\equiv \pm a^{(m+1)/4}\ \pmod m</math>
    5 KB (726 words) - 10:38, 6 February 2019
  • ...es} 1 & \text{if } p \equiv 1 \pmod{4} \\ -1 & \text{if } p \equiv 3 \pmod{4} \end{cases}</math> ...property is known as the [[law of quadratic reciprocity]]. The properties 4 and 5 are traditionally known as the ''supplements'' to quadratic reciproci
    2 KB (348 words) - 18:57, 28 September 2023
  • *If at least one of <math>p</math> or <math>q</math> are congruent to 1 mod 4: <math>p</math> is a quadratic residue modulo <math>q</math> if and only if *If both of <math>p</math> or <math>q</math> are congruent to 3 mod 4: <math>p</math> is a quadratic residue modulo <math>q</math> if and only if
    1 KB (208 words) - 18:19, 2 October 2022
  • This holds true for the first 4 terms: However this does not hold true for next 4 terms:
    4 KB (655 words) - 14:50, 19 September 2021
  • :<math>45^2\,\equiv \,2^4*7^0*13^1</math>
    10 KB (1,763 words) - 02:56, 12 March 2019
  • ...strated that a composite <math>N</math> will only pass at most <math>(N-1)/4</math> such tests. ...35. Since the exponent is 4, the sequence will use exponents from zero to 4.
    3 KB (432 words) - 15:33, 28 January 2019
  • 3*2^41+1 is a Factor of xGF(38,4,3)!!!! (0.000000 seconds)
    5 KB (774 words) - 07:39, 27 May 2024
  • ...hunting on his media server to "give back" to the project. After less than 4 months and on just his fourth try, he discovered the new prime number. By w
    987 bytes (147 words) - 01:27, 15 January 2024
  • | 4 || [[Team Prime Rib|Ars Technica Team Prime Rib]] || 4457877
    2 KB (206 words) - 09:56, 7 March 2019
  • :<math>s_1\ =\ 18\ -\ 10\ =\ 8,\ \sigma(8)\ =\ 1\ +\ 2\ +\ 4\ +\ 8</math> :<math>2^4\ *\ 31</math>
    6 KB (914 words) - 19:49, 21 February 2023
  • The divisors of 12 are <math>(1, 2, 3, 4, 6, 12)</math>, so :<math>\sigma(12)\ =\ 1+2+3+4+6+12\ =\ 28</math>
    671 bytes (92 words) - 00:34, 30 January 2019
  • ...whole numbers from 2 to P plus the number 1. In other words, Q = (2 x 3 x 4 x 5 ... x P) + 1. From the form of the number Q, it is obvious that no inte :the remainder r can be only 0, 1, 2, 3, 4, or 5
    2 KB (447 words) - 00:22, 10 July 2023
  • | align="right" | 10<sup>4</sup> || align="right" | 205
    2 KB (255 words) - 06:08, 21 February 2023
  • 4
    581 bytes (64 words) - 19:18, 5 April 2023

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