Amazing Properties of Bionomial Coefficients

8/10/2019 Amazing Properties of Bionomial Coefficients

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Amazing properties of binomial coeffcients 1

Amazing properties of binomial coefficients

Several research topics will be set to you at the conference. Your aim is the maximal advance in one of these topics. You canco-operate in the solving of problems, arbitrary teams are allowed (i.e. the team may consist of participants from differentcities). If you solve problems in different topics you may take part in different teams. The only thing you should avoid is tosign up the solutions of those problems that you really were not solving (this may happen if the team is too big and not allof its members solve the problems of some topic actively).

The following is the introductory set of problems about binomial coefficients. You may hand in the (written) solutions toKokahs K. (coach 15, seat 17) In Teberda the set of problems will be enlarged a lot and you may hand in your solutions of

this set of problems, except 1.2, in Teberda, too. You can hand in the solutions of the problem 1.2 in train only.

1 Problems for solving in train

1.1. Prove that a)p1

k

(1)k (modp); b) 2nn (4)np12n (modp) n p12 .1.2.Prove that the number of odd binomial coefficients in n-th row of Pascal triangle is equal to2r, wherer is the number of 1s in the binary expansion ofn.

1.3. Fix a positive integer m. By a m-arithmetical Pascal triangle we mean a triangle in which binomialcoefficients are replaced by their residues modulo m. We will also consider similar triangles with the

arbitrary residues a instead of 1s along the lateral sides of the triangle. The operation of the multiplyingby a number and addition of triangles of equal size are correctly defined. We will consider these operationsmodulo m.

aa a

a 2a aa 3a 3a a

bb b

b 2b bb 3b 3b b

+

a+ba+b a+b

a+b 2(a+b) a+ba+b 3(a+b) 3(a+b) a+b

= x

aa a

a 2a aa 3a 3a a

=

axax ax

ax 2ax axax 3ax 3ax ax

Let all the elements ofs-th row ofm-arithmetical Pascal triangle except the first and the last one be equalto 0. Prove that the triangle has a form depicted on fig. 1. Shaded triangles consist of zeroes, triangles knconsist ofs rows and satisfy the following relations

1) k1n + kn= kn+1; 2) kn= Ckn 00 (modm).The well known puzzle Tower of Hanoi consists of three rods, and a number of disks of different sizes which can slide onto

any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thusmaking a conical shape. The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:1) nly one disk may be moved at a time; 2) each move consists of taking the upper disk from one of the rods and sliding itonto another rod, on top of the other disks that may already be present on that rod; 3) no disk may be placed on top of asmaller disk.

Letn be the number of disks. Let T Hn be a graph, whose vertices are all possible correct placements of disks onto 3 rodsand edges connect placements that can be obtained one from another by 1 move. Consider also graph Pn, whose vertices are1s located in the first 2n rows of the 2-arithmetical Pascal triangle and edges connect neighboring 1s (i.e. two adjacent 1sin the same row or neighboring 1s by a diagonal in two adjacent rows )

1.4. prove that graphs T Hn and Pn are isomorphic.

1.5. Prove that that first 106 rows of 2-arithmetical Pascal triangle contain less than 1 % of 1s.1.6. Prove that ifn is divisible by p 1, then np1+ n2(p1)+ n3(p1)+. . .+ nn 1 (modp). Or,even better prove the general statement: if1 j, k p 1 n k (modp 1), then

n

j

+

n

(p 1) +j

+

n

2(p 1) +j

+

n

3(p 1) +j

+. . .

k

j

(modp).

00

01 11

02 12 22

03 13

23

33

04 14

24

34

44

. 1: . 2:


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Amazing properties of binomial coefficients 2

The official theoretical source for this set of problems is Vinbergs article [1]. Particularly the following theorems areconsidered to be known.

1. Wilsons theorem. For any prime p (and for primes only) the equivalence holds (p 1)! 1 (mod p).

2. Lukas theorem. Write the numbers n and k in base p :

n= ndpd + nd1p

d1 + . . .+ n1p + n0, k= kdpd + kd1p

d1 + . . . + k1p+ k0. (1)

Then

n

k

ndkd

nd1kd1

. . .

n1k1

n0k0

(modp) .

3. Kummers theorem. The exponent ordp

n

k is equal to the number of carries when we add k and = n k in

basep.

4. Wolstenholmes theorem. Ifp 5 then

2p

p 2 (mod p3), or, that is the same,

2p1

p1 1 (mod p3).

Remind that

0

0 = 1,

n

k = 0 for k > n and for k 2> .. . > r, then we can rewrite the last expression in the form

31 + 2 32 + 22 33 +. . .+ 2r1 3r .

2.3. Consider n-th row of Pascal triangle modulo 2 as binary expansion of some integer Pn. Prove that

Pn= Fi1 . . . Fis ,

wherei1, . . . , is are numbers of positions where 1s occur in the binary expansion ofn, and Fi= 22i + 1 is

i-th Fermat number.2.4. Prove that the number of non-zero elements in n-th row of p-arithmetical Pascal triangle equalsd

i=0(ni+ 1).

2.5.a) All the binomial coefficientsnk

, where0 < k < n, are divisible byp if and only ifn is a power ofp.

b) All the binomial coefficientsnk

, where 0 k n, are not divisible by p if and only if n+ 1 is

divisible by pd, in other words, all the digits ofn, except the leftmost, in base pare equal to p 1.2.6. Let 0 < k < n+ 1. Prove that if

nk1 ...p and nk ...p, then n+1k ...p, exceptthe case, whenn + 1 is

divisible by p.


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3 Generalization of Wilsons and Lukas theorems

3.1. Prove that ordp(n!) =n (nd+. . .+n1+n0)

p 1 .

3.2. Prove the following generalizations of Wilsons theorem. a)(1)[n/p](n!)p n0! (modp);b) Prove that for p 3

(pq !)p

1 (modpq) ,

and for p = 2, q 3 (pq!)p 1 (mod pq).c)

n!

p (1)n0!n1! . . . nd! (mod p), where = ordp(n!)

3.3. Generalized Lukas theorem. Let r= n k, = ordp(nk

). Then

1

p

n

k

(1)

n0!k0!r0!

n1!k1!r1!

. . . nd!

kd!rd!

(modp)

3.4. a) Prove that (1 +x)pd 1 +xpd (mod p)for all x = 0, 1, . . . , p 1.

b) Prove Laukas theorem algebraically.

3.5. a) Let m, n, k be nonnegative integers, and (n, k) = 1. Prove that Ckmn 0 (mod n).b) Prove that ifn

...pk, m ...p, then

nm

...pk.3.6. Let fn,a=

nk=0

(nk

)a. Prove that fn,a

di=0

fni,a (modp).

4 Variations on Wolstenholmes theorem

4.1. Prove that 1

1+

1

2+ . . .+

1

p 1 0 (mod p2).

4.2. Let p = 4k+ 3 be a prime number. Find 102 + 1

+ 112 + 1

+ . . .+ 1(p 1)2 + 1 (mod p).

4.3. a) Let k be a nonnegative integer such that for any prime divisor p of the number m k is not

divisible by(p 1). Prove that 11k

+ 1

2k+ . . . +

1

(p 1)k 0 (modm)(summation over all fractions whosedenominators are coprime to m).

b) Let k be odd and (k+ 1) ...(p 1). Prove that 1

1k+

1

2k +. . .+

1

(p 1)k 0 (mod p2).

4.4. Prove that the equivalence (12) from Vinbergs article holds in fact modulo p4.

4.5. Prove that the following properties are equivalent 1)

2p1p1

1 (mod p4);

2)

1

1+

1

2+ . . .+

1

p 1 0 (mod p3

); 3)

1

12 +

1

22 +. . .+

1

(p 1)2 0 (modp2

).

4.6. a) Prove algebraically that for any prime p and arbitrary k and n (pkpm

km) ... p2. (In Vinbergsarticle this fact is proven combinatorially.b) Prove the statement (9) form Vinbergs article: for any primep 5and arbitrarykandn (

pkpm

km)...p3.4.7. Let p 5. Prove that a)

p2

p

p1 (mod p5); b) ps+1p ps (modp2s+3).4.8. Prove that

p3p2

p2p (mod p8).


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Amazing properties of binomial coefficients 4

Additional problems to previous topics

4.11. Let m be a non negative integer, p 5be a prime. Prove that

1

mp+ 1

+ 1

mp+ 2

+

+

1

mp+ (p 1)0 (mod p2).

4.12. Let p and q be primes. Prove that2pq1pq1

1 (mod pq) if and only if 2p1p1 1 (mod q) and2q1q1

1 (modp).5 Sums of binomial coefficients

5.1. a) Prove that the sum3a1

k=0 2kk

is divisible by 3; b) is divisible by 3a.

5.2.Let Ck = 1k+12kk

be Catalan numbers. Prove that

nk=1

Ck 1 (mod 3)if and only if the number n + 1contains at least one digit 2 in base 3.

5.3. Let p 3, k= [2p/3]. Prove that the sump1

+p2

+. . .+

pk

is divisible by p2.

5.4. Let n...(p 1), where p is an odd prime. Prove that

n

p 1

+

n

2(p 1)

+

n

3(p 1)

+. . . 1 +p(n+ 1) (modp2).

5.5. Prove that if0 j p 1< n andq= n1p1 ] then

m:mj (mod p)(1)m

n

m

0 (mod pq).

5.6. Let p be an odd roime. Prove that n...(p+ 1) if and only if

n

j

n

j+ (p 1)

+

n

j+ 2(p 1)

n

j+ 3(p 1)

+. . . 0 (modp)

for all j = 1, 3, . . . , p 2.


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Solutions

1 Problems for solving in train

1.1. a) S o l u t i o n 1.p1

k

=

(p 1)(p 2) . . . (p k)1 2 k

(1)(2) . . . (k)1 2 k (1)

k (modp).

S o l u t i o n 2. It is evident by the formula for binomial coefficients that pi is divisible by p when1 i p 1. Since p1k1+ p1k = pk and p10 = 1 1 (modp), then(p10 + p11 )...p, and thereforep1

1

1 (modp). But p11 + p12 is divisible by p also, hence p12 1 (mod p)etc.) This problem is taken from [3, problem 162]. Since the fractions

2n+2n+1

/2nn

and

p12

n+1

/ p1

2n

are

highly reducible, the statement can be easily proven by induction. But we suggest a direct calculationfrom [3].

It easy to see that 2n

n

= 2n 1 3 (2n 1)

n!and

1

3

(2n

1) = (

1)n(

1)(

3)

(

2n+ 1)

(

1)n(p

1)(p

3)

(p

2n+ 1) =

= (1)n2np 1

2

p 32

p 2n+ 1

2

= (1)n2n

p 12

p 12

1

p 1

2 n+ 1

=

= (1)n2n (p12 )!

(p12 n)!(modp).

Therefore

2n

n

(1)n4n (

p12 )!

n!(p12 n)!= (4)n

p12

n

(mod p).

1.2. It follows directly from self-similar structure of an arithmetical Pascal triangle, that is described inthe next problems. It follows from Lucas theorem also, you can read the proof in [1].

1.3. We restrict ourselves with small contemplation, the full solution can be found in [3, problem 133].

Since thes-th row contains a long sequence of zeroes, then below these zeroes in(s + 1)-th row we havethe sequence of zeroes, too, (it is one element shorter than the upper sequence); in (s + 2)-th row there arethe sequence of zeroes also (it is one element shorter again) and so on. This explains the presence of thegrey triangle below 00 (fig. 1).

Further, the non-zero elements of the s-th row are equal to 1, hence the numbers situated along thesloped sides of the grey triangle all are 1s (due to the recurrence for binomial coefficients). So all thenumbers along the sloped sides of the triangles 01and

11are 1s, and therefore both triangles are identical

to 00.Now it is clear, what is the (2s)-th row of the triangle. The left- and the rightmost elements are 1s, all

other elements equal 0, except the central element that is equal to 2, because it is a sum of the two upper1s. Thus we obtain that two grey triangles are situated below 2s-th row, the triangles 02 and

22 to the

left and to the right of them are identical to00, and the triangle 12with 2s along its sloped sides is equalto 2 00.

And so on.

1.4. This statement we found in [21], several facts about binomial coefficients are proven there via Towerof Hanoi and the graph T Hn.

Let a be the diameter of the upper disc on the first rod, b be the diameter of the upper disc on thesecond rod and c be the diameter of the upper disc on the third rod. W.l.o.g. a < b < c, then we have 3possible moves in this configuration: from a to b or c and from b toc, we analogously have 3 moves if onerod is without discs. If all the discs are placed on one rod then we have 2 possible moves only; let A1, A2,A3 denote the configurations of this type.

Observe that by the problem 1.2 all the elements of2s-th row of Pascal triangle are 1s. Therefore graphPn has the rotational symmetry of the third order, because the recurrence

nk1

+nk

=n+1k

, that allows

us to construct the triangle from top to bottom, is equivalent in arithmetic modulo 2 to the recurrences nk1

=nk

+n+1

k

and

nk

= nk1

+n+1

k

, that allows us to construct the triangle from the low left


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A1

A3 A2

A2 A1

A3

A1 A3

A2

2

1

3

. 3:

corner in the upper right direction and from the low right corner in the upper left direction. It follows alsothat the triangle of the double size contains 3 copies of the initial triangle.

Now let us prove by induction that there exists a bijection betweenT Hn and Pn, such that the verticesof the triangle Pn correspond to the configurations A1, A2, A3. The basen = 1 is evident.

Proof of the step of induction. Assume that the bijection between T Hn and Pn has been constructed.The 2-arithmetical Pascal triangle with the side length 2n+1 contains 3 copies of the triangle with the side

length 2n. Number the copies and mark its vertices as shown on fig.3. Consider all the configurations ofthe Tower of Hanoi for which the (n + 1)-th (biggest) disc is placed on rod i. If we fix the placement of thisdisc then displacements of other discs correspond to the graph that is isomorphic to T Pn. By inductionhypothesis we can choose a bijection between this graph and the graphPn in thei-th copy of the triangle,such that the configurations Aj correspond to the vertices of the triangles with the same marks. When wemove the biggest disc, say, from the first rod to the second, all other discs must be on the 3rd rod. Thismove correspond to the edge connecting two neighboring vertices A3 on the left sloped side of big triangle.The same reasons concern other moves of the biggest disc. Therefore we obtain an isomorphism betweenT Pn+1 andPn.

1.5. The bijection with Tower of Hanoi gives us a formula (when the number of rows is a power of 2): thefirst2k rows of 2-arithmetical Pascal triangle contain 3k 1s. The formula can be also proved by induction

via recurrence from the problem 1.3. Using this formula we can obtain an estimation. Since 106


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sum is plus . Now transform the sum from the problem statement:

n

(p 1) +j

m

j

m

j 1

+. . .

+

m

p 1 +j

m

p 1 +j 1

+. . .+

m

j

+

+ m

2(p1) +j m

2(p1) +j 1+. . .+ m

2(p1) +j+. . .=

mi=0

(1)i

m

i

+

m

(p1) +j

(modp).

the first sum is equal to 0, the second sum is equivalentkj

(mod p) by the induction hypothesis.

Solution 2 ([J], [T]). Induction by n. The base n p 1 is trivial: both sides contain the sameterm. Prove the step of induction.

n

j

+

n

(p 1) +j

+. . .=

n 1

j

+

n 1j 1

+

n 1

(p 1) +j

+

n 1

(p 1) +j 1

+. . .=

=n 1j + n 1(p 1) +j+. . .+ n 1j 1+ n 1(p 1) +j 1+. . .

k 1j

+

k 1j 1

=

k

j

(modp).

But it should be accurate in cases when p 1 divides j or k, because the induction hypothesis does nothold for j = 0 or k = 0 (it uses the value p 1 instead of 0). Therefore we must consider more carefullythe cases when j = 1 or k = 1. We restrict ourselves by consideration of one partial case only. Let p= 5,

j = 1 and we fulfill step ton = 13. Then we have1

1

?

13

1

+

13

6

+

13

11

=

12

1

+

12

6

+

12

11

+

12

0

+

12

5

+

12

10

.

By induction hypothesis the sum in the first parentheses has a residue41

(and not01

as the previouscalculation shows). In the second parentheses the induction hypothesis covers all the terms except the firstone, so the sum has residue

120

+40

. Writingp 1instead of 4 for clarity, we obtain that the whole sum

is equivalent ton10

+p11

+p10

11 (modp), as required.S o l u t i o n 3 (algebraical reasoning with Lukas theorem, [18]). Induction by n. Base n p 1 is

trivial. Now let n p, write all parameters in base p, let p(m) denotes the sum of digits ofm. It is clearthat ifm j (mod p), thenp(m) j (modp). The sum under consideration is equal by Lukas theoremto n0

m0

n1m1

. . .

ndmd

(modp) ,

where the summation is over all m= md. . . m1m0 n, for which p(m)

j (mod p). This sum is equal

to the sum of coefficients ofxj , xj+p1, xj+2(p1), . . . in the expression(1 +x)n0(1 +x)n1 . . . (1 +x)nd = (1 +x)p(n) .

But it is evident that this sum of coefficients equals1rp(n)

rj (mod p1)

p(n)

r

,

which satisfy the induction hypothesis because 1 p(n) n1, and supply the desired equivalence sincep(n) n j (modp).

S o l u t i o n 4 (linear algebra, [D]). The polynomialsx, x2, . . . , xp1 are linearly independent over Zpand form a basis in the space of functions f: Zp Zp, f(0) = 0. By Fermats little theorem (1 + x)n (1 +x)k (modp). Applying the relations xi+a(p1) xi to the left hand side, we obtain that our sum asan element ofZp is equal to the coefficient ofx

j in the right hand side, i. e.kj

.


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2 Arithmetical triangle and divisibility

2.1. a) This result is due to Roberts [27]. By ak denote the number of 1s in the first 3k rows, and by bk

denote the number of 2s. Due to the recurrence from problem 1.3 we obtain

ak+1= 5ak+bk, bk+1= 5bk+ak.

Now the statement of problem follows by induction.

b) A n s w e r: 12 5k(5k + 1) 15k. By ak denote the number of nonzero elements in the first 5k rows.As in previous problem we have a recurrence

ak+1= 15ak+ 10 5k(5k 1)

2 .

Since the whole triangle consists of 5k(5k+1)

2 elements, it is natural to change variablesak = 5k(5k+1)

2 bk.Then we can rewrite the previous relation in terms ofbk as bk+1= 15bk.

c) A n s w e r:p(p+1)

2

k. This is Fines result [13]. It can be obtained by induction by means of recurrence

of the problem 1.3.

2.2. Solution 1. Induction by 1. The base 1 = 0, 1 can be easily checked. Let the statement hasbeen proven for all 1 < a. Prove it for 1 = a. Evedently m 21 < 21 . Let s = 21 (in notations ofproblem 1.3). Consider the m-th row in the triangle 00, where m= 2

2 + 23 + . . . + 2r . By the inductionhypothesis the number of 1s in this row and above it equals

32 + 2 33 +. . .+ 2r2 3r . (2)Then for the number m = m + 21 we have a row that intersects the triangles 10 and

11 (due to

2-arithmetics they are both identical to triangle 00). The part of Pascal triangle from top to this rowcontains triangle 00 (containing 3

1 1s by induction hypothesis) and partially triangles 10 and 11 (the

number of 1s in them is given by (2)). So the total number of 1s is

31 + 2(32 + 2 33 +. . .+ 2r2 3r ).

S o l u t i o n 2 (combinatorial sense of coefficients, [T]).

Lemma 1. Let the k-th row contains 2r 1s (or, equivalently, k contains r 1s in base 2) and let1> 2> > m, 2m > k. Then the row with number 21 + 22 +. . .+ 2m +k contains 2m+r 1s.

P r o o f. It is clear that the number21 + 22 +. . .+ 2m +k in base 2 contains m+r 1s and hencethe corresponding row contains2m+r 1s.

L e m m a 2. The rows with the following numbers

21 + 22 +. . .+ 2m1 , 21 + 22 +. . .+ 2m1 + 1, . . . , 21 + 22 +. . .+ 2m1 + 2m 1,

contain 2k3m 1s.

P r o o f. By lemma 1 the row with number 21 + 22 +. . .+ 2m1 +i contains 2kxi 1s, where xi isthe number of 1s in i-th row. Then the total number of 1s in these rows equals 2k

xi. But

xi is the

number of 1s in the first 2m 1 rows of Pascal triangle, this number is equal to 3m (it is known, forexample, by problem 1.4).

The statement of problem follows from lemma 2.

2.3.The problem is from [1], the solution is from [18]. The problem statement follows from Lukas theoremdue to the following observation (it is also mentioned in [1]): a binomial coefficient

nk

is odd if and only

if the set of 1s in the binary expansion of k is the subset of the set of 1s in the binary expansion ofn.Therefore Pn = 2k, where the summation is over all k described in the previous phrase. For p = 2 letSn= {i: ni= 1} in notations of formula (1). Then

Pn=ISn

iI

22i

=iSn

Fi.


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11/23


In particular 2n

k

2n

k 1

+

2n

k 2

. . .

2n

m+ 1

0 (mod 2n) .

Calculate the exponentord22n

r

by Kummers theorem. Iford2r= a then we havenacarries in addition

r and 2n r (it is clear by the standard algorithm of addition), hence ord22n

r

= n a. In particular

2n | 2n

r for odd r , that allows us to consider only one half of summands: 2n

k 1

+

2n

k 3

+. . .+

2n

m+ 1

0 (mod 2n) .

Now all the2n

i

in the left hand side have even parameter i, therefore ord2

2n

x

< n.

We will prove that this congruence is impossible and obtain a contradiction. Choose x with minimalord2

2nx

. Since ord2

2nx

< n and the whole sum is divisible by 2n, there exists y, for which ord2

2nx

=

ord22ny

. Then the binary representations ofx and y end with equal number of 0s, and hence there exists

z betweenx and y which binary representation ends with bigger number of 0s. Thenord22n

z


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2nn+k2

, . . . ,

2nn+k100

. Then GCD of all denominators in the right hand sides of the equalities does not

exceed (n +1)(n + 2) . . .

n + [

n ]

2n and (2n)100

n = 2

n log2n+

n.

For each there existsNsuch that for all n > Nwe have the equality n2 > n log2n + n. If we reduce2nn

by GCD for these n, the quotient is at least 2n/2.

2.13. a) The problem was presented at Leningrad olympiad, 1977.S o l u t i o n 1 (without Kummers theorem). This is solution from the excellent book [4]. Assume that

all these numbers are divisible by m. Then the numbersn+k 1

k 1

=

n+k

k

n+k 1k

,

n+k 2k 1

=

n+k 1

k

n+k 2k

,

. . . n

k 1

=

n+ 1

k

n

k

are also divisible by m. Then analogously m divides all the numbersn+i

j

, where i j are arbitrary

nonnegative integers. Butn0

(i= j = 0) is not divisible by m. A contradiction.

Solution 2 (Kummers theorem). Let p be a prime divisor of m. Prove that at least one of thenumbers

nk

,n+1k

, . . . ,

n+kk

is not divisible by p. By Kummers theorem if we choose (n k n)

such that the addition k+ fulfills in base p without carries then the binomial coefficientk+

k

is not

divisible by p.We will explain how to choose by giving a concrete example. Let p = 7, k = 133. We will write all

the numbers in base 7. Since we try to choose in the set ofk + 1 numbers, we can always choose suchthat k + to be one of the following numbers

. . . 133, . . . 233, , . . . , . . . 633.

(Remind that 6 is the greatest digit in our example.) It is clear that the addition k+ fulfills withoutcarries.

b) We found this problem in [2]. It is not difficult to construct n by Kummers theorem. Let ordpm= s,andk haved + 1digits in base p. Letn

...pd+s+1. Then the representations of numbers n k,n k + 1, . . . ,

n 1 contain digits (p 1) in positions from (d+ 2) to (d+s+ 2). When we add k to these numbers wehave carries in these positions. Therefore by Kummers theorem all the corresponding binomial coefficients

are divisible by ps.Since it is not difficult to combine our reasoning for distinct p, the statemetn is proven.

3 Generalizations of Wilsons and Lukas theorems

3.1. It is well known that ordp(n!) =

k

npk

. Ifn = ndp

d +nd1pd1 +. . .+n1p+n0 (representation inbasep), then

npk

=ndp

dk +nd1pdk1 + . . .+ nk+1p +nk and we can rewrite the formula for ordp(n!)in the form

ordp(n!) =

dk=1

di=k

nipik

=

di=1

ni(pi1 +pi2 +. . .+p+ 1) =

di=1

nipi

1

p 1 =

d

i=0

nipi

d

i=0

ni

p 1 .

This is exactly what we need.


13/23


3.2. a) Split the factors ofn! on groups of(p 1)factors:

(n!)p =

[np]1

k=0

(kp + 1) (kp + 2) (kp +p1) [np ]p + 1[np ]p + 2. . . [np ]p + n0 (1)[np ]n0! (mod p) .

) This statement can be found in Gauss works [15]. The product (pq!)p contains factors in pairs: afactor and its inverse modulo pq, the product of each pair is 1 modulo pq. So we need to watch on those

factors m which equals to its inverse, this factors satisfy the congruence

m2 1 (modpq).

For odd prime p the congruence has 2 solutions:1. For p = 2, q 3 the congruence has two moresolutions:2q1 1.

c) Since n! = (n!)p p[np][np ]!, the statement can be proven by induction by means of the congruence

of statement a) of this problem.

3.3. We found this problem on the web-page of A.Granville [17]. It well known Legendres formula for thenumber is that

= ordp

nk

=

np

kp

rp

+

np2 k

p2 r

p2

+. . . (4)

Denote n = [n/p] for brevity and so forth, and collect all terms divisible by p in the the formula fora binomial coefficient:

n

k

=

(n!)p(k!)p(r!)p

p[n/p]

p[k/p] p[r/p] n!

k! r! .

By generalized Wilsons theorem (problem 3.2, b) the first fraction equals n0!k0!r0! (modp), the third fractionallows us to apply induction, and the middle fraction (together with the sign of the first fraction) supplyall the expressions containing by the formula (4).

3.4.

a) Expand brackets in (1 +x)

pd

use the fact that p | pd

k

for 1

k

p

d

1 by Kummers theorem.b) Let n = np+n0, k = kp+k0. By the previous statement (1 +x)pn

(1 +xp)n (modp). Then

(1 +x)n = (1 +x)pn

(1 +x)n0 (1 +xp)n(1 +x)n0 (modp).

This congruence means that we transform the coefficients of the polynomial modulo p. The coefficient ofxk at the l.h.s. equals

nk

. All the exponents in the first brackets at the r.h.s. are divisible by p, hence the

only way to obtain the term xpk+k0 is multiplying thexpk

from the first bracket and xk0 from the second.Thus we obtain

n

k

n0k0

and so

nk

=n

k

n0k0

. Now Lukas theorem follows by induction.

3.5. a, b) It follows from Kummers theorem.

3.6.[9]. In the following calculation we use that niki = 0 for ki> ni; this allows us to apply Lukas theoremand truncate a lot of summands:

fn,a=n

k=0

n

k

a

ndkd=0

nd1kd1=0

n0

k0=0

di=0

niki

a

di=0

niki=0

niki

a

di=0

fni,a (modp).

4 Variations on Wolstenholmes theorem

4.1. This is an exercise on reading an article. The statement is proven in article [1]. Observe that

2

p1i=1

1

i =

p1i=1

1

i +

1

p i =pp1i=1

1

i(p i).


14/23


Hence the sum under consideration is divisible by p. Since 1i 1pi (mod p), it remains to check thatp1i=1

1

i2 0 (modp).

But 112

, 122

, . . . , 1(p1)2 modulo p is the same set as

1, 12, 22, . . . , (p 1)2. Therefore it is sufficient toprove that

p1

i=1

i2

0 (modp). (5)

Letp1i=1

i2 s (mod p). It p > 5 we can always choose a, such that a2 1 (modp). Then the sets{1, 2, . . . , p 1} and{a, 2a , . . . , (p 1)a} coincide (the proof is the same as in the footnote) and

s p1i=1

i2 =

p1i=1

(ai)2 =a2p1i=1

i2 a2s (mod p) .

Thuss 0 (mod p).4.2. A n s w e r: 2k+ 2. This problem of A. Golovanov was presented at Tuimaada-2012 olympiad. Observe

that forp = 4k + 3the equationx2 + 1 = 0has no solutions in the set of residues modulo p, and hence thedenominators of all fractions are non zero.

S o l u t i o n 1. Let ai= i2 + 1, i= 0, . . . , p 1. Then the expression equals

p1(a0, a1, . . . , ap1)p(a0, a1, . . . , ap1)

,

wherei is an elementary symmetrical polynomial of degree i. Find the polynomial for which the numbersai are its roots:

p1

i=0

(x 1 i2).

Change the variablex 1 =t2 and obtainp1i=0

(t2 i2) =p1i=0

(t i)p1i=0

(t+i) (tp t)(tp t) =t2p 2tp+1 +t2.

Now apply the inverse change of variables and obtain for p= 4k+ 3

p1i=0

(x 1 i2) (x 1)p 2(x 1)p+12 + (x 1) =xp +. . .+ (p+ 2 p+12 + 1)x 4.

By Vietes theorem p 4 (modp), p1 2 (modp), therefore p1p 12 2k+ 2 (mod p).Solution 2. Split all nonzero residues modulo p, except1, on pairs of reciprocal. We obtain 2k

pairs and in each pair (i, j)

ij 1 i2j2 1 (ij)2 +i2 +j2 + 1 i2 +j2 + 2 (modp).

Therefore,

1 (ij)2 +i2 +j2 + 1

(i2 + 1)(j2 + 1) i

2 +j2 + 2

(i2 + 1)(j2 + 1)=

1

i2 + 1+

1

j2 + 1 (mod p).

So, the sum is equal to 102+1

+ 112+1

+ 1(1)2+1 + 2k 2k+ 2.

1 These sets coincide because they contain p 1 element each, and it is clear that all the reminders in each set are nonzero and pairwise distinct.


15/23


S o l u t i o n 3. By Fermats little theorem the operations x x1 and x xp2 modulo p coincide.So it is sufficient to calculate the sum

p1x=0

(x2 + 1)p2 =p1x=0

p2m=0

p 2

m

x2m =

p2m=0

p 2

m

S2m, (6)

where S2m =p1

x=0

x2m. Evidently S2m 1 (modp) for m = p12 . Prove that S2m 0 (mod p) for allother m p 1. Indeed, for each m we can choose a non zero residue a such that a2m 1 (modp) andafter that we can reason as in (5). For the sum (6) we have

p2m=0

p 2

m

S2m

p 2p12

=

4k+ 1

2k+ 1

= (4k+ 1) 4k . . . (2k+ 1)

1 2 . . . (2k+ 1)

(2) (3) . . . (2k+ 2)1 2 . . . (2k+ 1) 2k+ 2 (modp).

4.3. We found these statements in [16].a) For each prime divisor p| m choose ap such that p (akp 1). By the Chinese reminder theorem

choosea such that a ap (mod p) for all p. Then the result can be proven by reasoning as in (5).b) Observe that for odd k by the binomial formula we have ik + (p ik) kik1p (mod p2). Then

2

p1i=1

1

ik =

p1i=1

1

ik +

1

(p i)k

=

p1i=1

ik + (p i)kik(p i)k

p1i=1

kik1pik(i)k kp

p1i=1

1

ik+1 (mod p2).

The sum in the r.h.s is divisible by p by the statement a).

4.4. The congruence holds even modulo p7 (see [24]), but it goes a bit strong. We can reason as in [1],tracing all powers till p4, and obtain

p 12p

1 =

(2p 1)(2p 2) . . . (p+ 1)p!

= 2p

1 1

2p

2 1 . . .

2p

p

1 1

1 2pp1i=1

1

i+4p2

p1i,j=1i


16/23


So the statement 1) is equivalent to the congruence

p1i=1

1

i(p i) 4p1i,j=1i


17/23


the fraction modulo p2 and obtain that it is congruent to 0. For the remaining part of the expression wecan apply the induction hypothesis and obtain

k

m 1

k m+ 1m

=

k

m

(modp2).

b) S o l u t i o n 1 (combinatorial). As it has been suggested in [1], consider samples ofkp objects fromthe set ofpn objects. Let the initial set be split on blocks ofp objects. The number of block samples equalsnk

. Hence it remains to check that non block samples is divisible by p3. But the number of non block

samples with 3 or more blocks is divisible by p3 (see [1]). For k >1 every non block sample consists of atleast 3 blocks, so in this case the statement is true. It remains to consider a case when k = 1and we countthe number of non block samples ofp objects from the set of2p objects. This number equals

2pp

2, byWolstenholmes theorem it is divisible by p3.

Solution 2. In the formulaab

= a(a1)...(ab+1)

b(b1)...1 split the numerator and the denominator onto blocksofp terms, reduce the first terms in each block, and collect the quotients in a separate expression:

mp

kp

=

mp (mp 1) . . . mp (p1)kp (kp 1) . . . kp (p1)

(m1)p

(m1)p 1 . . . (m1)p (p1)(k1)p (k1)p 1 . . . (k1)p (p1)

. . .

(mk+1) p

(mk+1)p 1 . . . (mk+1)p (p1)p (p 1) . . . 1 =

=

m

k

(mp 1) . . .

mp (p1)

(kp 1) . . . kp (p1) . . .

(mk+1)p 1 . . . (mk+1)p (p1)(p 1) . . . 1 .

It remains to check that the product of fractions is congruent to 1 (modp3). For this prove the congruence

(np 1) . . . np (p1)(rp 1) . . . rp (p1) 1 (mod p3)

or, even, it would be better to prove the following congruence

(np 1) . . . np (p1)(p 1)!

(rp 1) . . . rp (p1)(p 1)! (modp

3) .

This is true because both parts are congruent to 1 (mod p3), that can be shown analogously to the proofof Wolstenholmes theorem.

4.7. a) [5, theorem 2.14]. Transform the differencep2

p

p

1

=

p2(p2 1) . . . (p2 (p 1))1 2 . . . (p 1)p p=

p

(p 1)!

(1p2)(2p2) . . . ((p1)p2)12. . . (p1)

.

It remains to check that

(1 p2

)(2 p2

) . . . ((p 1) p2

) 1 2 . . . (p 1) (modp4

).

Expand brackets in the l.h.s.:

(1p2)(2p2) . . . ((p1)p2) = 1 2 . . . (p1)+p2

1 +1

2+ . . .+

1

p 1

(p1)!+terms divisible by p4.

By the problem 4.1 the second summand is divisible by p4.

b) Observe thatps+1

p

=ps ps+11p1 , hence it is sufficient to prove that ps+11p1 1 (modps+3).

ps+1 1p 1

=

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

ps+1

1 1

ps+1

2 1

. . .

ps+1

p 1 1

(1)p1 +ps+11 +12

+ . . .+ 1p 1

(mod ps+3).

Since(1)p1 = 1 and 1 + 12 + . . .+ 1p1 0 (modp2) we are done.


18/23


4.8. The problem is from [1], we present solution [T].

p3

p2

p2

p

=p

p3 1p2 1

p2 1p 1

=

=p

p3

1 1

p3

2 1

. . .

p3

p2 1 1

p2

1 1

p2

2 1

. . .

p2

p 1 1

=

=p

p2

1 1

p2

2 1

. . .

p2

p 1 1p

2

1k=1pk

p3

k 1

1

.

It is sufficient to prove that the last bracket is divisible by p7. Transform the product:

p21k=1pk

p3

k1

=

p212

k=1pk

p3

k1

p3

p2k 1

=

p212

k=1pk

p6 p5k(p2k) +1

1+p5(p1)

p212

k=1pk

1

k(p2k) (mod p7).

Now we have to check that the last sum is divisible by p2. This is true because by problem 4.3a)

p212

k=1pk

1

k(p2k) p212

k=1pk

1

k2 0 (modp2).

4.9. The statement is taken from [6, theorem 5], its generalization can be found in [7].

S o l u t i o n 1 ([5, proposition 2.19]). Use the fact that the difference2k+1

2k

2k2k1 is equal to thecoefficient ofx2

kin the polynomial

(1 +x)2k+1

(1 x2

)2k

= (1 +x)2k

(1 +x)2k

(1 x)2k

=

=

1 +

2k

1

x+

2k

2

x2 +. . .+x2

k

2

2k

1

x+

2k

3

x3 +. . .+

2k

2k 1

x2k1

.

Since the second polynomial contains odd exponents only, the coefficient ofx2k

in the product equals

2

2k

1

2k

2k 1

+

2k

3

2k

2k 3

+. . .+

2k

2k 1

2k

1

.

By problem 3.5 b)2k divides each binomial coefficient in this expression, moreover each term occurs twice

in the sum, and the sum itself is multiplied by 2. Thus all the expression is divisible by 22k+2

.

Solution 2 ([CSTTVZ]). Since2n+1

2n

= 2

2n+112n1

, it is sufficient to prove that

2n+1 1

2n 1

2n 12n1 1

(mod 22n+1).

Similarly to (3) we obtain2n+1 1

2n 1

=

2n+1

1 1

2n+1

3 1

. . .

2n+1

2n 1 1

2n 12n1 1

.

It is sufficient to prove that

L=

2n+1

1 1

2n+1

3 1

. . .

2n+1

2n 1 1

1 (mod 22n+1) .


19/23


This is true because

L (1)2n1 2n+1

1

1+

1

3+

1

5+ . . .+

1

2n 1

1 2n+1

2n

1 (2n 1)+ 2n

3 (2n 3)+ . . .+ 2n

(2n1 1)(2n1 + 1)

1 (mod 22n+1) .

4.10. This is theorem of Morley [26].S o l u t i o n 1 (authors proof, 1895). It goes a bit beyond the school curriculum.Take the formula which expresses cos2n+1 x via cosines of multiple angles,1 or, as they were saying in

that times, write cos2n+1 x in the form handy for integrating:

22ncos2n+1x= cos(2n+1)x+(2n+1) cos(2n1)x+ (2n+1) 2n1 2 cos(2n3)x+. . .+

(2n+1) 2n . . . (n+2)n!

cos x.

Now integrate it2 over the interval [0, 2 ]:

22n

cos2n+1 x dx=sin(2n+ 1)x

2n+ 1 +

2n+ 1

2n 1sin(2n 1)x+. . . ,

22n/20

cos2n+1 x dx= ()n

12n+ 1

2n+ 12n 1+ . . .

.

Every first grade student of university knows that it is convenient to use integration by parts for calculatingthis integral:

I2n+1 =

/20

cos2n+1 x dx=

/20

cos2n x cos x dx= cos2n x sin x

/2

0

+2n

/20

cos2n1 x sin2 x dx=

= 0 + 2n

/2

0

cos2n1 x(1

cos2 x) dx= 2n

I2n1

2n

I2n+1

,

thereforeI2n+1= 2n2n+1 I2n1. Since I1= 1, we can apply the formula n times and obtain

/20

cos2n+1 x dx= 2n (2n 2) . . . 2(2n+ 1)(2n 1) . . . 3.

Equating of these two results give us the formula

22n 2n (2n 2) . . . 2

(2n+ 1)(2n

1) . . . 3

= ()n

1

2n+ 12n+ 1

2n

1

+ . . .+(2n+1) 2n . . . (n+2)

n!

.

Letp = 2n + 1be a prime number. We obtain the desired congruence by multiplying the last formula by p:

22n 2n (2n 2) . . . 2

(2n 1)(2n 3) . . . 3 ()n (mod p2) .

S o l u t i o n 2 ([CSTTVZ]). We will use the following notations:

A=

p12

i=1

1

i, B =

1i


20/23


21/23


b) Solution of [D]. The sum is a coefficient ofx3a1 in the polynomial

x3a1

1 +(x+ 1)2

x +

(x+ 1)4

x2 +. . .+

(x+ 1)2(3a1)

x3a1

=

(x+1)23a

x3a 1

(x+1)2

x 1 x3a1 =(x+ 1)

23a x3ax2 +x+ 1

=

= x23a +

23a1

x23a1 +

23a2

x23a2 +. . .+ 1 x3a

x3

1

(x 1) .

In order to find this coefficient we will perform the long division of the numerator by the denominatorand then multiply the result by (x 1). We do not need to find the quotient at whole, it is sufficient toperform the division till the moment when the coefficient ofx3

a2 will be found, remind that we are tryingto find this coefficient modulo 3a only. Since for b ...3 all the binomial coefficient

23ab

are divisible by 3a

(by Kummers theorem), we can collect all these coefficient in a separate sum. When we divide this sumby x3 1 all the coefficients of the quotient are divisible by 3a therefore we can discard this sum. Theremaining expression is

x23a +23a3

x23a3 + 23a6 x23a6 +. . .+ 1 x3ax3 1 (x 1).

All the exponents in the numerator are divisible by 3, hence after division by x3 1 all the exponents ofthe quotient are divisible by 3, too, and after the multiplying it by x 1, there will be no exponents of theform 3k+ 2. So the coefficient that we seek equals 0 (mod 3a).

5.2. This problem was published in Monthly [25]. Since2n+ 2

n+ 1

4

2n

n

= 2 2n+ 1

n+ 1

2n

n

4

2n

n

= 2Cn,

thenCn2n+2n+1

2nn (mod 3). Therefore this sum is telescopic modulo 3:n

k=1

Ck

2n+ 2

n+ 1

2n

n+ 2n

n

2n 2n 1 +. . . =

2n+ 2

n+ 1+ 1 (mod 3).

So by Kummers theorem we have to clarify when we have at least one carry in the addition of the number(n+ 1) with itself in base 3. It it clear that it happens only ifn+ 1 contains at least one 2 in base 2.

5.3. This is problem A5 of Putnam Math. Competition, 1998. Since 1ppn

(1)n1n (modp), we havek

n=1

1

p

p

n

kn=1

(1)n1n

=k

n=1

1

n 2

[k/2]n=1

1

2n

kn=1

1

n+

p1n=p[k

2]

1

n=

p1n=1

1

n 0 (modp).

The summation in the sum to the left of asterisk really starts from n = k+ 1 (it is easy to check: for

p= 6r+ 1 we havek= 4r and p [k

2 ] = 4r+ 1 =k+ 1, similarly for p = 6r+ 5).5.4. This statement is from [11]. Solution [CSTTVZ]. Induction on n. The base is trivial. Prove theinduction step fromn = n (p 1)ton. Let q= np1 . Since

n+p 1x(p 1)

=

p1i=0

p 1

i

n

x(p 1) i

,

we can rewrite the sum under consideration in the form

n

p

1

+

n

2(p

1)

+

n

3(p

1)

+. . .=

q

x=1p1

i=0 p1

i n

x(p

1)

i

=

=

p1i=0

p1

i

qx=1

n

x(p1) i

. (11)


22/23


By the problem 1.1 a) we havep1

i

(1)i (modp); let p1i =ap + (1)i. By the problem 1.6 we haveq

x=1

nx(p1)i

p1i (1)i (modp)for i = 0, 1, . . . , p2; let qx=1

nx(p1)i

=bp+ (1)i. Then

p1

i

qx=1

n

x(p1) i

=

ap+ (1)ibp+ (1)i 1 + (1)i(ap+bp) == 1+(

1)ip1

i+

q

x=1

n

x(p1)i 2 (1)i = (1)

ip1i+

q

x=1

n

x(p1)i1 (mod p2).

Remind that these transformations hold for 0 i p 2. We can continue equality (11), by separatingthe summand for i = p 1:p1i=0

p1

i

qx=1

n

x(p1)i

p2i=0

(1)i

p1

i

+

qx=1

n

x(p1)i

1

+

q1x=0

n

x(p1)

=

=

p2i=0

(1)i

p1i

+

p2i=0

(1)i

qx=1

n

x(p1) i

(p 1) +

n

0

+

q1x=1

n

x(p1)

.

The first sum here equals1, because p10 p11 + p12 +. . . = 0. By the same reasons the second(double) sum together with the summand

n0

equals 0. The last sum equals1 +p(n+ 1)by the induction

hypothesis. Therefore the whole expression equals1 + 0 p + 1 + 1 +p(n+ 1) = 1 +pn. This is exactlywhat we need because 1 +p(n+ 1) = 1 +p(n+p 1 + 1) 1 +pn (modp2).5.5. This is result of Fleck, 1913, it is cited in [18]. Solution [CSTTVZ].

For p = 2 the sum is not alternating and the result is trivial. Let p be odd. We use the inductionon q. The base follows from the statement 2.5 a). Prove the induction step fromn = n (p1) to n. Theexpression

x below denotes the summation over x in natural bounds (i.e. in bounds for which all the

binomial coefficients are correctly defined). We have

m:mj (mod p)(1)mn

m

=x (1)

xn+p 1

xp+j

=x (1)

xp1i=0

p 1

i n

xp+j i ==

p1i=0

p 1

i

x

(1)x

n

xp+j i

.

By the induction hypothesispq1

x(1)x nxp+ji; by the problem 1.1 a) p1i (1)i (modp). Therefore

p1i=0

p 1

i

x

(1)x

n

xp+j i

p1i=0

(1)ix

(1)x

n

xp+j i

(modpq) .

The last (double sum equals n0 n

1 + n2 n

3 +. . .= 0.5.6. The result of Bhaskaran (1965), it is cited in [18], solution [CSTTVZ].

Induction on n. Let

f(n, j) =

n

j

n

j+ (p 1)

+

n

j+ 2(p 1)

n

j+ 3(p 1)

+. . .

The base n = p+ 1 is trivial, but observe thatp+1

i

1 (mod p) for i = 0, 1, p , p+ 1, otherwise thisbinomial coefficient is divisible by p. Prove the step of induction from n = n (p+ 1) to n. By theobservation above we have

n+ (p+ 1)

j+ (p

1)k =p+1

i=0

n

j+ (p

1)k

ip+ 1

i i{0,1,p,p+1}

n

j+ (p

1)k

i ==

n

j+ (p 1)k

+

n

j 1 + (p 1)k

+

n

j 1 + (p 1)(k 1)

+

n

j 2 + (p 1)(k 1)

(mod p) .


23/23


Since f(n, j) =

k(1)k nj+k(p1)

is an alternating sum, the underlined summands cancel (except the

first and the last, but these summands are equal to 0 due to incorrect binomial coefficietns). So we obtainthe equalities

f(n, j) f(n, j) f(n, j 2) j >1 , f(n, 1) f(n, 1) +f(n, p 2) .Now the part only if of the problem statement follows from the induction hypothesis, and the part if,too: iff(n, j) 0 (mod p)for j = 1, 3, . . . , p 2, then

f(n, p 2) f(n, p 4) . . . f(n, 1) f(n, p 2) ,from where f(n, j) 0 (mod p)for all required j , and then n ...(p+ 1), hence n...(p+ 1).

References

The authors of many solutions are participants of the conference:[D] Didin Maxim;[] Krekov Dmitri;[J] Jastin Lim Kai Ze;

[T] Teh Zhao Yang Anzo;[CSTTVZ] Cevid Domagoj, Stokic Maksim, Tanasijeviic Ivan, Trifunovic Petar, Vukorepa Borna, Zikelic ore

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Amazing Properties of Bionomial Coefficients