American Mathematical Monthly Problem 11391 by Marian Tetiva (edited)
Letpbe a prime number, andsa positive integer. Letk be an integer, and letnbe an integer such that n≥k ≥ps−ps−1. For every i∈ {1,2, ..., n},letxi be an integer.
Assume that (p, s, k)6= (2,2,2) and (s, k)6= (1, p−1). Prove that
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt ≡0 modp )
≡0 modps and
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt 6≡0 modp )
≡0 modps.
Solution by Darij Grinberg.
First, we forget about the problem and show an important (and probably well- known) fact:
Theorem 1. Let S be a finite set. Let R be a commutative ring with 1.
For every i ∈S, let ai be an element of R. Let u be a nonnegative integer such that u≤ |S|. Then,
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| . Proof of Theorem 1. We will prove Theorem 1 by induction over |S|. Induction base: If |S|= 0, then S =∅, so that
X
T⊆S
(−1)|T| X
t∈T
at
!u
= X
T⊆∅
(−1)|T| X
t∈T
at
!u
= (−1)|∅|
| {z }
=(−1)0=1
X
t∈∅
at
| {z }
=0
u
(since the only subset of ∅ is∅)
= 1·0u = 0u and
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| =
0, if u <0;
(−1)|S|
| {z }
=(−1)0=1
0!
|{z}
=1
·Y
i∈∅
ai
| {z }
=1
, if u= 0 =
0, if u <0;
1, if u= 0 = 0u, so that
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| .
Thus, Theorem 1 is proven in the case |S|= 0. This completes the induction base.
Induction step: Letnbe a positive integer. Assume that Theorem 1 holds for every finite setS satisfying |S|=n−1.We have to show that Theorem 1 also holds for every finite set S satisfying |S|=n.
We assumed that Theorem 1 holds for every finite set S satisfying |S|=n−1. In other words, we assumed the following assertion:
Assertion A: Let S be a finite set satisfying |S|=n−1. Let R be a commutative ring with 1.For every i∈S,letai be an element of R. Letu be a nonnegative integer such that u≤ |S|. Then,
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| . Upon renaming S intoS0 and uinto v, this assertion rewrites as:
Assertion A0: LetS0 be a finite set satisfying|S0|=n−1. LetRbe a commutative ring with 1.For every i∈S0, letai be an element ofR. Letv be a nonnegative integer such that v ≤ |S0|.Then,
X
T⊆S0
(−1)|T| X
t∈T
at
!v
=
( 0, if v <|S0|; (−1)|S0||S0|!· Q
i∈S0
ai, if v =|S0| .
Now, we have to show that Theorem 1 also holds for every finite set S satisfying
|S|=n. In other words, we have to prove the following assertion:
Assertion B: LetS be a finite set satisfying|S|=n. Let R be a commutative ring with 1. For everyi∈S,letai be an element of R. Letube a nonnegative integer such that u≤ |S|.Then,
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| . Proof of Assertion B. The conditions u≤ |S| and |S|=n yieldu≤n.
Letα be an element ofS (such anα exists, since|S|=n >0 yieldsS 6=∅). Then,
|S\ {α}|=|S| −1 = n−1. Now, X
T⊆S
(−1)|T| X
t∈T
at
!u
= X
T⊆S such that
α∈T
(−1)|T| X
t∈T
at
!u
+ X
T⊆S such that
α /∈T
(−1)|T| X
t∈T
at
!u
= X
T⊆S such that
α∈T
(−1)|(T\{α})∪{α}|
X
t∈(T\{α})∪{α}
at
u
+ X
T⊆S such that
α /∈T
(−1)|T| X
t∈T
at
!u
(1)
here we have replaced T by (T \ {α})∪ {α} in the first sum, since every set T such that α∈T satisfiesT = (T \ {α})∪ {α}
.
Now, the mapping from{T ⊆S such thatα ∈T}to{T0 ⊆S\ {α}}which maps every T to T \ {α} is a bijection1; hence, we can substitute T0 for T \ {α} in the sum
P
T⊆S such that
α∈T
(−1)|(T\{α})∪{α}| P
t∈(T\{α})∪{α}
at
!u
and obtain
X
T⊆S such that
α∈T
(−1)|(T\{α})∪{α}|
X
t∈(T\{α})∪{α}
at
u
= X
T0⊆S\{α}
(−1)|T0∪{α}|
X
t∈T0∪{α}
at
u
.
Thus, (1) becomes X
T⊆S
(−1)|T| X
t∈T
at
!u
= X
T0⊆S\{α}
(−1)|T0∪{α}|
X
t∈T0∪{α}
at
u
+ X
T⊆S such that
α /∈T
(−1)|T| X
t∈T
at
!u
= X
T0⊆S\{α}
(−1)|T0∪{α}|
X
t∈T0∪{α}
at
u
+ X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u
(since {T ⊆S such that α /∈T}={T ⊆S\ {α}})
= X
T0⊆S\{α}
(−1)|T0|+1 aα+X
t∈T0
at
!u
+ X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u
since T0 ⊆S\ {α} yieldsα /∈T0 and thus |T0 ∪ {α}|=|T0|+ 1
and P
t∈T0∪{α}
at =aα+ P
t∈T0
at
!
= X
T⊆S\{α}
(−1)|T|+1 aα+X
t∈T
at
!u
+ X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u
(here we renamed T0 into T in the first sum)
= X
T⊆S\{α}
(−1)|T|+1
| {z }
=−(−1)|T|
aα+X
t∈T
at
!u
+ (−1)|T| X
t∈T
at
!u
= X
T⊆S\{α}
(−1)|T| − aα+X
t∈T
at
!u
+ X
t∈T
at
!u!
1In fact, its inverse mapping is the mapping from {T0⊆S\ {α}} to {T ⊆S such thatα∈T} which maps everyT0 to T0∪ {α}.
= X
T⊆S\{α}
(−1)|T| −
u
X
m=1
u m
amα X
t∈T
at
!u−m
+ X
t∈T
at
!u!
+ X
t∈T
at
!u!
since the binomial theorem yields
aα+P
t∈T
at
u
=
u
P
m=0
u m
amα
P
t∈T
at
u−m
=
u
P
m=1
u m
amα
P
t∈T
at u−m
+ u
0
| {z }
=1
a0α
|{z}
=1
X
t∈T
at
!u−0
| {z }
= P
t∈T
at
!u
=
u
P
m=1
u m
amα
P
t∈T
at u−m
+
P
t∈T
at u
= X
T⊆S\{α}
(−1)|T| −
u
X
m=1
u m
amα X
t∈T
at
!u−m!
=− X
T⊆S\{α}
(−1)|T|
u
X
m=1
u m
amα X
t∈T
at
!u−m
=−
u
X
m=1
u m
amα X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u−m
. (2)
Now, for every m ∈ {1,2, ..., u}, we can apply Assertion A0 to S0 = S \ {α} and v =u−m(these satisfy|S0|=n−1,since|S0|=|S\ {α}|=n−1,and v ≤ |S0|,since u≤ |S|and m ≥1 yieldu−m ≤ |S| −1 =n−1 =|S0|, so thatv ≤ |S0|), and obtain
X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u−m
=
0, if u−m <|S\ {α}|; (−1)|S\{α}||S\ {α}|!· Q
i∈S\{α}
ai, if u−m=|S\ {α}| . Since |S\ {α}|=n−1,this becomes
X
T⊆S\{α}
(−1)|T| X
t∈T
at
!u−m
=
( 0, if u−m < n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−m =n−1 .
Thus, (2) rewrites as X
T⊆S
(−1)|T| X
t∈T
at
!u
=−
u
X
m=1
u m
amα
( 0, if u−m < n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−m=n−1
=−
1
X
m=1
u m
amα
( 0, if u−m < n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−m=n−1
here we replaced the
u
P
m=1
sign by an
1
P
m=1
sign, since all addends form >1 are zero (because m >1 leads to u−m < u−1≤n−1 (since u≤n) and thus
( 0, if u−m < n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−m=n−1 = 0)
=− u
1
| {z }
=u
a1α
|{z}
=aα
( 0, if u−1< n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−1 = n−1
=−uaα
( 0, if u−1< n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u−1 =n−1
=−uaα
( 0, if u < n;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u=n
(sinceu−1< n−1 is equivalent to u < n, and since u−1 =n−1 is equivalent to u=n)
=
=0
z }| {
−uaα·0, if u < n;
− u
|{z}
=n
aα(−1)n−1(n−1)!· Q
i∈S\{α}
ai, if u=n
=
0, if u < n;
−(−1)n−1
| {z }
=(−1)n
·n(n−1)!
| {z }
=n!
·aα Y
i∈S\{α}
ai
| {z }
=Q
i∈S
ai, sinceα∈S
, if u=n
=
( 0, if u < n;
(−1)nn!· Q
i∈S
ai, if u=n =
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| (sincen =|S|).
2 Thus, Assertion B is proven. In other words, Theorem 1 holds for every finite set S satisfying |S|=n. This completes the induction step.
Therefore, the induction proof of Theorem 1 is complete.
Next, a straightforward lemma:
2Strictly speaking, this argument does not function in the caseu= 0,but in this case we can prove
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, ifu <|S|; (−1)|S||S|!· Q
i∈S
ai, ifu=|S|
Lemma 2. Let p be a prime number, and s a positive integer. Let k be an integer such that k ≥ ps−ps−1. Assume that (p, s, k) 6= (2,2,2), (p, s, k)6= (2,2,3) and (s, k)6= (1, p−1). Then, k!≡0 modps.
Proof of Lemma 2. Since p is a prime, we have p≥2.
First we consider the cases= 1.In this case,ps=p1 =pandps−1 =p1−1 =p0 = 1, so that k ≥ps−ps−1 =p−1, but k 6= p−1 (because s = 1, but (s, k)6= (1, p−1)), so that k ≥ p, and thus k! = Q
i∈{1,2,...,k}
i =p Q
i∈{1,2,...,k}\{p}
i ≡0 modp,what rewrites as k!≡0 modps (since ps=p). Thus, Lemma 2 holds in the case s= 1.
Hence, for the rest of the proof of Lemma 2, we may assume thats >1.So assume that s >1.Thus, s≥2.
If s ≥ 3, then k! = 1·2·...·k is divisible by (ps−ps−1) (ps−ps−1−ps−2) (since ps−ps−1 andps−ps−1−ps−2 are two distinct factors in the product 1·2·...·k,because k ≥ps−ps−1 > ps−ps−1−ps−2 >0 3), but (ps−ps−1) (ps−ps−1−ps−2) is divisible byps (since
ps−ps−1
| {z }
=ps−1(p−1)
ps−ps−1−ps−2
| {z }
=p(ps−1−ps−2−ps−3)
=ps−1p(p−1) ps−1 −ps−2−ps−3
=ps(p−1) ps−1−ps−2−ps−3
), so that k! is divisible by ps, and thus k! ≡ 0 modps. Thus, Lemma 2 holds in the case s≥3.
Hence, for the rest of the proof of Lemma 2, we may assume that s < 3. Together with s ≥2, this yieldss = 2. Thus,k ≥ps−ps−1 becomes k ≥p2−p2−1 =p2−p1 = p2−p=p(p−1).
If p >2, then k! = 1·2·...·k is divisible by p(p−1)·p (since p(p−1) and p are two distinct factors in the product 1·2·...·k, becausek ≥p(p−1)> p >0 4), but p(p−1)·p is divisible byps (since p(p−1)·p=p2(p−1) = ps(p−1)), so thatk! is divisible by ps,and thus k!≡0 modps.Thus, Lemma 2 holds in the case p >2.
Hence, for the rest of the proof of Lemma 2, we may assume that p≤ 2. Together with p ≥ 2, this yields p = 2. Thus, k ≥ p(p−1) becomes k ≥ 2 (2−1)
| {z }
=1
= 2. But k 6= 2 (sincep= 2 ands= 2,but (p, s, k)6= (2,2,2)) andk 6= 3 (sincep= 2 ands = 2, but (p, s, k) 6= (2,2,3)). Hence, k ≥ 4. Thus, k! =
k
Q
i=1
i = 1·2·3·4
| {z }
≡0 mod 4
·
k
Q
i=5
i ≡ 0 mod 4.
But 4 = 22 =ps, so this becomes k!≡0 modps. Hence, Lemma 2 is proven.
differently: By (2), we have
X
T⊆S
(−1)|T| X
t∈T
at
!u
=−
u
X
m=1
u m
amα
0, ifu−m < n−1;
(−1)n−1(n−1)!· Q
i∈S\{α}
ai, ifu−m=n−1
| {z }
an empty sum, sinceu=0
=−0 = 0 =
( 0, ifu <|S|; (−1)|S||S|!· Q
i∈S
ai, ifu=|S| (sinceu= 0< n=|S|).
3Here,ps−ps−1−ps−2>0 follows fromps−ps−1=ps−1 (p−1)
| {z }
≥1,sincep≥2
≥ps−1> ps−2.
4Here,p(p−1)> p follows fromp−1>1 (sincep >2).
Theorem 1 and Lemma 2 quickly yield:
Corollary 3. Let p be a prime number, and s a positive integer. Let k be an integer, and ua nonnegative integer such that u≤k, and let S be a finite set such that|S|=k≥ps−ps−1. For everyi∈S,letai be an integer.
Assume that (p, s, k)6= (2,2,2), (p, s, k)6= (2,2,3) and (s, k) 6= (1, p−1). Then,
X
T⊆S
(−1)|T| X
t∈T
at
!u
≡0 modps.
Proof of Corollary 3. Notice that u≤k rewrites as u≤ |S|.
Lemma 2 yieldsk!≡0 modps,thus|S|! =k!≡0 modps. Now, Theorem 1 (applied toR =Z) yields
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
0, if u <|S|; (−1)|S| |S|!
|{z}
≡0 modps
· Q
i∈S
ai, if u=|S| ≡
0, if u <|S|;
0, if u=|S| = 0 modps, and Corollary 3 is proven.
Next, we introduce a notation:
For any assertion U, we denote by [U] the Boolean value of the assertion U (that is, [U] =
1, ifU is true;
0, if U is false ).
It is then clear that if B is a set, ifU(X) is an assertion for every subset X of B, and if α(X) is a number for every subset X of B, then
X
X⊆B
[U(X)]α(X) = X
X⊆B such that U(X) is true
α(X). (3)
5 In particular, if B is a set, and U(X) is an assertion for every subset X of B, then X
X⊆B
[U(X)] =|{X ⊆B | U(X) is true}|. (4)
5In fact,
X
X⊆B
[U(X)]α(X) = X
X⊆B
1, if U(X) is true;
0, if U(X) is false ·α(X) = X
X⊆B such that U(X) is true
1·α(X) + X
X⊆B such that U(X) is false
0·α(X)
| {z }
=0
= X
X⊆B such that U(X) is true
1·α(X)
| {z }
=α(X)
= X
X⊆B such that U(X) is true
α(X).
6
Also, every assertion U satisfies [not U] = 1−[U]. For any two assertionsU and V, we have [U and V] = [U]·[V].
We come to another easy fact:
Lemma 4. Letp be a prime number; let s be a positive integer, and a an integer. Then,
aps−ps−1 ≡[a6≡0 modp] modps.
Proof of Lemma 4. Since p is a prime, we have p ≥ 2. Thus, ps − ps−1 = ps−1 (p−1)
| {z }
≥1,sincep≥2
≥ps−1 ≥2s−1 (sincep≥2), so thatps−ps−1 ≥s(due to the well-known inequality 2s−1 ≥s for any integer s≥1). Hence, as|aps−ps−1.
Ifa ≡0 modp,then p|a, and thus ps|as,what (together with as |aps−ps−1) yields ps|aps−ps−1,and thusaps−ps−1 ≡0 modps,what yieldsaps−ps−1 ≡0 = [a6≡0 modp] modps (since [a6≡0 modp] = 0, because a 6≡ 0 modp is false, since a ≡ 0 modp). Thus, Lemma 4 holds in the case a≡0 modp.
Hence, for the rest of the proof of Lemma 4, we may assume that a6≡0 modp. So assume that a6≡0 modp.Then,p-a, so thata is coprime tops (sincep is prime), and thus aφ(ps) ≡1 modps by Euler’s Totient Theorem. Since φ(ps) =ps−1(p−1) = ps− ps−1, this becomes aps−ps−1 ≡1 modps. Since [a6≡0 modp] = 1 (because a 6≡ 0 modp is true), this yields aps−ps−1 ≡1 = [a6≡0 modp] modps. Thus, Lemma 4 is proven.
From Corollary 3 and Lemma 4, we can easily conclude:
Corollary 5. Let p be a prime number, and s a positive integer. Let k be an integer, and let S be a finite set such that |S| = k ≥ ps − ps−1. For every i ∈ S, let ai be an integer. Assume that (p, s, k) 6= (2,2,2) and (s, k)6= (1, p−1). Then,
X
T⊆S
(−1)|T|
"
X
t∈T
at6≡0 modp
#
≡0 modps (5)
and
X
T⊆S
(−1)|T|
"
X
t∈T
at ≡0 modp
#
≡0 modps. (6)
6In fact, setα(X) = 1 for everyX⊆B; then,
X
X⊆B
[U(X)]
| {z }
=[U(X)]·1
=[U(X)]α(X)
= X
X⊆B
[U(X)]α(X) = X
X⊆B such that U(X) is true
α(X) (by (3))
= X
X⊆B such that U(X) is true
1 =|{X ⊆B | U(X) is true}|
Proof of Corollary 5. For everyT ⊆S,we have
P
t∈T
at
ps−ps−1
≡
P
t∈T
at 6≡0 modp
modps (by Lemma 4, applied to a= P
t∈T
at). Hence,
X
T⊆S
(−1)|T|
"
X
t∈T
at6≡0 modp
#
≡X
T⊆S
(−1)|T| X
t∈T
at
!ps−ps−1
modps. (7) Now, we will consider the case (p, s, k) = (2,2,3).In this case, Theorem 1 (applied tou=ps−ps−1,which satisfiesu≤ |S|because u=ps−ps−1 = 22−22−1 = 22−21 = 4−2 = 2<3 = k =|S|) yields
X
T⊆S
(−1)|T| X
t∈T
at
!ps−ps−1
=
( 0, if ps−ps−1 <|S|; (−1)|S||S|!· Q
i∈S
ai, if ps−ps−1 =|S| = 0 since ps−ps−1 <|S|
≡0 modps.
Next, we will consider the case (p, s, k) 6= (2,2,3). In this case, we can apply Corollary 3 to u=ps−ps−1 (since ps−ps−1 ≤k), and obtain
X
T⊆S
(−1)|T| X
t∈T
at
!ps−ps−1
≡0 modps.
Thus, in both cases (p, s, k) = (2,2,3) and (p, s, k) 6= (2,2,3), we have obtained the congruence
X
T⊆S
(−1)|T| X
t∈T
at
!ps−ps−1
≡0 modps. Combining this congruence with (7), we thus get
X
T⊆S
(−1)|T|
"
X
t∈T
at6≡0 modp
#
≡0 modps
in both cases (p, s, k) = (2,2,3) and (p, s, k)6= (2,2,3). Hence, (5) is proven in both cases (p, s, k) = (2,2,3) and (p, s, k)6= (2,2,3). In other words, (5) always holds.
We can also apply Theorem 1 to u= 0 (since 0≤ |S|), and obtain X
T⊆S
(−1)|T| X
t∈T
at
!0
=
( 0, if 0<|S|; (−1)|S||S|!· Q
i∈S
ai, if 0 =|S| = 0
(since 0<|S|, since |S| ≥ps−ps−1 > ps−1−ps−1 = 0). Since
P
t∈T
at 0
= 1 for every T ⊆S,this becomes
X
T⊆S
(−1)|T|·1 = 0. (8)
Now, X
T⊆S
(−1)|T|
"
X
t∈T
at≡0 modp
#
=X
T⊆S
(−1)|T| "
not X
t∈T
at 6≡0 modp
#!
=X
T⊆S
(−1)|T| 1−
"
X
t∈T
at 6≡0 modp
#!
=X
T⊆S
(−1)|T|·1
| {z }
=0 by (8)
−X
T⊆S
(−1)|T|
"
X
t∈T
at6≡0 modp
#
| {z }
≡0 modps by (5)
≡0−0 = 0 modps.
This proves (6). Thus, Corollary 5 is proven.
A final lemma before we come to the problem:
Lemma 6. Let k be an integer, and let n be an integer such that n ≥ k.
LetT be a subset of{1,2, ..., n}. Then,
|{S ⊆ {1,2, ..., n} | |S|=k and T ⊆S}|=
n− |T| k− |T|
.
Proof of Lemma 6. The mapping from {S ⊆ {1,2, ..., n} | |S|=k and T ⊆S} to {S0 ⊆ {1,2, ..., n} \T | |S0|=k− |T|} which maps every S to S\T is a bijection7; hence,
|{S ⊆ {1,2, ..., n} | |S|=k and T ⊆S}|=|{S0 ⊆ {1,2, ..., n} \T | |S0|=k− |T|}|. But by the combinatorial interpretation of binomial coefficients,
|{S0 ⊆ {1,2, ..., n} \T | |S0|=k− |T|}|=
|{1,2, ..., n} \T| k− |T|
=
n− |T| k− |T|
(since T ⊆ {1,2, ..., n} yields|{1,2, ..., n} \T|=|{1,2, ..., n}|
| {z }
=n
− |T|=n− |T|). Thus,
|{S ⊆ {1,2, ..., n} | |S|=k and T ⊆S}|=
n− |T| k− |T|
, and Lemma 6 is proven.
Now, let us finally solve the problem:
7In fact, its inverse mapping is the mapping from {S0⊆ {1,2, ..., n} \T | |S0|=k− |T|} to {S ⊆ {1,2, ..., n} | |S|=k andT ⊆S} which maps everyS0 to S0∪T.
We have
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt ≡0 modp )
=
k
X
j=0
(−1)j
n−k+j j
X
T⊆{1,2,...,n}
"
|T|=k−j and X
t∈T
xt≡0 modp
#
| {z }
=[|T|=k−j]·
"
P
t∈T
xt≡0 modp
#
(by (4))
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
# k X
j=0
(−1)j
n−k+j j
[|T|=k−j]
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
# k X
j=0
(−1)k−j
n−k+ (k−j) k−j
| {z }
=
n−j k−j
[|T|=k−(k−j)]
| {z }
=[|T|=j]
(here we substitutedk−j for j in the second sum)
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
# k X
j=0
(−1)k−j
n−j k−j
[|T|=j]
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
# n X
j=0
(−1)k−j
n−j k−j
[|T|=j]
here we replaced the
k
P
j=0
sign by an
n
P
j=0
sign, since all addends forj > k are zero (ask−j <0 and therefore
n−j k−j
= 0 for j > k)
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
# |T|
X
j=|T|
(−1)k−j
n−j k−j
[|T|=j]
here we replaced the
n
P
j=0
sign by an
|T|
P
j=|T|
sign, since all addends forj 6=|T| are zero (as [|T|=j] = 0 for j 6=|T|) and since |T| ∈ {0,1, ..., n} (sinceT ⊆ {1,2, ..., n})
= X
T⊆{1,2,...,n}
"
X
t∈T
xt ≡0 modp
#
(−1)k−|T|
| {z }
=(−1)k·(−1)−|T|
=(−1)k·(1/(−1))|T|
=(−1)k·(−1)|T|
n− |T| k− |T|
[|T|=|T|]
| {z }
=1
= (−1)k X
T⊆{1,2,...,n}
(−1)|T|
"
X
t∈T
xt≡0 modp
#
n− |T| k− |T|
= (−1)k X
T⊆{1,2,...,n}
(−1)|T|
"
X
t∈T
xt≡0 modp
#
|{S ⊆ {1,2, ..., n} | |S|=k and T ⊆S}|
since
n− |T| k− |T|
=|{S ⊆ {1,2, ..., n} | |S|=k and T ⊆S}| by Lemma 6
= (−1)k X
T⊆{1,2,...,n}
(−1)|T|
"
X
t∈T
xt≡0 modp
#
X
S⊆{1,2,...,n}
[|S|=k and T ⊆S]
| {z }
=[|S|=k]·[T⊆S]
= (−1)k X
S⊆{1,2,...,n}
[|S|=k] X
T⊆{1,2,...,n}
[T ⊆S]·(−1)|T|
"
X
t∈T
xt ≡0 modp
#
| {z }
= P
T⊆{1,2,...,n}
such that T⊆S
(−1)|T|
"
P
t∈T
xt≡0 modp
# by (3)
= (−1)k X
S⊆{1,2,...,n}
[|S|=k] X
T⊆{1,2,...,n}
such that T⊆S
(−1)|T|
"
X
t∈T
xt≡0 modp
#
= (−1)k X
S⊆{1,2,...,n}
[|S|=k]X
T⊆S
(−1)|T|
"
X
t∈T
xt ≡0 modp
#
| {z }
= P
S⊆{1,2,...,n}
such that|S|=k P T⊆S
(−1)|T|
"
P
t∈T
xt≡0 modp
# by (3)
= (−1)k X
S⊆{1,2,...,n}
such that|S|=k
X
T⊆S
(−1)|T|
"
X
t∈T
xt≡0 modp
#
| {z }
≡0 modpsby (6), applied toat=xt
≡(−1)k X
S⊆{1,2,...,n}
such that|S|=k
0 = 0 modps.
Thus, we have proven
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt ≡0 modp )
≡0 modps. The same argument, with ≡0 modp replaced by 6≡0 modp and with (6) replaced by
(5), yields
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt 6≡0 modp )
≡0 modps. Thus, the problem is solved.
Remark. We can generalize Corollary 5:
Corollary 7. Let p be a prime number, and s a positive integer. Let k be an integer, and let S be a finite set such that |S|= k ≥ps−ps−1. For every i∈S,letai be an integer. Letc∈Z. Assume that (p, s, k)6= (2,2,2) and (s, k)6= (1, p−1). Then,
X
T⊆S
(−1)|T|
"
X
t∈T
at 6≡cmodp
#
≡0 modps (9)
and
X
T⊆S
(−1)|T|
"
X
t∈T
at≡cmodp
#
≡0 modps. (10)
Proof of Corollary 7. For every T ⊆S,we have X
t∈T
at−c
!ps−ps−1
≡
"
X
t∈T
at−c6≡0 modp
#
modps (by Lemma 4, applied to a= P
t∈T
at−c). Hence,
X
T⊆S
(−1)|T|
"
X
t∈T
at 6≡cmodp
#
=X
T⊆S
(−1)|T|
"
X
t∈T
at−c6≡0 modp
#
≡ X
T⊆S
(−1)|T| X
t∈T
at−c
!ps−ps−1
= X
T⊆S
(−1)|T|
ps−ps−1
X
u=0
ps−ps−1 u
X
t∈T
at
!u
(−c)ps−ps−1−u
since
P
t∈T
at−c
ps−ps−1
=
ps−ps−1
P
u=0
ps−ps−1 u
P
t∈T
at u
(−c)ps−ps−1−u by the binomial theorem
=
ps−ps−1
X
u=0
ps−ps−1 u
(−c)ps−ps−1−uX
T⊆S
(−1)|T| X
t∈T
at
!u
modps. (11)
Now, we will consider the case (p, s, k) = (2,2,3).In this case, Theorem 1 yields X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S|
for every u∈ {0,1, ..., ps−ps−1} (since u≤ps−ps−1 = 22−22−1 = 22−21 = 4−2 = 2<3 = k=|S|), so that
X
T⊆S
(−1)|T| X
t∈T
at
!u
=
( 0, if u <|S|; (−1)|S||S|!· Q
i∈S
ai, if u=|S| = 0 (sinceu <|S|)
≡0 modps for every u∈ {0,1, ..., ps−ps−1}.
Next, we will consider the case (p, s, k)6= (2,2,3). In this case, Corollary 3 yields X
T⊆S
(−1)|T| X
t∈T
at
!u
≡0 modps for every u∈ {0,1, ..., ps−ps−1} (sinceu≤ps−ps−1 ≤k).
Thus, in both cases (p, s, k) = (2,2,3) and (p, s, k) 6= (2,2,3), we have obtained the congruence
X
T⊆S
(−1)|T| X
t∈T
at
!u
≡0 modps. (12)
for every u ∈ {0,1, ..., ps−ps−1}. Thus, this congruence (12) always holds. Hence, (11) becomes
X
T⊆S
(−1)|T|
"
X
t∈T
at 6≡cmodp
#
=
ps−ps−1
X
u=0
ps−ps−1 u
(−c)ps−ps−1−uX
T⊆S
(−1)|T| X
t∈T
at
!u
| {z }
≡0 modps by (12)
≡0 modps, so that (9) is proven.
Besides, X
T⊆S
(−1)|T|
"
X
t∈T
at ≡cmodp
#
=X
T⊆S
(−1)|T| "
not X
t∈T
at 6≡cmodp
#!
=X
T⊆S
(−1)|T| 1−
"
X
t∈T
at6≡cmodp
#!
=X
T⊆S
(−1)|T|·1
| {z }
=0 by (8)
−X
T⊆S
(−1)|T|
"
X
t∈T
at 6≡cmodp
#
| {z }
≡0 modpsby (9)
≡0−0 = 0 modps.
This proves (10). Thus, Corollary 7 is proven.
This quickly entails a generalization of the problem:
Theorem 8. Let pbe a prime number, and s a positive integer. Let k be an integer, and let n be an integer such thatn ≥k ≥ps−ps−1. For every i∈ {1,2, ..., n}, letxi be an integer. Let c∈Z.
Assume that (p, s, k)6= (2,2,2) and (s, k)6= (1, p−1). Then,
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt≡cmodp )
≡0 modps and
k
X
j=0
(−1)j
n−k+j j
(
T ⊆ {1,2, ..., n} | |T|=k−j and X
t∈T
xt6≡cmodp )
≡0 modps.
Proof of Theorem 8. The proof of Theorem 8 is the same as our above solution of the problem, with 0 modp replaced by cmodp, and with (5) and (6) replaced by (9) and (10), respectively.
Note that a weaker version of Corollary 7 (namely, a version requiring k≥ps) was discussed in
http://www.mathlinks.ro/Forum/viewtopic.php?t=188350 .
