Binom eng dvi

Volstenholm teoremasining o'zgarishi

1. 
   Bu maqolani o'qish uchun mashqdir. Bayonot [1] maqolada isbotlangan. Shuni kuzating
   

_Lp−1 ₁

2

_Lp−1 _{1 1}

=+

_Lp−1 ₁

= p.

i

i=1

i

i=1

p − i

_i=1 i(p − i)

Demak, ko'rib chiqilayotgan yig'indi p ga bo'linadi. 1 ≡ − 1 dan boshlab

(mod p), buni tekshirish qoladi

_Lp−1 ₁

ip −i

i=1

_i2 ≡ 0 (mod p).

Lekin 1, 1,. . . , 1 modul p 1, chto 12, 22, bilan bir xil to'plamdir. . . , (p - 1)2. Shuning uchun bu etarli

₁2 ₂2

(p−1)²

buni isbotlash

_Lp−1
i=1
i^{2 ≡} 0 (mod p). (5)

Mayli

p 1


−
_i2


2
i=1

≡ s (mod p). Bu p > 5 bo'lsa, biz har doim a tanlashimiz mumkin, shunday qilib a

≡ 1 (mod p). Keyin to'plamlar

{1, 2, . . . , p− 1} va {a, 2a, . . . , (p− 1) a} mos keladi (isbot izohdagi bilan bir xil) va

Shunday qilib, s ≡ 0 (mod p).

_Lp−1

s ≡

i=1
i^{2 =}

_Lp−1
i=1
(ai)2 = a2

_Lp−1
i=1

i^{2 ≡ a2}s (mod p).

2. 
   Javob: 2k + 2. A. Golovanovning bu muammosi Tuimaada-2012 olimpiadasida taqdim etilgan. E'tibor bering, p = 4k + 3 uchun x2 + 1 = 0 tenglama p moduli qoldiqlar to'plamida yechimga ega emas va shuning uchun barcha kasrlarning maxrajlari nolga teng emas.
   

S yechim 1. ai = i2 + 1, i = 0, bo'lsin. . . , p - 1. U holda ifoda teng bo'ladi

sp₋₁(a0, a1, ..., ap₋₁) _,

sp(a0, a1, ..., ap₋₁)

bu yerda si - darajaning elementar simmetrik polinomi i. Raqamlari bo'lgan ko'phadni toping

ai uning ildizlari:

p −1

(x − 1 − i^2).

X - 1 = t2 o'zgaruvchisini o'zgartiring va oling

i=0

p −1

(t2 − i^{2) =}

i=0

p −1

(t − i)

i=0

p −1
i=0
(t + i) ≡ (tp − t)(tp − t) = t^2p − 2tp⁺¹ + t².

Endi o'zgaruvchilarning teskari o'zgarishini qo'llang va p = 4k + 3 uchun oling

p −1

(x − 1 − i²⁾ ≡ (x − 1) p − 2(x − 1)

i=0

p+1


p ^p+1
2 + (x − 1) = x + . . . + (p + 2· 2 + 1)x − 4.

tomonidan Vieteniki nazariyam s

≡ 4 (mod p), s

≡ 2 (mod p), shuning uchun ≡≡ 2k + 2 (mod p).

^sp−1 1

^sp

2

p

p−1
2. ± 1 dan tashqari barcha nolga teng bo'lmagan qoldiqlarni modulli p o'zaro juftliklarga bo'ling. Biz 2k olamiz

juft va har bir juftlikda (i, j)

ij ≡ 1 ⇔ i^{2j2 ≡} 1 ⇔ (ij)2 + i2 + j2 + 1 ≡ i^{2 + j2 +} 2 (mod p).

Shuning uchun,
(ij)2 + i2 + j2 + 1

i^{2 + j2 +} 2 1 1

^{1 ≡} (i² + 1)(j² + 1) ^≡ (i² + 1)(j² + 1) = i² + 1 + j² + 1 ^{(mod p).}

Shunday qilib, yig'indi 1 + 1 + ga teng ¹ + 2k ≡ 2k + 2.

0²+11 ²+1( −1)² +1

¹ Bu to'plamlar bir-biriga mos keladi, chunki ular o'z ichiga oladi p − Har birida 1 ta element mavjud va har bir toʻplamdagi barcha eslatmalar boʻlmasligi aniq nol va juftlik farqlanadi.

S o l u t i o n 3. Fermaning kichik teoremasi bo'yicha x → x−1 va x → xp−2 modul p amallar mos keladi.

Shuning uchun summani hisoblash kifoya

p −1 2<sub>m

L</sub>p−1
x=0
(x2 + 1) p⁻² =

_Lp−1 _Lp−2 _{p − 2}

m

x=0 m=0

_x2m <sub>=

L</sub>p−2 _{p − 2}

m

m=0
p−1
S_2m, (6)

bu erda S2m =

x

x=0

. Shubhasiz S2m ≡ -1 (mod p) m = uchun

_{2018-03-22 S2m ≡ 0 (mod p) hamma uchun ekanligini isbotlang}

boshqa m p - 1. Darhaqiqat, har bir m uchun a2m ≡ 1 (mod p) bo‘ladigan nol bo‘lmagan qoldiqni tanlashimiz mumkin va shundan keyin (5) dagi kabi fikr yuritishimiz mumkin. Bizda (6) yig'indisi bor

_Lp−2

m=0

p − 2

m
^S2m


−
p 2


−
≡ − _{p 1} = −

4k + 1

2k + 1

₌ (4k + 1) · 4k · . . .· (2k + 1)


— ≡
1 · 2 · . . .· (2k + 1)


≡ − ≡
(−2) · (−3) . . . (2k + 2) _{2k + 2 (mod p).}

1 · 2 · . . .· (2k + 1)

3. 
   Biz bu bayonotlarni [16] da topdik.
   
   1. 
      
      p
      Har bir tub bo'luvchi uchun p | m pf (ak - 1) ni shunday tanlang. Xitoy eslatma teoremasiga ko'ra, hamma p uchun ≡ ap (mod p) bo'lsin. Keyin natija (5) dagi kabi mulohaza yuritish orqali isbotlanishi mumkin.
      
   2. 
      E'tibor bering, binomial formula bo'yicha toq k uchun bizda ik + (p - ik) ≡ kik−1p (mod p2) mavjud. Keyin
      

_Lp−1 ₁

2 _ik

i=1

_Lp−1 ₁

= _ik

i=1

1

+ =

(p − i)<sup>k

Lp−1</sup> i^k + (p − i)^k


i^k (p i)^k


−
i=1

_Lp−1

≡

i=1

_kik−1_p i^k (−i)^k
≡ −kp

_Lp−1 <sub>1

i</sub>k+1

i=1
(mod p2).

Rhsdagi yig'indi a) ifodasi bilan p ga bo'linadi.

4. 
   Muvofiqlik hatto p7 moduliga ham mos keladi (qarang [24]), lekin u biroz kuchliroq. Biz [1] dagi kabi fikr yuritib, p4 ga qadar barcha kuchlarni kuzatib, olishimiz mumkin
   

p − 1

₌ (2p − 1)(2p − 2) · . . .· (p + 1) ₌



2p 2p

— 1 − 1

· . . .·

2p

— 1 ≡

2p − 1 p!

12

_Lp−1 _1
Lp−1 ₁

p − 1
_Lp−1 ₁

2 3 4

≡ 1 − 2p

Oxirgi summani quvvat yig'indilari orqali ifodalash mumkin:

i=1

+4p

i
i, j=1

i

_ij − 8p
i, j, k=1 i(mod p). (7)

ijk

_Lp−1 ₁

SS SS ³

_Lp−1 ₁

= 3 − 1 2 + 1 , qaerda Sk = .

i, j, k=1 i
ijk 32 6

_ik

i=1

Endi biz S1 va S3 p2 ga bo'linamiz (oxirgisi 4.3b muammosi tufayli). Shuning uchun (7) formuladagi oxirgi atama o'tkazib yuborilishi mumkin.

5. 
   Muammo [1] dan, o'zgarishlarni [14] da topish mumkin. beri
   

_Lp−1 ₁ 2 _k2

k=1

_Lp−1_{( 1}

= _k2 +

k=1

1 ⁾

(p − k)²

^Lp−1 k² + (p k)2


−

−
⁼ k²(pk )²

k=1
≡ −2

_Lp−1 <sub>1

k=1</sub> k(p − k)
(mod p2),

gap 3) mos kelishiga ekvivalent

_Lp−1 ₁

≡ 0 (mod p). 2) bayonoti ekvivalentdir

2
_k=1 k(p − k)

bir xil muvofiqlikka, chunki 2

_Lp−1 <sub>1


L</sub>p−1_{( 1}

= 2+

₁ )

= 2p

_Lp−1

1

. Nihoyat biz dan bilamiz

oldingi muammo bu

k

k=1

k

k=1

2
_Lp−1

p − k

_Lp−1

_k=1 k(p − k)

2p − 1

p − 1

≡ 1 − p

¹ + 4p2

_i=1 i(p − i)

1

ij

i, j=1

i

(mod p4).

Demak, 1) ifodasi mos kelishiga ekvivalent

_Lp−1 <sub>1

L</sub>p−1


1
2

i=1

i(p − i) ^{≡ 4}

i, j=1

i

(mod p). (8)

ij

Rhsdagi ifodani qayta yozing:

_Lp−1 ₁

^Lp−1 <sub>1 2

L</sub>p−1 ₁

^Lp−1 <sub>1 2

L</sub>p−1 ₁

4= 2

ij

i, j=1

i

i

k=1

— 2

k=1

_k2 ≡ 2

i

k=1

+ 2

k=1

.

k(p − k)

Qavs ichidagi yig'indi p ga bo'linadi, uning kvadrati p2 ga bo'linadi va biz bu atamani o'tkazib yuborishimiz mumkin. Keyin dan

(8) 1) gapning mos kelishiga ekvivalent ekanligini ko‘ramiz

_Lp−1 ₁

≡ 0 (mod p).

2
_k=1 k(p − k)

6. 
   a) S yechim 1 ([5, taklif 2.12]). Induktsiya n bo'yicha. Tenglikda qavslarni kengaytiring
   

(a + b) pⁿ = (a + b)p(ⁿ⁻¹⁾(a + b)^p

Esifat the koesamarali of apmbp^(n−m):

pn ₌ p(n − 1) p

₊ p(n − 1) p
+ . . .+

p(n − 1)

p ₊ p(n − 1) p .

pmpm 0

pm − 11

pm − p + 1

p − 1 pm − p p

Hammasi jamlamalar except birinchi ad last are diko'rinadigan by p², chunkise by Lukas' nazariyam each binomiyal koesamarali

p ga bo'linadi. Shuning uchun

pn

_≡ p(n − 1) ₊ p(n − 1)
(mod p2).

pm

Induksiya gipotezasiga ko'ra

pm p(m − 1)

p(n − 1) ₊

pm

p(n − 1)

p(m − 1)


≡
n − 1 ₊

m

n 1 n


−
m − 1 ^≡ m
(mod p2).

S variant 2 ([D]). kp ≡ k (mod p2) ni m dagi induksiya orqali isbotlang.

mpm

2

−
m = 1 asosini isbotlash uchun biz buni tekshirishimiz kerak

pk k

p 1

≡ 0 (mod p). Bizda ... bor

pk k

p ⁻ 1

₌ pk(pk − 1) . . . (pk− p + 1) _{k =}


−
p!


−
(pk − 1) (pk − 1) . . . (pk− p + 1) ₁

(p − 1)!
. (9)

Numeratordagi ko'paytiruvchilarni juftlarga ajrating:

(pk − i)(pk − p + i) ≡ pi^{2 − i2} (mod p2).

Har bir juftlikning p2 mahsulot moduli k ga bog'liq emasligini ko'ramiz. Shuning uchun farq (9) moduli p2

k ga ham bog'liq emas. k = 1 uchun u l ga 0 ga teng bo'lgani uchun barcha k uchun 0 ga teng.