Diofant tenglamalarni yechish
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SOLVE THE DIOPHANTES EQUATIONS
- Bu sahifa navigatsiya:
- 3.Yuqori darajali diofant tenglamalarini yechish.
- 4-misol
- 5-misol.
- 6-misol.
- Илмий хабарнома. Физика-математика тадқиқотлари, 2020, №1(3)
- 8-misol.
- 4. Mustaqil yechish uchun tenglamalar.
- SOLVE THE DIOPHANTE`S EQUATIONS T.T. Ibaydullayev 1 , A. L.Abdulvohidov 1
- Ibaydullayev To`lanboy Tursunboyevich
- Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
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3-misol. 127𝑥𝑥 − 52𝑦𝑦 + 1 = 0 tenglamani yeching. Yechish: 127
52 = 2 + 1 2 + 1 3+ 1 1+15 ; 2 + 1 2 +
1 3+ 1 1+0 =
22 9 ;
127
52 − 22 9 = 1143 − 1144 52 ∙ 9
= − 1 52 ∙ 9 ;
bundan 127 − 52 ∙ 22 + 1 = 0 berilgan tenglama bilan taqqoslab 𝑥𝑥 0 = 2;
𝑦𝑦 0 = 22 va shunga ko`ra uning yechimlari 𝑥𝑥 = 9 + 52𝑡𝑡 , 𝑦𝑦 = 22 + 127𝑡𝑡 , 𝑡𝑡 = 0, ±1, ±2, … .. formula bilan berilishini ko`ramiz. 3.Yuqori darajali diofant tenglamalarini yechish. Yuqori darajali aniqmas tenglamalarni butun sonlarda yechishning konkret usullari bo`lmasa-da, biz ba`zi usullar: qoldiqlar nazariyasidan, qisqa ko`paytirish formulalaridan hamda mantiqiy fikrlardan foydalanamiz: a) qoldiqlar nazariyasidan foydalanish: har qanday sonning kvadratini 3 ga yoki 4 ga bo`lishda qoldiqda 0 yoki 1 sonlari hosil bo`ladi. Haqiqatan ham: (2𝑚𝑚) 2 = 4𝑚𝑚
2 ; (2𝑚𝑚 + 1) 2 = 4(𝑚𝑚
2 + 𝑚𝑚) + 1; (3𝑚𝑚 − 1) 2 = 3(3𝑚𝑚 2 − 2𝑚𝑚) + 1; (3𝑚𝑚) 2 = 3 ∙ 3𝑚𝑚 2 ; (3𝑚𝑚 + 1) 2 = 3(3𝑚𝑚
2 + 2𝑚𝑚) + 1 ayniyatlar fikrimizni tasdiqlaydi. Endi tenglamalarni yechish uchun namunalar keltiramiz. 4-misol. 99𝑥𝑥 2 − 97𝑦𝑦 2 = 2005 tenglamani yeching. Yechish: Berilgan tenglamani 4(25𝑥𝑥 2 − 24𝑦𝑦 2 − 501) = 𝑥𝑥 2 + 𝑦𝑦
2 + 1 ko`rinishida yozamiz. Sonning kvadratini 4 ga bo`lishda qoldiqda 0 yoki 1 qolgani uchun 𝑥𝑥 2 + 𝑦𝑦 2 + 1 ifodani 4 ga bo`lishda qoldiqda 1, 2, 3 sonlari hosil bo`ladi. Bunday tenglikning bo`lishi mumkin emas, demak, berilgan tenglama yechimga ega emas.
5-misol. 𝑥𝑥 3 − 𝑥𝑥 = 3𝑦𝑦 2 + 1 tenglamani natural sonlarda yeching. Yechish: Berilgan tenglamani 𝑥𝑥(𝑥𝑥 − 1)(𝑥𝑥 + 1) = 3𝑦𝑦 2 + 1 ko`rinishida yozsak, tenlamaning chap tomonida doimo 3 ga karrali, o`ng tomonida esa 3 ga bo`lishda doimo 1 qoldiq bo`ladi. Bunday tenglikning bo`lishi mumkin emas va berilgan tenglama yechimga ega emas. b) qisqa ko`paytirish formulalaridan foydalanish: bu holda 𝑎𝑎 2 − 𝑏𝑏 2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏); 𝑎𝑎 3 + 𝑏𝑏
3 = (𝑎𝑎 +
𝑏𝑏)( 𝑎𝑎 2 − 𝑎𝑎𝑏𝑏 + 𝑏𝑏 2 ); 𝑎𝑎
3 − 𝑏𝑏
3 = (𝑎𝑎 − 𝑏𝑏)(𝑎𝑎 2 + +𝑎𝑎𝑏𝑏 + 𝑏𝑏 2 ) formulalaridan foydalanib misollar yechiladi. 6-misol. 𝑥𝑥 3 + 91 = 𝑦𝑦 3 tenglamani natural sonlarda yeching. Yechish: Tenglamani 𝑦𝑦 3 − 𝑥𝑥 3 = 91, ya`ni (𝑦𝑦 − 𝑥𝑥)(𝑦𝑦 2 + 𝑦𝑦𝑥𝑥 + 𝑥𝑥 2 ) = 7 ∙ 13 ko`rinishida yozib,
{
𝑦𝑦 2 + 𝑦𝑦𝑥𝑥 + 𝑥𝑥 2 = 91
va { 𝑦𝑦 − 𝑥𝑥 = 7 𝑦𝑦 2
2 = 13
sistemalarini hosil qilamiz. Bularni yechib, 𝑥𝑥 = 5, 𝑦𝑦 = 6 yechimni topamiz. v) zanjirli kasrlarga ajratish bilan yechiladigan misollardan namunalar keltiramiz. 7-misol. 55(𝑥𝑥 3 𝑦𝑦 3 + 𝑥𝑥
2 + 𝑦𝑦
2 ) = 229(𝑥𝑥𝑦𝑦 3 + 1) tenglamani yeching. Yechish: Tenglamani
𝑥𝑥 3 𝑦𝑦 3 + 𝑥𝑥 2 + 𝑦𝑦 2 𝑥𝑥𝑦𝑦
3 + 1
= 229
55
shaklda yozib, zanjirli kasrlarga ajratamiz. Natijada ketma-ket:
𝑥𝑥 2 + y 2 𝑥𝑥𝑦𝑦
3 + 1 = 4 + 9 55 ; 𝑥𝑥 2 + 1 𝑥𝑥𝑦𝑦 + 1 y 2 = 4 +
1 6 +
1 9
tengliklarga kelamiz. Oxirgi tenglikdan 𝑥𝑥 2 = 4 , 𝑥𝑥𝑦𝑦 = 6, 𝑦𝑦 2 = 9, ya`ni tenglama (±2 ; ±3 ) yechimga ega ekanligini ko`ramiz. MATHEMATICS 68 Илмий хабарнома. Физика-математика тадқиқотлари, 2020, №1(3)
g) mantiqiy fikrlashlar tenglamalarning butun sonlardagi yechimini topishning eng asosiy yo`lidir. Masalan, mashhur “Ferma masalasi”: kishilar 2000-yilgacha har qanday n≥3 natural soni uchun 𝑥𝑥 𝑛𝑛 + 𝑦𝑦
𝑛𝑛 = 𝑧𝑧
𝑛𝑛
tenglama natural sonlarda yechimga ega emas ekanligini isbotlash masalasiga asosiy e’tiborni qaratganlar. Hozirda bu masala yechimi ma`lum. Biz “Ferma muammosi”ning xususiy holi bo`lgan quyidagi misolni qaraymiz. 8-misol. 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑧𝑧
2 tenglamani natural sonlarda yeching. Yechish. Agar (𝑥𝑥, 𝑦𝑦) = 𝑑𝑑 deb faraz qilsak, 𝑥𝑥 = 𝑑𝑑𝑥𝑥 1 , 𝑦𝑦 = 𝑑𝑑𝑥𝑥 2 bo`lib, 𝑧𝑧 = 𝑑𝑑𝑧𝑧 1 ekanini ko`ramiz. Shuning uchun (𝑥𝑥, 𝑦𝑦) = 1 deb faraz qilamiz. 𝑥𝑥 2 == 𝑧𝑧 2 — 𝑦𝑦
2 = (𝑧𝑧 + 𝑦𝑦)(𝑧𝑧 − 𝑦𝑦) dan 𝑧𝑧 + 𝑦𝑦 = 𝑎𝑎𝑑𝑑 1 , (𝑎𝑎, 𝑏𝑏) = 1 deb olsak, 𝑥𝑥 2 = 𝑎𝑎𝑏𝑏𝑑𝑑 1 2 . Bundan esa 𝑎𝑎 = 𝑢𝑢 2 , 𝑏𝑏 = 𝑣𝑣 2
1 , 𝑑𝑑
1 = 1 deb olinsa, 𝑥𝑥 = 𝑢𝑢𝑣𝑣, 𝑧𝑧 + 𝑦𝑦 = 𝑢𝑢 2 , 𝑧𝑧 − 𝑦𝑦 = 𝑣𝑣 2 va bundan esa tenglamaning (𝑢𝑢𝑣𝑣, 𝑢𝑢 2
2 2 , 𝑢𝑢 2 +𝑣𝑣 2 2 ) , 𝑢𝑢 > 𝑣𝑣 yechimlarini topamiz. Agar 𝑢𝑢 = 2𝑚𝑚, 𝑣𝑣 = 2𝑚𝑚 deb olinsa, yechim (4𝑚𝑚𝑚𝑚, 2(𝑚𝑚 2 − 𝑚𝑚 2 ), 2(𝑚𝑚
2 + 𝑚𝑚
2 )) ko`rinishini oladi, bu yerda 𝑚𝑚 > 𝑚𝑚 natural sonlardir. Endi o`quvchilar mustaqil bajarishlari uchun quyidagi misollarni taklif etamiz. 4. Mustaqil yechish uchun tenglamalar. 1. Qoldiqlar nazariyasidan foydalanib, quyidagi tenglamalarni butun sonlarda yeching. 1) 𝑥𝑥 2
2 = −1; 10) 𝑥𝑥 3 − 3𝑦𝑦
2 = 17;
2) 𝑥𝑥 2 − 3𝑦𝑦 2 = 17; 11) 2𝑥𝑥 2 − 5𝑦𝑦
2 = 7;
3) 3𝑥𝑥 2 + 2 = 𝑦𝑦 2 ; 12) 𝑦𝑦 2 = 5𝑥𝑥
2 + 6;
4) 3𝑥𝑥 2 + 8 = 𝑦𝑦 2 ; 13) 15𝑥𝑥 2 − 7𝑦𝑦
2 = 9;
5) 𝑥𝑥 2 + 4𝑥𝑥 − 8𝑦𝑦 = 11; 14) 𝑥𝑥 2 − 4𝑦𝑦 = 1; 6) 3𝑥𝑥 2
2 = 13; 15) 𝑥𝑥 3 − 𝑥𝑥 = 3𝑦𝑦 2 + 1;
7) 19𝑥𝑥 2 + 28𝑦𝑦 2 = 729; 16) 6𝑥𝑥 2 + 5𝑦𝑦
2 = 74;
8) 3𝑥𝑥 2 + 1 = 5𝑦𝑦; 17) 2003𝑥𝑥 2 − 2001𝑦𝑦
2 = 1999;
9) 2𝑥𝑥 2 − 1 = 5𝑦𝑦; 18) 2005𝑥𝑥 2 + 2004𝑦𝑦
2 − 2005𝑧𝑧
2 = 2001;
19) 20𝑥𝑥 2 + 35𝑦𝑦 2 = 2004.
2. Quyidagi tenglamalarni natural sonlarda yeching. 1) 𝑥𝑥 2
2 = 91; 5) 𝑥𝑥 3 − 𝑦𝑦
3 = 2005
2) 𝑥𝑥 2 − 𝑦𝑦 2 = 2005; 6) 𝑥𝑥 2 − 656𝑥𝑥𝑦𝑦 − 65𝑦𝑦2 = 1983; 3) 𝑥𝑥 3 + 91 = 𝑦𝑦 3 ; 7) 𝑥𝑥 3 + 𝑦𝑦
3 = 1972;
4) 𝑥𝑥 3 + 𝑦𝑦 3 = 2005; 8) 49𝑥𝑥 2 − 36𝑦𝑦
2 = 625.
3. Quyidagi tenglamalarni natural sonlarda yeching. 1) 1! + 2! + 3! + ⋯ … + 𝑥𝑥! = 𝑦𝑦𝑧𝑧; 2) (𝑥𝑥
1 2 + 1)(𝑥𝑥 2 2 + 2 2 ) … … (𝑥𝑥 𝑛𝑛 2
2 ) = 2
𝑛𝑛 𝑚𝑚! 𝑥𝑥
1 𝑥𝑥 2 … . . 𝑥𝑥 𝑛𝑛 ; 3) 31(𝑥𝑥𝑦𝑦𝑧𝑧𝑥𝑥 + 𝑥𝑥𝑦𝑦 + 𝑥𝑥𝑥𝑥 + 𝑧𝑧𝑥𝑥 + 1) = 40(𝑦𝑦𝑧𝑧𝑥𝑥 + 𝑦𝑦 + 1); 4) 𝑦𝑦𝑧𝑧(𝑧𝑧𝑥𝑥 − 10) + 𝑧𝑧(𝑥𝑥 + 𝑧𝑧) − 10 = 0; 5) 55(𝑥𝑥 3 𝑦𝑦 3 + 𝑥𝑥
2 + 𝑦𝑦
2 ) = 229(𝑥𝑥𝑦𝑦 3 + 1);
6) 𝑥𝑥 2 + 2𝑦𝑦 2 = 𝑧𝑧
2 ; 7) 𝑥𝑥 2 − 4𝑥𝑥𝑦𝑦 + 5𝑦𝑦 2 = 169.
Adabiyotlar:
1. Ayupov Sh.A., Rixsiyev B.B., Qo`chqorov O.Sh. Matematika olimpiadalari masalalari. 1-qism. – Toshkent: Fan, 2004. – 55 b.; 2-qism. – Toshkent: Fan, 2004. – 52 b. 2. Агаханов Н.X., Купцов Л.П., Нестеренко Ю.В., Резниченко С.В., Слинько А.М. Maтематические олимпиады школьников. – Moсква: Просвещение, 1997. – 208 с. 3. Васильев Н.Б., Гутенмаxер В.Л., Раббот Ж.П., Тоом А.Л. Заочные математические олимпиады. – Moсква: Наука, 1981. – 128 с. 4. Гальперин Г.А., Толпыго А.К. Московские математические олимпиады. – Moсква: Просвещение, 1986. – 303 с. 5. Tохиров А., Мўминов Ғ. Математикадан олимпиада масалалари. – Тошкент: Ўқитувчи, 1996. – 120 б. 6. Мo`minov G`. Nishonov T. Маtеmatika olimpiadalari masalalari. – Andijon: Hayot, 2017. – 144 б. 7. Botirov Sh. Matematika. Mavzulashtirilgan testlar to`plami. – Buxoro: “Buxoro” нашриёти, 2009. – 224 б. 8. Усманов М. Математикадан мисол ва масалалар тўплами. – Тошкент: Navro`z, 2016. – 654 б. 9. Mirzaahmedov M.A. va boshq. Matematika, 10. I qism. – Toshkent: Extremum press, 2017. – 191 б. 10. Mirzaahmedov M.A. va boshq. Matematika, 11. II qism. – Toshkent: Zamin nashr, 2018. – 143 б.
SOLVE THE DIOPHANTE`S EQUATIONS T.T. Ibaydullayev 1 , A. L.Abdulvohidov 1 Ilmiy xabarnoma. Fizika-matematika tadqiqotlari – Scientific Bulletin. Physical and Mathematical Research. 2020. 1(3). 62 – 69. 1 Andijan State University, Andijan, 170100, Str. University, 129 (Uzbekistan). E-mail: agsu_info@edu.uz Keywords: diophantine equations, integer solution, theory of divisibility, theory of remainders, equivalent fraktions, Euclidean algorithm.
This article is based on the lectures for gifted students of the faculty of Physics and Mathematics on the solution of Diophantine equations in science circles. If the number of unknowns involved in a system of equations exceeds the number of equations, such equations are called Diophantine equations or indeterminate equations. Specifically, equations of the form
3𝑥𝑥 − 5𝑦𝑦 = 8, 𝑥𝑥 2 + 3𝑥𝑥𝑦𝑦 − 𝑦𝑦 2 = 12,
𝑥𝑥 3 + 𝑦𝑦 2 − 3𝑥𝑥 + 5 = 0, 𝑥𝑥 3 + 𝑦𝑦
3 = 𝑧𝑧
3 , … are indefinite equations. Many of the equation or system of equations determine all the numbers to find solutions to the most common examples. Short multiplication formulas, theory serve as the primary means of logical thinking in the solution of mathematical equation. However, in the theory of division of equations of the form such as 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐, there are formulas for solving the corresponding fractions. Firstly, we presented the basic properties of solving equations of this type in integer numbers based on the theory of division, formulas for solving them, and finally described the examples of solving them by appropriate fractions. In the next step, we give examples of solving high-level indeterminate equations. The importance of studying the solution of linear and high-level indeterminate equations in whole numbers is obvious in solving many practical problems in such areas as finance, economics, technology. Therefore, these issues are included in the programs of entrance exams to higher education institutions. We believe that theoretical information in this article and as well as specific examples with solutions can be used by applicants while preparing for entrance exams to higher education institutions and their teachers, as well as teachers mathematics who organize extracurricular activities in secondary schools. For this purpose, at the end of the article there are a number of independent works that can be solved on the basis of examples and theoretical data, as well as to strengthen knowledge on these issues. The set of examples and questions provided by the State Test Center under the Cabinet of ministers of the Republic of Uzbekistan, intended for entrance exams to higher education institutions in different years is was widely being used in covering the topic of the article.
1. Ayupov Sh.A., Rihsiyev B.B., Qo`chqorov O.Sh. (2004) Matematika olimpiadalari masalalari [Problems of mathematical olympiads]. Part 1-2. Tashkent: Fan. 2. Agahanov N.Kh., Kuptsov L.P., Nesterenko Yu.V., Reznichenko S.V., Slinko A.M. (1997) Matematicheskiye olimpiady dlya shkolnikov [Mathematical Olympiads for pupils]. Moscow: Prosveshcheniye. 3
[Correspondence Mathematical Olympiads]. Moscow: Nauka. 4. Gal'perin G.A., Tolpygo A.K . (1996) Moskovskiye matematicheskiye olimpiady [Moscow Mathematical Olympiads]. Moscow: Prosveshcheniye. 5. Tohirov А., Mo`minov G`. Matematikadan olimpiada masalalari [Olympiad problems in mathematics]. Tashkent: O`qituvchi. 6. Мo`minov G`. Nishonov T. (2017) Маtеmatika olimpiadalari masalalari [Problems of Mathematical Olympiads]. Andijon: Hayot. 7. Botirov Sh. (2009) Matematika. Mavzulashtirilgan testlar to`plami [A set of thematic tests]. Buxoro: Publishing House "Buxoro". 8. Usmonov M. (2016) Matematikadan misol va mavzular to`plami [A set of examples and problems from mathemat- ics]. Tashkent: Navro`z. 9. Mirzaahmedov M.A., and others. (2017) Matematika. [Mathematic]. Schoolbook for 10th class. Part 1. Tashkent: Extremum press. 10. Mirzaahmedov M.A., and others. (2018) Matematika [Mathematic 11]. Schoolbook for 11th class. Part 2. Tashkent: Zamin nashr.
Matematika kafedrasi dotsenti. E-mail: ibaydullayev73@mail.ru Abdulvoxidov Alisher Latipovich – Andijon davlat universiteti Matematika kafedrasi tayanch doktoranti. E-mail: abdulvoxidov_a@mail.ru
МАТЕМАТИКА 69 Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
g) mantiqiy fikrlashlar tenglamalarning butun sonlardagi yechimini topishning eng asosiy yo`lidir. Masalan, mashhur “Ferma masalasi”: kishilar 2000-yilgacha har qanday n≥3 natural soni uchun 𝑥𝑥 𝑛𝑛 + 𝑦𝑦
𝑛𝑛 = 𝑧𝑧
𝑛𝑛
tenglama natural sonlarda yechimga ega emas ekanligini isbotlash masalasiga asosiy e’tiborni qaratganlar. Hozirda bu masala yechimi ma`lum. Biz “Ferma muammosi”ning xususiy holi bo`lgan quyidagi misolni qaraymiz. 8-misol. 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑧𝑧
2 tenglamani natural sonlarda yeching. Yechish. Agar (𝑥𝑥, 𝑦𝑦) = 𝑑𝑑 deb faraz qilsak, 𝑥𝑥 = 𝑑𝑑𝑥𝑥 1 , 𝑦𝑦 = 𝑑𝑑𝑥𝑥 2 bo`lib, 𝑧𝑧 = 𝑑𝑑𝑧𝑧 1 ekanini ko`ramiz. Shuning uchun (𝑥𝑥, 𝑦𝑦) = 1 deb faraz qilamiz. 𝑥𝑥 2 == 𝑧𝑧 2 — 𝑦𝑦
2 = (𝑧𝑧 + 𝑦𝑦)(𝑧𝑧 − 𝑦𝑦) dan 𝑧𝑧 + 𝑦𝑦 = 𝑎𝑎𝑑𝑑 1 , (𝑎𝑎, 𝑏𝑏) = 1 deb olsak, 𝑥𝑥 2 = 𝑎𝑎𝑏𝑏𝑑𝑑 1 2 . Bundan esa 𝑎𝑎 = 𝑢𝑢 2 , 𝑏𝑏 = 𝑣𝑣 2
1 , 𝑑𝑑
1 = 1 deb olinsa, 𝑥𝑥 = 𝑢𝑢𝑣𝑣, 𝑧𝑧 + 𝑦𝑦 = 𝑢𝑢 2 , 𝑧𝑧 − 𝑦𝑦 = 𝑣𝑣 2 va bundan esa tenglamaning (𝑢𝑢𝑣𝑣, 𝑢𝑢 2
2 2 , 𝑢𝑢 2 +𝑣𝑣 2 2 ) , 𝑢𝑢 > 𝑣𝑣 yechimlarini topamiz. Agar 𝑢𝑢 = 2𝑚𝑚, 𝑣𝑣 = 2𝑚𝑚 deb olinsa, yechim (4𝑚𝑚𝑚𝑚, 2(𝑚𝑚 2 − 𝑚𝑚 2 ), 2(𝑚𝑚
2 + 𝑚𝑚
2 )) ko`rinishini oladi, bu yerda 𝑚𝑚 > 𝑚𝑚 natural sonlardir. Endi o`quvchilar mustaqil bajarishlari uchun quyidagi misollarni taklif etamiz. 4. Mustaqil yechish uchun tenglamalar. 1. Qoldiqlar nazariyasidan foydalanib, quyidagi tenglamalarni butun sonlarda yeching. 1) 𝑥𝑥 2
2 = −1; 10) 𝑥𝑥 3 − 3𝑦𝑦
2 = 17;
2) 𝑥𝑥 2 − 3𝑦𝑦 2 = 17; 11) 2𝑥𝑥 2 − 5𝑦𝑦
2 = 7;
3) 3𝑥𝑥 2 + 2 = 𝑦𝑦 2 ; 12) 𝑦𝑦 2 = 5𝑥𝑥
2 + 6;
4) 3𝑥𝑥 2 + 8 = 𝑦𝑦 2 ; 13) 15𝑥𝑥 2 − 7𝑦𝑦
2 = 9;
5) 𝑥𝑥 2 + 4𝑥𝑥 − 8𝑦𝑦 = 11; 14) 𝑥𝑥 2 − 4𝑦𝑦 = 1; 6) 3𝑥𝑥 2
2 = 13; 15) 𝑥𝑥 3 − 𝑥𝑥 = 3𝑦𝑦 2 + 1;
7) 19𝑥𝑥 2 + 28𝑦𝑦 2 = 729; 16) 6𝑥𝑥 2 + 5𝑦𝑦
2 = 74;
8) 3𝑥𝑥 2 + 1 = 5𝑦𝑦; 17) 2003𝑥𝑥 2 − 2001𝑦𝑦
2 = 1999;
9) 2𝑥𝑥 2 − 1 = 5𝑦𝑦; 18) 2005𝑥𝑥 2 + 2004𝑦𝑦
2 − 2005𝑧𝑧
2 = 2001;
19) 20𝑥𝑥 2 + 35𝑦𝑦 2 = 2004.
2. Quyidagi tenglamalarni natural sonlarda yeching. 1) 𝑥𝑥 2
2 = 91; 5) 𝑥𝑥 3 − 𝑦𝑦
3 = 2005
2) 𝑥𝑥 2 − 𝑦𝑦 2 = 2005; 6) 𝑥𝑥 2 − 656𝑥𝑥𝑦𝑦 − 65𝑦𝑦2 = 1983; 3) 𝑥𝑥 3 + 91 = 𝑦𝑦 3 ; 7) 𝑥𝑥 3 + 𝑦𝑦
3 = 1972;
4) 𝑥𝑥 3 + 𝑦𝑦 3 = 2005; 8) 49𝑥𝑥 2 − 36𝑦𝑦
2 = 625.
3. Quyidagi tenglamalarni natural sonlarda yeching. 1) 1! + 2! + 3! + ⋯ … + 𝑥𝑥! = 𝑦𝑦𝑧𝑧; 2) (𝑥𝑥
1 2 + 1)(𝑥𝑥 2 2 + 2 2 ) … … (𝑥𝑥 𝑛𝑛 2
2 ) = 2
𝑛𝑛 𝑚𝑚! 𝑥𝑥
1 𝑥𝑥 2 … . . 𝑥𝑥 𝑛𝑛 ; 3) 31(𝑥𝑥𝑦𝑦𝑧𝑧𝑥𝑥 + 𝑥𝑥𝑦𝑦 + 𝑥𝑥𝑥𝑥 + 𝑧𝑧𝑥𝑥 + 1) = 40(𝑦𝑦𝑧𝑧𝑥𝑥 + 𝑦𝑦 + 1); 4) 𝑦𝑦𝑧𝑧(𝑧𝑧𝑥𝑥 − 10) + 𝑧𝑧(𝑥𝑥 + 𝑧𝑧) − 10 = 0; 5) 55(𝑥𝑥 3 𝑦𝑦 3 + 𝑥𝑥
2 + 𝑦𝑦
2 ) = 229(𝑥𝑥𝑦𝑦 3 + 1);
6) 𝑥𝑥 2 + 2𝑦𝑦 2 = 𝑧𝑧
2 ; 7) 𝑥𝑥 2 − 4𝑥𝑥𝑦𝑦 + 5𝑦𝑦 2 = 169.
Adabiyotlar:
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SOLVE THE DIOPHANTE`S EQUATIONS T.T. Ibaydullayev 1 , A. L.Abdulvohidov 1 Ilmiy xabarnoma. Fizika-matematika tadqiqotlari – Scientific Bulletin. Physical and Mathematical Research. 2020. 1(3). 62 – 69. 1 Andijan State University, Andijan, 170100, Str. University, 129 (Uzbekistan). E-mail: agsu_info@edu.uz Keywords: diophantine equations, integer solution, theory of divisibility, theory of remainders, equivalent fraktions, Euclidean algorithm.
This article is based on the lectures for gifted students of the faculty of Physics and Mathematics on the solution of Diophantine equations in science circles. If the number of unknowns involved in a system of equations exceeds the number of equations, such equations are called Diophantine equations or indeterminate equations. Specifically, equations of the form
3𝑥𝑥 − 5𝑦𝑦 = 8, 𝑥𝑥 2 + 3𝑥𝑥𝑦𝑦 − 𝑦𝑦 2 = 12,
𝑥𝑥 3 + 𝑦𝑦 2 − 3𝑥𝑥 + 5 = 0, 𝑥𝑥 3 + 𝑦𝑦
3 = 𝑧𝑧
3 , … are indefinite equations. Many of the equation or system of equations determine all the numbers to find solutions to the most common examples. Short multiplication formulas, theory serve as the primary means of logical thinking in the solution of mathematical equation. However, in the theory of division of equations of the form such as 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐, there are formulas for solving the corresponding fractions. Firstly, we presented the basic properties of solving equations of this type in integer numbers based on the theory of division, formulas for solving them, and finally described the examples of solving them by appropriate fractions. In the next step, we give examples of solving high-level indeterminate equations. The importance of studying the solution of linear and high-level indeterminate equations in whole numbers is obvious in solving many practical problems in such areas as finance, economics, technology. Therefore, these issues are included in the programs of entrance exams to higher education institutions. We believe that theoretical information in this article and as well as specific examples with solutions can be used by applicants while preparing for entrance exams to higher education institutions and their teachers, as well as teachers mathematics who organize extracurricular activities in secondary schools. For this purpose, at the end of the article there are a number of independent works that can be solved on the basis of examples and theoretical data, as well as to strengthen knowledge on these issues. The set of examples and questions provided by the State Test Center under the Cabinet of ministers of the Republic of Uzbekistan, intended for entrance exams to higher education institutions in different years is was widely being used in covering the topic of the article.
1. Ayupov Sh.A., Rihsiyev B.B., Qo`chqorov O.Sh. (2004) Matematika olimpiadalari masalalari [Problems of mathematical olympiads]. Part 1-2. Tashkent: Fan. 2. Agahanov N.Kh., Kuptsov L.P., Nesterenko Yu.V., Reznichenko S.V., Slinko A.M. (1997) Matematicheskiye olimpiady dlya shkolnikov [Mathematical Olympiads for pupils]. Moscow: Prosveshcheniye. 3
[Correspondence Mathematical Olympiads]. Moscow: Nauka. 4. Gal'perin G.A., Tolpygo A.K . (1996) Moskovskiye matematicheskiye olimpiady [Moscow Mathematical Olympiads]. Moscow: Prosveshcheniye. 5. Tohirov А., Mo`minov G`. Matematikadan olimpiada masalalari [Olympiad problems in mathematics]. Tashkent: O`qituvchi. 6. Мo`minov G`. Nishonov T. (2017) Маtеmatika olimpiadalari masalalari [Problems of Mathematical Olympiads]. Andijon: Hayot. 7. Botirov Sh. (2009) Matematika. Mavzulashtirilgan testlar to`plami [A set of thematic tests]. Buxoro: Publishing House "Buxoro". 8. Usmonov M. (2016) Matematikadan misol va mavzular to`plami [A set of examples and problems from mathemat- ics]. Tashkent: Navro`z. 9. Mirzaahmedov M.A., and others. (2017) Matematika. [Mathematic]. Schoolbook for 10th class. Part 1. Tashkent: Extremum press. 10. Mirzaahmedov M.A., and others. (2018) Matematika [Mathematic 11]. Schoolbook for 11th class. Part 2. Tashkent: Zamin nashr.
Matematika kafedrasi dotsenti. E-mail: ibaydullayev73@mail.ru Abdulvoxidov Alisher Latipovich – Andijon davlat universiteti Matematika kafedrasi tayanch doktoranti. E-mail: abdulvoxidov_a@mail.ru MATHEMATICS Download 1.1 Mb. Do'stlaringiz bilan baham: 
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