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  • ...h>a,b,c,n</math> with <math>n \geq 3</math>, there are no solutions to the equation <math>a^n + b^n = c^n</math>. ...Theorem for <math>n \ge 7</math>. It is one of the most famous still-open problems in number theory, and despite various attempts to prove the conjecture, as
    3 KB (453 words) - 10:13, 9 June 2023
  • A '''Diophantine equation''' is an [[equation]] relating [[integer]] (or sometimes [[natural number]] or [[whole number]] ...metric form]] is used to express the relation between the variables of the equation.
    9 KB (1,434 words) - 00:15, 4 July 2024
  • ...quations]], testing whether certain large numbers are prime, and even some problems in cryptology. If we were solving this as an equation over the reals, we would immediately conclude that either <math>x - 5</math
    14 KB (2,317 words) - 18:01, 29 October 2021
  • {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}} ==Solution 1 (Diophantine Equation)==
    3 KB (442 words) - 02:13, 8 August 2022
  • ...> using the integers <math>0,1,2,\ldots,n^2</math> as digits. That is, the equation So we solve the [[diophantine equation]] <math>a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3</math>.
    2 KB (410 words) - 00:37, 25 August 2024
  • ...irs. Please check on "Comment of S2" below to see how to use [[Diophantine equation]] to make a simple deduction. ~ Will_Dai By the property of [[Diophantine equation]], given a problem to find integers x and y so that ax + by = c for some in
    3 KB (564 words) - 03:47, 4 August 2023
  • ...421</math> as the smallest integers satisfying this quadratic Diophantine equation. Then, since <math>cos B</math> = <math>\frac{29}{\sqrt{29^2 + k^2}}</math> [[Category:Intermediate Geometry Problems]]
    3 KB (536 words) - 08:54, 5 November 2024
  • ==The Basics: Twenty-Seven Problems== ...cs.tex Click here for the source file]. This file contains a collection of problems and solutions. Our beginning and newly intermediate-level problem solvers w
    5 KB (687 words) - 01:03, 4 February 2020
  • ...em has no solutions in the integers mod 19. Therefore the solution has no equation in the integers. <math>\blacksquare</math> ...the exponent lemma to examine the power of 3 that divides each side of the equation when <math>m > 0</math> to obtain
    7 KB (1,053 words) - 09:38, 12 August 2015
  • ...\dots , n_{14})</math> if any, apart from permutations, of the Diophantine Equation <math>n_1^4+n_2^4+\cdots +n_{14}^4=1599</math>. ...</math>, and thus there are no integral solutions to the given Diophantine equation.
    1 KB (183 words) - 03:45, 21 January 2023
  • A '''Pell equation''' is a type of [[diophantine equation]] in the form <math>x^2-Dy^2 = \pm1</math> for a [[natural number]] <math>D ...ve this it must first be shown that there is a single solution to the pell equation.
    6 KB (1,075 words) - 19:26, 16 November 2024
  • Here is a list of '''Olympiad Books''' that have Olympiad-level problems used to train students for future [[mathematics]] competitions. *''Inequalities An Approach Through Problems - '''B. J. Venkatachala'''
    19 KB (2,581 words) - 13:25, 1 November 2024
  • Problems from the '''1979 [[United States of America Mathematical Olympiad | USAMO]] ...\dots , n_{14})</math> if any, apart from permutations, of the Diophantine Equation <math>n_1^4+n_2^4+\cdots +n_{14}^4=1599</math>.
    2 KB (321 words) - 12:04, 24 December 2015
  • ===Solution 1 (Diophantine PoP)=== ...</math> with respect to circle <math>A.</math> This yields the diophantine equation
    5 KB (848 words) - 09:16, 3 December 2024
  • ...problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that <math>x</math>... voilà! ...dd the two systems in <math>y+x</math> and <math>y-x+1</math> to get a new equation in terms of <math>2y+1.</math> In order for <math>y</math> to be an integer
    8 KB (1,434 words) - 00:23, 10 November 2024
  • ...e|[[2019 AMC 10B Problems#Problem 25|2019 AMC 10B #25]] and [[2019 AMC 12B Problems#Problem 23|2019 AMC 12B #23]]}} ...th> must satisfy <math>2x+3y=18</math>. We recognize this as a Diophantine equation. Taking <math>\pmod{2}</math> yields <math>y=0 \pmod{2}</math>. Since <math
    13 KB (2,175 words) - 15:16, 12 December 2024
  • ...ions, we see that <math>21k - 60\ell = 57</math>. Solving this Diophantine equation gives us that <math>k = 20a + 17</math> and <math>\ell = 7a + 5</math>. Sin ...inition, <math>b = 7x + 3.</math> Substituting this back into our original equation, we get <cmath>n = 60(7x + 3) + 57 = 420x + 237.</cmath>
    17 KB (2,522 words) - 18:33, 31 August 2024
  • Now, we find integer solution(s) of this equation with <math>c \leq b</math>. Multiplying this equation by 4, we get
    16 KB (2,730 words) - 01:56, 4 January 2023
  • ...ate|[[2022 AMC 10A Problems/Problem 1|2022 AMC 10A #1]] and [[2022 AMC 12A Problems/Problem 1|2022 AMC 12A #1]]}} ...ath>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}</ma
    3 KB (425 words) - 23:45, 14 September 2024