Binom eng dvi

k
Induksiya bosqichi. Mayli

kp

(m−1)p

^≡ m−1

(mod p2). Bizda ... bor

kpkp

=

mp (m − 1) p

=

_· (p(k − m) + 1)(p(k − m) + 1) . . . (p(k− m) + p) ₌


pm(soat − 1) . . . (soat− p + 1)

— − − − −


kp (p(k m) + 1)(p(k m) + 1) . . . (p(k m) + p 1) k m + 1

·· . (10)

(m − 1)p (soat − 1) . . . (soat− p + 1) m
E'tibor bering, ikkala kasr ham p2 moduli bilan to'g'ri belgilangan. Baza isbotida bo'lgani kabi, katta kasrning numeratoridagi ifoda (modul p2) k ga bog'liq emas. Keyin hisoblash uchun k = 0 ni qo'yishimiz mumkin

fraktsiya moduli p2 va uning 0 ga mos kelishini oling. Ifodaning qolgan qismi uchun biz induksiya gipotezasini qo'llashimiz va olishimiz mumkin.

_≡ k _· k − m + 1 ₌ k
(mod p2).

m − 1 mm
b) S yechim 1 (kombinator). [1] da taklif qilinganidek, kp ob'ektlarining namunalarini ko'rib chiqing

the o'rnatish of pn obob'ektlar. Let the initial o'rnatish be Split on blokks of p obob'ektlar. The raqamr of blokk namunalar esifatlar


k
ⁿ . Shuning uchun blok bo'lmagan namunalar p3 ga bo'linishini tekshirish kerak. Lekin bloklanmaganlar soni


2p
3 yoki undan ortiq blokli namunalar p3 ga bo'linadi (qarang [1]). K > 1 uchun har bir bloksiz namuna kamida 3 ta blokdan iborat, shuning uchun bu holda bayonot to'g'ri bo'ladi. Biz k = 1 va d hisoblagan holatni ko'rib chiqish uchun qoladi

2p ob'ektlar to'plamidan p ob'ektning bloksiz namunalari soni. Bu raqam teng

Wolstenholme nazariyam it is diko'rinadigan by p³.

_p − 2, tomonidan

atamalar, har bir blokdagi birinchi shartlarni kamaytiring va ko'rsatkichlarni alohida ifodada to'plang:
S o l u t i o n 2. a = a(a−1)...(a−b+1) formulada hisob va maxrajni bloklarga ajrating.

ning p

bb (b−1)...1

mp ^{m p · (mp − 1) . . . mp − (p−1)}

_· (m−1) p ·

(m−1) p − 1

. . . (m−1) p − (p−1)

=

^kp k p · (kp − 1) . . .

kp − (p−1)



(k−1) p · (k

−1) p −

·×
1 . . .

(k−1) p − (p−

. . .

1)

_× (m−k+1) p · (m−k+1) p − 1 . . . (m−k+1) p − (p−1) =

p · (p − 1) . . . 1

m _· ^{(mp − 1) . . . mp − (p−1)}

(m−k+1) p − 1 . . . (m−k+1) p − (p−1)

=

^k (kp − 1) . . .

kp − (p−

. . .

1)

.

· ·
(p − 1) . . . 1

Fraksiyalarning mahsuloti ekanligini tekshirish qoladi 1 ga mos keladi (mod p3). Buning uchun muvofiqlikni isbotlang

(np − 1) . . . np − (p−1)

≡ 1 (mod p3)

(rp − 1) . . . rp − (p−1)
yoki, hatto, quyidagi muvofiqlikni isbotlash yaxshiroqdir

(np − 1) . . . np − (p−1) _≡ (rp − 1) . . . rp − (p−1)
(mod p3).

(p − 1)! (p − 1)!

Bu to'g'ri, chunki ikkala qism ham 1 ga (mod p3) mos keladi, buni isbotiga o'xshash tarzda ko'rsatish mumkin. Wolstenholme teorema.

7. 
   a) [5, teorema 2.14]. Farqni o'zgartiring
   

_p2

222 ⁽⁾

_— p ₌ p (p − 1) . . . (p − (p − 1)) _{−p =} p

(1−p²⁾⁽²⁻p²) . . . ((s−1)−p^2)−1·2·. . .·(p−1) .

p 1 1 · 2 · . . .· (p − 1)p (p − 1)!

Buni tekshirish qoladi

(1 − p^{2)(2 − p2) . . . ((p - 1) - p2) ≡ 1 · 2 · . . . · (p -}1) (mod p4).

lhsdagi qavslarni kengaytiring:

222

₂ ( ₁₁ ) ₄

(1− p )(2 − p ). . . ((s− 1) − p ) = 1· 2 · . . .·(p − 1) +s

1 + + . . . +

2 p − 1

(p − 1)! +p ga bo'linadigan atamalar.

4.1 masala bo'yicha ikkinchi yig'indi p4 ga bo'linadi.

b) Obxizmat qilish that ^ps+1 = ps · ^ps+1⁻¹ , hece it is yetarli to prove that ^ps+1⁻¹ ≡ 1 (mod ps+3).

pp −1 p−1

_ps+1 _{− 1}

(ps⁺¹ − 1)(ps⁺¹ − 2) . . . (ps⁺¹ − (p − 1))

_ps+1 _ps+1

_ps+1

=

p − 1

=

1 · 2 · · · (p − 1)

₁ − 1 ₂ − 1

. . .

_{p − 1} − 1 ≡

1
(

p−1 s+1

^{1 )} s+3

≡ (−1)+ p

1 + + . . . +

2 p − 1

(mod p ).

Since (−1) p⁻¹ = 1 ad 1 + ¹ + . . . + ¹ ≡ 0 (mod p2) biz tugatdik.

2 p−1

8. 
   Muammo [1] dan, biz [T] yechimini taqdim etamiz.
   

2

p
_p3

− = p

₃

p − 1 <sub>−

2</sub>

p − 1 ₌

p² p

p² − 1

p − 1

<sub>p3 p3

p3

p2 p2

p2</sub>

= p ₁ − 1

₂ − 1

. . .

_{p2 − 1} − 1 −

₁ − 1

₂ − 1

. . .

_{p 1} − 1=


−

<sub>

p</sub>2

_p3


_p2 _p2

p²−1 _


= p ₁ − 1

₂ − 1

. . .

p − 1 ^{− 1}





k=1 pfk

_k − 1

— 1^ .

Oxirgi qavsning p7 ga bo'linishini isbotlash kifoya. Mahsulotni o'zgartiring:

₂ p²−1

p²−1

7
p²−1

p −1 _p3



5
² _p³ _p<sup>3


2</sup> p⁶ − p<sup>5

L2</sup> 1

k=1 pfk

_k −1

=

k=1 pfk

_k −1

p² −k ⁻¹

=

k=1 pfk

k(p² −k) +1

≡ 1+p (s−1)

k=1 pfk

_{k(p2 −k)} (mod p).

Endi biz oxirgi yig'indining p2 ga bo'linishini tekshirishimiz kerak. Bu to'g'ri, chunki 4.3a muammosi bo'yicha)

p²−1

_L2 ₁

p²−1


1
_L2

2

k=1 pfk

k(p² −k) ^{≡ −}
k=1 pfk

_k2 ≡ 0 (mod p ).

9. 
   Bayonot [6, teorema 5] dan olingan, uning umumlashtirilishini [7] da topish mumkin. S yechim 1 ([5, taklif 2.19]). Farq borligidan foydalaning²k+1 − ²k
   

ga teng

polinomdagi x2k koeffitsienti

(1 + x)2k+1 − (1 − x^{2) 2}k = (1 + x)2k

()

(1 + x)2k − (1 − x)2k =

₂k ₂k−1

= 1+

₂k ₂k


2
x + x

12

₂k 2k

+ . . . +x · 2 ₁ x +

₂k


3
x + . . . +

3

² _x2k−1 _.

k
2^k − 1

Ikkinchi ko'phad faqat toq darajalarni o'z ichiga olganligi sababli, mahsulotdagi x2k koeffitsienti teng

<sub>k k

k k</sub>

kk

222

2+

12 ^k − 1 3

2

2^k − 3

+ . . .+

22

.

2^k − 11

3.5-masala bo'yicha b) 2k har bir binomial koeffitsientga bo'linadi bu ifodada, bundan tashqari, har bir atama ikki marta uchraydi the summa, ad the so'm buo'zini is ko'paytirmoqd by 2. Payss all the eifodalash is diko'rinadigan by 22 ming^+2.

2-chi variant ([CSTTVZ]). 2n+1 dan boshlab ²n+1⁻¹ , buni isbotlash kifoya

_2n = 2

2ⁿ−1

₂n+1 _{− 1}

2n − 1
2n+1

Xuddi shunday (3) ga erishamiz

2ⁿ − 1

^≡ ₂n−1 _{− 1}

(mod 2) .

₂n+1 <sub>− 1

2</sub>n+1 ₂n+1

₂n+1

2n − 1

=

2ⁿ − 1

Buni isbotlash kifoya

₁ − 1

₃ − 1

. . .

2ⁿ − 1 ^{− 1}

^· ₂n−1 _{− 1 .}

₂n+1

L =

1

₂n+1

— 1 ₃

— 1 . . .

₂n+1 2ⁿ − 1


2n+1
— 1 ≡ 1 (mod 2) .

Bu haqiqat, chunki

L ≡ (−1)2n−1 − 2ⁿ⁺¹
1 1 1 1

+++ . . .+ ≡

135

2ⁿ − 1

₂n ₂n ₂n

n+1

2n+1

≡ 1 − 2

1 · (2ⁿ − 1) + 3 · (2ⁿ − 3) + . . . + (2ⁿ⁻¹ − 1)(2ⁿ⁻¹ + 1)

≡ 1 (mod 2) .

10. 
    Bu Morli [26] teoremasi.
    

S o l u t i o n1 (muallifning isboti, 1895). Bu maktab o'quv dasturidan biroz tashqariga chiqadi.

Take the formula qaysih expresslar cos2ⁿ⁺¹ x orqali cosinuslar of ko'pe burchaks, 1 yoki o'sha paytlarda aytganidek, yozing cos2ⁿ⁺¹ x in the shakl qo'ly uchun integratsiyalash:

22ncos2^n+1x = cos(2n+1)x+(2n+1) cos(2n−1)x+(2n+1^{) · 2n} cos(2n−3)x+. . .+(2n+1^{) · 2n . . . (n+2)} cos x.

2
Endi it2 ni [0, p ] oraliqda integrallang:

1 · 2 n!

₂2n

cos2ⁿ⁺¹ x dx = gunoh(2n + 1)x + 2n + 1 gunoh(2n − 1)x + . . . ,

p/2

2n + 1

2n − 1

₂2n

cos2ⁿ⁺¹ x dx = (−)ⁿ

1 2n + 1

− + . . . .

2n + 1

0

2n − 1

Universitetning har bir birinchi sinf talabasi hisoblash uchun qismlar bo'yicha integratsiyadan foydalanish qulayligini biladi bu integral:

p/2

p/2

p/2

p/2

^I2n+1 ⁼

0

cos2n+1 x dx =

cos x cos x dx = cos x gunoh x
0

2n 2n

0

p/2

+2n

0

cos2n⁻¹ x sin2 x dx =

= 0 + 2n

0

cos2ⁿ⁻¹ x(1 − cos2 x) dx = 2n · I_2n−1 − 2n · I_2n+1 ,

2n+1
Manaoldin I_2n+1 = 2ⁿ · I_2n−1. Since I₁ = 1, we taxminann ilovay the formula n vaqts ad obtain



p/2

_{cos2n+1 x dx =} 2n · (2n − 2) . . . 2 _.

(2n + 1)(2n − 1) . . . 3

0

Ushbu ikki natijani tenglashtirish bizga formulani beradi

_22n 2n · (2n − 2) . . . 2 _{= (−)n} 1 ₋ 2n + 1 _{+ . . . +}(2n+1) · 2n. . . (n+2) .

(2n + 1)(2n − 1) . . . 3

2n + 1

2n − 1 n!

p = 2n + 1 tub son bo'lsin. Oxirgi formulani p ga ko'paytirish orqali kerakli muvofiqlikni olamiz:


≡ −
_22n 2n · (2n − 2) . . . 2 _{( )n (mod p2).}

(2n − 1)(2n − 3) . . . 3

2-chi variant ([CSTTVZ]). Biz quyidagi belgilardan foydalanamiz: