Diofant tenglamalarni yechish
Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
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SOLVE THE DIOPHANTES EQUATIONS
- Bu sahifa navigatsiya:
- Isboti
- 2-misol.
- Илмий хабарнома. Физика-математика тадқиқотлари, 2020, №1(3)
- 3.Yuqori darajali diofant tenglamalarini yechish.
- 4-misol
- 5-misol.
- 6-misol.
- Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
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Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
Demak, 𝑦𝑦 − 𝑦𝑦 0 = 𝑎𝑎𝑎𝑎, 𝑎𝑎- ixtiyoriy butun son. Bu holda 𝑥𝑥 − 𝑥𝑥 0 = −𝑏𝑏𝑎𝑎, 𝑦𝑦 − 𝑦𝑦 0 = 𝑎𝑎𝑎𝑎 bo`lib, (6) formula orqali (1) uchun barcha yechimlarning ifodalanishi kelib chiqadi. 6-teorema: Agar (𝑎𝑎, 𝑏𝑏) = 1 bo`lsa, u holda (1) tenglamaning barcha butun sonli yechimlarini (5) tenglamaning ( 𝑥𝑥 0 , 𝑦𝑦
0 ) yechimlari orqali
𝑥𝑥 = 𝑐𝑐𝑥𝑥 0 + 𝑏𝑏𝑎𝑎 , 𝑦𝑦 = 𝑐𝑐𝑦𝑦 0 – 𝑎𝑎𝑎𝑎 (7) formula bilan ifodalash mumkin. Isboti: 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑎𝑎(𝑐𝑐𝑥𝑥 0 + 𝑏𝑏𝑎𝑎) + 𝑏𝑏(𝑐𝑐𝑦𝑦 0 – 𝑎𝑎𝑎𝑎) = 𝑐𝑐(𝑎𝑎𝑥𝑥 0 + 𝑏𝑏𝑦𝑦
0 ) = 𝑐𝑐 ∙ 1 = 𝑐𝑐 . 2-misol. 37𝑥𝑥 − 256𝑦𝑦 = 3 tenglamani butun sonlarda yeching. Yechish: 256 = 37 ∙ 6 + 34; 37 = 34 ∙ 1 + 3; 34 = 3 ∙ 11 + 1 bo`lgani uchun
1 = 34 − 3 ∙ 11 = 34 − (37 − 34 ∙ 1) ∙ 11 = 34 ∙ 12 − 37 ∙ 11 == (256 − 37 ∙ 6) ∙ 12 − 37 ∙ 11 = = 37 ∙ (−83) − 256 ∙ (−12),
ya`ni 37 ∙ (−83) − 256 ∙ (−12) = 1 dan 𝑥𝑥 0 = −83 , 𝑦𝑦 0 = −12 va 𝑐𝑐 = 3 bo`lganligi uchun (7) ga ko`ra berilgan tenglamaning yechimlari 𝑥𝑥 = −249 − −256𝑎𝑎 ; 𝑦𝑦 = −36 − 37𝑎𝑎 dan iborat. 2. Endi 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 tenglamaning yechimlarini topish uchun 𝑎𝑎 𝑏𝑏 kasrni munosib kasrlar orqali amalga oshirish mumkinligini ko`rsatamiz. Evklid algoritmidan foydalanamiz:
𝑎𝑎 = 𝑏𝑏𝑞𝑞 1 + 𝑟𝑟
1 , 𝑟𝑟 1
𝑏𝑏 = 𝑟𝑟
1 𝑞𝑞 2 + 𝑟𝑟 2 , 𝑟𝑟 2 < 𝑟𝑟 1 , 𝑟𝑟 1 = 𝑟𝑟 2 𝑞𝑞 3 + 𝑟𝑟 3, 𝑟𝑟 3 < 𝑟𝑟 2 , ……………………………………. (4) 𝑟𝑟
𝑛𝑛−2 = 𝑟𝑟
𝑛𝑛−1 𝑞𝑞 𝑛𝑛 + 𝑟𝑟 𝑛𝑛 , 𝑟𝑟 𝑛𝑛 < 𝑟𝑟 𝑛𝑛−1
𝑟𝑟 𝑛𝑛−1 = 𝑟𝑟 𝑛𝑛 𝑞𝑞 𝑛𝑛+1 + 0 𝑟𝑟 𝑛𝑛 = 0
bo`lish jaroyonida hosil bo`lgan qoldiqlar
𝑏𝑏 > 𝑟𝑟
1 > 𝑟𝑟
2 > 𝑟𝑟
3 > ⋯ … ≥ 0
tengsizliklarni qanoatlantiradi. Hosil bo`lgan tengliklarni 𝑎𝑎 𝑏𝑏 = 𝑞𝑞 1 + 1 𝑏𝑏 𝑟𝑟 2
𝑏𝑏 𝑟𝑟 2 = 𝑞𝑞 2 + 1 𝑟𝑟 2 𝑟𝑟 3
… … … … … … … 𝑟𝑟 𝑛𝑛−2
𝑟𝑟 𝑛𝑛−1
= 𝑞𝑞 𝑛𝑛−1
+ 1 𝑟𝑟 𝑛𝑛−1 𝑟𝑟 𝑛𝑛
𝑟𝑟 𝑛𝑛−1 𝑟𝑟 𝑛𝑛 = 𝑞𝑞 𝑛𝑛
tengliklar bilan almashtirib, 𝑎𝑎 𝑏𝑏
𝑎𝑎 𝑏𝑏 = 𝑞𝑞
1 + 1 𝑞𝑞 2 + 1 𝑞𝑞 3 + 1 𝑞𝑞4+ ….. … + 1 𝑞𝑞 𝑛𝑛−1 + 1 𝑞𝑞𝑛𝑛
ifodani hosil qilamiz. Zanjirli kasrlning ba`zi bog`inlarini qoldirib, qolgan bog`inlarini tashlab yuborishdan hosil bo`lgan kasrlarni, ya`ni munosib kasrlarni hosil qilamiz.
𝑆𝑆 1 = 𝑞𝑞 1
𝑎𝑎 𝑏𝑏 , 𝑆𝑆 3 = 𝑞𝑞
1 + 1 𝑞𝑞 2 + 1 𝑞𝑞 3 < 𝑎𝑎 𝑏𝑏 𝑆𝑆 2 = 𝑞𝑞 1 + 1 𝑞𝑞 2 > 𝑎𝑎 𝑏𝑏 , 𝑆𝑆 4 = 𝑞𝑞 1 + 1 𝑞𝑞 2 + 1 𝑞𝑞 3 + 1 𝑞𝑞4 > 𝑎𝑎 𝑏𝑏 va hokazo munosib kasrlarni hosil qilamiz. Hosil qilinishiga ko`ra:
𝑆𝑆
< 𝑆𝑆 3
2𝑘𝑘−1
𝑎𝑎 𝑏𝑏 ; 𝑆𝑆 2 > 𝑆𝑆 4 >……> 𝑆𝑆
2𝑘𝑘 > 𝑎𝑎 𝑏𝑏 ,
𝑘𝑘- munosib kasr, 𝑆𝑆 𝑘𝑘 ni
𝑆𝑆 𝑘𝑘 = 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘 , 1 ≤ 𝑘𝑘 ≤ 𝑛𝑛 ko`rinishida olib, munosib kasrlarning surat va maxrajlarini hosil qilish qoidalarini ko`rsatmiz.
𝑆𝑆
= 𝑞𝑞 1 = 𝑞𝑞 1 1 = 𝑃𝑃 1 𝒬𝒬 1 ; 𝑃𝑃 1 = 𝑞𝑞
1 ; 𝒬𝒬 1 = 1
𝑆𝑆 2 = 𝑞𝑞 1 + 1 𝑞𝑞 2 = 𝑞𝑞 1 𝑞𝑞 2 + 1
𝑞𝑞 2 = 𝑃𝑃 2 𝒬𝒬 2 ; 𝑃𝑃 2 = 𝑞𝑞
1 𝑞𝑞 2 + 1 ; 𝒬𝒬 2 = 𝑞𝑞 2
𝑆𝑆 3 = 𝑞𝑞 1 + 1 𝑞𝑞 2 + 1 𝑞𝑞 3 = 𝑞𝑞 1 𝑞𝑞 2 𝑞𝑞 3 + 𝑞𝑞
1 + 𝑞𝑞
3 𝑞𝑞 2 𝑞𝑞 3 + 1 = 𝑃𝑃 3 𝒬𝒬 3 ; 𝑃𝑃 3 = 𝑞𝑞
1 𝑞𝑞 2 𝑞𝑞 3 + 𝑞𝑞 1 + 𝑞𝑞
3 ;
𝒬𝒬 3 = 𝑞𝑞 2 𝑞𝑞 3 + 1.
Bu tengliklardan 𝑃𝑃 3 = 𝑃𝑃
2 𝑞𝑞 3 +𝑃𝑃 1 ; 𝒬𝒬 3 = 𝒬𝒬
2 𝑞𝑞 3 +𝒬𝒬 1
munosabatlarni hosil qilamiz. Matematik induksiya metodidan foydalanib, barcha k≥3 lar uchun
𝑃𝑃
= 𝑃𝑃 𝑘𝑘−1
𝑞𝑞 𝑘𝑘 + 𝑃𝑃 𝑘𝑘−2 ; 𝒬𝒬 𝑘𝑘 = 𝒬𝒬
𝑘𝑘−1 𝑞𝑞 𝑘𝑘 + 𝒬𝒬 𝑘𝑘−2
munosabatlar bajarilishini ko`rsatamiz. Munosib kasrlarning aniqlanishiga ko`ra 𝑆𝑆 𝑘𝑘 munosib kasrda 𝑞𝑞 𝑘𝑘 ni 𝑞𝑞
𝑘𝑘 + 1 𝑞𝑞 𝑘𝑘 +1 ga almashtish natijasida 𝑆𝑆 𝑘𝑘 munosib kasr 𝑆𝑆 𝑘𝑘+1 munosib kasrga o`tadi. Induktiv mulohazalarga ko`ra,
𝑆𝑆
= 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘 = 𝑃𝑃 𝑘𝑘−1
𝑞𝑞 𝑘𝑘 + 𝑃𝑃 𝑘𝑘−2 𝒬𝒬 𝑘𝑘−1 𝑞𝑞 𝑘𝑘 + 𝒬𝒬 𝑘𝑘−2
dan
𝑆𝑆 𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘+1 𝒬𝒬 𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘−1 (𝑞𝑞 𝑘𝑘 + 1 𝑞𝑞 𝑘𝑘+1 ) + 𝑃𝑃 𝑘𝑘−2
𝒬𝒬 𝑘𝑘−1(
𝑞𝑞 𝑘𝑘 + 1 𝑞𝑞 𝑘𝑘+1 ) + 𝒬𝒬 𝑘𝑘−2
= 𝑃𝑃 𝑘𝑘 + 1 𝑞𝑞 𝑘𝑘+1 𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘 + 1 𝑞𝑞 𝑘𝑘+1 𝒬𝒬 𝑘𝑘−1
= 𝑃𝑃 𝑘𝑘 𝑞𝑞 𝑘𝑘+1
+ 𝑃𝑃 𝑘𝑘−1
𝒬𝒬 𝑘𝑘 𝑞𝑞 𝑘𝑘+1 + 𝒬𝒬
𝑘𝑘−1 ,
bundan esa
𝑃𝑃 𝑘𝑘 = 𝑃𝑃
𝑘𝑘 𝑞𝑞 𝑘𝑘+1 + 𝑃𝑃 𝑘𝑘−1
; 𝒬𝒬 𝑘𝑘 = 𝒬𝒬 𝑘𝑘 𝑞𝑞 𝑘𝑘+1 + 𝒬𝒬 𝑘𝑘−1
.
Matematik induksiya metodidan foydalanib, barcha k≥3 lar uchun 𝑃𝑃 𝑘𝑘 = 𝑃𝑃 𝑘𝑘−1
𝑞𝑞 𝑘𝑘 + 𝑃𝑃 𝑘𝑘−2 ; 𝒬𝒬 𝑘𝑘 = 𝒬𝒬
𝑘𝑘−1 𝑞𝑞 𝑘𝑘 + 𝒬𝒬 𝑘𝑘−2
MATHEMATICS 66 Илмий хабарнома. Физика-математика тадқиқотлари, 2020, №1(3)
munosabatlar o`rinli ekanligi ravshan bo`ldi. Shu bilan birga,
𝑆𝑆 𝑘𝑘 − 𝑆𝑆
𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘 − 𝑃𝑃 𝑘𝑘+1 𝒬𝒬 𝑘𝑘+1
= 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
.
Ammo 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 =(𝑃𝑃
𝑘𝑘−1 𝑞𝑞 𝑘𝑘 + 𝑃𝑃 𝑘𝑘−2
) 𝒬𝒬 𝑘𝑘−1
− (𝒬𝒬 𝑘𝑘−1
𝑞𝑞 𝑘𝑘 + 𝒬𝒬 𝑘𝑘−2 )𝑃𝑃
𝑘𝑘−1 = = −(𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘−2 − 𝑃𝑃 𝑘𝑘−2
𝒬𝒬 𝑘𝑘−1
)
bo`lgani uchun quyidagi tengliklar zanjirini hosil qilamiz: 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 = (−1)(𝑃𝑃 𝑘𝑘−1 𝒬𝒬
− 𝑃𝑃 𝑘𝑘−2
𝒬𝒬 𝑘𝑘−1
)=(−1) 2 (𝑃𝑃 𝑘𝑘−2 𝒬𝒬 𝑘𝑘−3 − −𝑃𝑃 𝑘𝑘−3
𝒬𝒬 𝑘𝑘−2
)= =…..=(−1) 𝑘𝑘−2 (𝑃𝑃
2 𝒬𝒬 1 − 𝑃𝑃 1 𝒬𝒬 2 ) = (−1)
𝑘𝑘−2 (𝑞𝑞
1 𝑞𝑞 2 + 1 − 𝑞𝑞 1 𝑞𝑞 2 ) = (−1)
𝑘𝑘−2 .
Shunga ko`ra,
𝑆𝑆 𝑘𝑘 − 𝑆𝑆
𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1 −𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
= (−1)
𝑘𝑘−2 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
= (−1)
𝑘𝑘 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
.
𝑎𝑎 𝑏𝑏 kasrning zanjirli kasrlar yoyilmasida barchasi bo`lib 𝑛𝑛 ta bog`inlar mavjud, 𝑛𝑛- munosib kasr berilgan 𝑎𝑎 𝑏𝑏
kasr bilan ustma-ust tushadi. Demak, 𝑘𝑘 = 𝑛𝑛 bo`lganda,
𝑎𝑎 𝑏𝑏 −𝑆𝑆
𝑛𝑛−1 = 𝑆𝑆
𝑛𝑛 − 𝑆𝑆
𝑛𝑛−1 = (−1) 𝑛𝑛 𝒬𝒬 𝑛𝑛 𝒬𝒬 𝑛𝑛−1
= (−1)
𝑛𝑛 𝑏𝑏𝒬𝒬
𝑛𝑛−1 ,
ya`ni
𝑎𝑎 𝑏𝑏 - 𝑃𝑃 𝑛𝑛−1 𝒬𝒬 𝑛𝑛−1
= (−1)
𝑛𝑛 𝑏𝑏𝒬𝒬
𝑛𝑛−1 .
Umumiy maxrajga keltirib,
𝑎𝑎𝒬𝒬 𝑛𝑛−1 − 𝑏𝑏𝑃𝑃
𝑛𝑛−1 = (−1)
𝑛𝑛
yoki (−1) 𝑛𝑛 𝑐𝑐 ga ko`paytirib,
𝑎𝑎[ (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 ] + 𝑏𝑏[ (−1) 𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1 ] = с
tenglikka kelamiz, bu yerda 𝑥𝑥
0 = (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 , 𝑦𝑦 0 = (−1)
𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1
tengliklar bilan aniqlanadigan (𝑥𝑥 0 , 𝑦𝑦 0 ) juftlik 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 tenglamani qanoatlantirishini ko`ramiz. Demak, quyidagi teorema o`rinli:
𝑥𝑥 0 = (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 , 𝑦𝑦 0 = (−1)
𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1
tengliklar bilan aniqlanuvchi (𝑥𝑥 0 , 𝑦𝑦 0 ) juftlik 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 tenglamaning yechimi bo`ladi. Uning barcha yechimlari esa
𝑥𝑥 = (−1) 𝑛𝑛 𝒬𝒬
𝑛𝑛−1 − 𝑏𝑏𝑏𝑏 , 𝑦𝑦 = (−1) 𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1 + 𝑎𝑎𝑏𝑏 , 𝑏𝑏 = 0, ±1, ±2, … … ..
formulalar orqali beriladi.
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. МАТЕМАТИКА 67 Scientific Bulletin. Physical and Mathematical Research, 2020, №1(3)
munosabatlar o`rinli ekanligi ravshan bo`ldi. Shu bilan birga,
𝑆𝑆 𝑘𝑘 − 𝑆𝑆
𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘 − 𝑃𝑃 𝑘𝑘+1 𝒬𝒬 𝑘𝑘+1
= 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
.
Ammo 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 =(𝑃𝑃
𝑘𝑘−1 𝑞𝑞 𝑘𝑘 + 𝑃𝑃 𝑘𝑘−2
) 𝒬𝒬 𝑘𝑘−1
− (𝒬𝒬 𝑘𝑘−1
𝑞𝑞 𝑘𝑘 + 𝒬𝒬 𝑘𝑘−2 )𝑃𝑃
𝑘𝑘−1 = = −(𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘−2 − 𝑃𝑃 𝑘𝑘−2
𝒬𝒬 𝑘𝑘−1
)
bo`lgani uchun quyidagi tengliklar zanjirini hosil qilamiz: 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
− 𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 = (−1)(𝑃𝑃 𝑘𝑘−1 𝒬𝒬
− 𝑃𝑃 𝑘𝑘−2
𝒬𝒬 𝑘𝑘−1
)=(−1) 2 (𝑃𝑃 𝑘𝑘−2 𝒬𝒬 𝑘𝑘−3 − −𝑃𝑃 𝑘𝑘−3
𝒬𝒬 𝑘𝑘−2
)= =…..=(−1) 𝑘𝑘−2 (𝑃𝑃
2 𝒬𝒬 1 − 𝑃𝑃 1 𝒬𝒬 2 ) = (−1)
𝑘𝑘−2 (𝑞𝑞
1 𝑞𝑞 2 + 1 − 𝑞𝑞 1 𝑞𝑞 2 ) = (−1)
𝑘𝑘−2 .
Shunga ko`ra,
𝑆𝑆 𝑘𝑘 − 𝑆𝑆
𝑘𝑘+1 = 𝑃𝑃 𝑘𝑘 𝒬𝒬 𝑘𝑘−1 −𝒬𝒬 𝑘𝑘 𝑃𝑃 𝑘𝑘−1 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
= (−1)
𝑘𝑘−2 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
= (−1)
𝑘𝑘 𝒬𝒬 𝑘𝑘 𝒬𝒬 𝑘𝑘−1
.
𝑎𝑎 𝑏𝑏 kasrning zanjirli kasrlar yoyilmasida barchasi bo`lib 𝑛𝑛 ta bog`inlar mavjud, 𝑛𝑛- munosib kasr berilgan 𝑎𝑎 𝑏𝑏
kasr bilan ustma-ust tushadi. Demak, 𝑘𝑘 = 𝑛𝑛 bo`lganda,
𝑎𝑎 𝑏𝑏 −𝑆𝑆
𝑛𝑛−1 = 𝑆𝑆
𝑛𝑛 − 𝑆𝑆
𝑛𝑛−1 = (−1) 𝑛𝑛 𝒬𝒬 𝑛𝑛 𝒬𝒬 𝑛𝑛−1
= (−1)
𝑛𝑛 𝑏𝑏𝒬𝒬
𝑛𝑛−1 ,
ya`ni
𝑎𝑎 𝑏𝑏 - 𝑃𝑃 𝑛𝑛−1 𝒬𝒬 𝑛𝑛−1
= (−1)
𝑛𝑛 𝑏𝑏𝒬𝒬
𝑛𝑛−1 .
Umumiy maxrajga keltirib,
𝑎𝑎𝒬𝒬 𝑛𝑛−1 − 𝑏𝑏𝑃𝑃
𝑛𝑛−1 = (−1)
𝑛𝑛
yoki (−1) 𝑛𝑛 𝑐𝑐 ga ko`paytirib,
𝑎𝑎[ (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 ] + 𝑏𝑏[ (−1) 𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1 ] = с
tenglikka kelamiz, bu yerda 𝑥𝑥
0 = (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 , 𝑦𝑦 0 = (−1)
𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1
tengliklar bilan aniqlanadigan (𝑥𝑥 0 , 𝑦𝑦 0 ) juftlik 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 tenglamani qanoatlantirishini ko`ramiz. Demak, quyidagi teorema o`rinli:
𝑥𝑥 0 = (−1)
𝑛𝑛 с 𝒬𝒬
𝑛𝑛−1 , 𝑦𝑦 0 = (−1)
𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1
tengliklar bilan aniqlanuvchi (𝑥𝑥 0 , 𝑦𝑦 0 ) juftlik 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 tenglamaning yechimi bo`ladi. Uning barcha yechimlari esa
𝑥𝑥 = (−1) 𝑛𝑛 𝒬𝒬
𝑛𝑛−1 − 𝑏𝑏𝑏𝑏 , 𝑦𝑦 = (−1) 𝑛𝑛−1 с 𝑃𝑃
𝑛𝑛−1 + 𝑎𝑎𝑏𝑏 , 𝑏𝑏 = 0, ±1, ±2, … … ..
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