Internat. J. Math. & Math. Sci.
VOL. II NO. 4 (1988) 743-750
743
SOME CONGRUENCE PROPERTIES OF BINOMIAL COEFFICIENTS
AND LINEAR SECOND ORDER RECURRENCES
NEVILLE ROBBINS
Department of athematlcs
San Francisco State University
.San Francisco, CA 94132
(Received May 21, 1987)
ABSTRACT.
-
0
prime,
Using elementary methods, the following results are obtalned:(1) If p is
<
n, 0
m
b
If r,s are the roots of
n
u
n
v
r-s
n
u
k2
n
x
ap
2
rn+s n,
n
(III) If p is odd and p
(IV)
<
(_1)Bu
k2
KEYWORDS AND PHRASES.
n
n-m,
(m)
ab, then
and p
Ax-B, where (A,B)
and k
D,
and
> O, then (II)
u
then
kp
(_D)
n
P
v
kpn
u
kp
n-I
D
n-m
(-I)P-I(ap b
A2_4B >
=- vkpn-1
(rood
(rood
pn).
0, if
(rood
pn).
pn);
n)
n-I (rood 2
Binomial coefficient, linear second order recurrence.
1980 AMS SUBJECT CLASSIFICATION CODE: lOAl0k IOA35.
I.
INTRODUCTION.
Following Lucas [I], let A, B be integers such that (A,B)
D--
Let
A2-4B >
n
0.
O.
Let the roots of
Let the sequences
u
Ax-B
x
u
n
v
n
be:
1/2(
r
),
v
).
(1.1)
r-s
n
s--I/2 (A-
be defined by:
n n
r -s
n
and
rn+sn
(1.2)
then
u
v
0
o
0
u
2, v
r+s=A
u
A,
v
n
n
Au
n-l-
for n
2
Av
n-I -Bv n-2 for n
2
Bu
n-2
(1.4)
744
N. ROBBINS
B
rs
Let p be prime.
<
If 0
0
If
a
<
k
p
a
If
x
If
0
v
(1.8)
2 _2B n
(I .9)
2Um+n UmVn + UnVm
(I.I0)
n
O (t)
p
PJ
j if
p 3+
t
t.
(1.11)
n
<
n
(n-k
(k)
(1.12)
pn)
(rood
k
V2n
n, then
(1.13)
2
2n),
(rood
+/- a
<
UnV n
Let
n-I
p
U2n
p, then
k
n
(1.6)
then x
a
2
(rood
2n+l
(1.14)
e
(I .5)
e
pe
and
pk,
then
Op((Pk))
(I.I) through (I.I0) appear in Lucas [I].
(1.15) is theorem 2 in Robblns [2].
REMARK.
elementary.
(I.II) through (1.14) are
MAI N RE SULTS.
2.
Lemma (2 I), If
n
(aP b
n
(pPm)
PROOF.
all
such
Now
factors
P1
m
i--O
m
bpm-ip
J
j
-I
where
a___m
n
a_
b(pm-l)
<
b
(mod
apn-j
bpm-j
j--0
b-1
bpm-I
P2
(-I)
bp
n, 0
m
n-m
(m)
then
0
pmlj,
b-I
|I
i=O
(_l)bpm_b
and
P2
_apn-m_i
b-i
(rood P n
n
(a
sn-m
(-I) b(
pro_ I)
(rood
pn).
ap
n-m
and p
a b,
pn).
PIP2
jip
aPm)
(bp
<
where
is
ap
P1
the
is the product of
product
n-m
b
)’
Therefore
while
of
all
such
factors
where
SOME CONGRUENCE PROPERTIES OF BINOMIAL COEFFICIENTS
THEOREM 2.1.
n
m
<
n, 0
<
(rood
pn).
b
ap
n-m
PROOF.
If
p
odd,
is
-
b(pm-l)
then
b(2m-l)
hypothesis, b is odd, so
LEMMA 2.2.
2n+l
A
nl
2n+l
A2n+l
k=l
LEMMA 2.3.
PROOF.
PROOF.
2n+l-k k
s
r
so (1.5) implies
2n+l)
k
P
vi+j
n, y
wy
PROOF.
so (1.2) and (1.6) imply
-
(rood p)
v
)
i
v
(rood p); (I.13) implles
p-- Ap
(rood p).
v
P
J, then
vi+j
Ap
-
A
(rood
BJv
viv j
(ri+si)(rJ+sJ)-(ri+J+si+J)
r
(rs) J (rl-J+s
isJ+r j s i
If
z (rood
xz (rood
pro),
pn).
w
x (mod
LEMMA 2.6. If
k
O, m )
and v
P
m
pn),
-
Hypothesis implies
Hypothesis also implies
PROOF.
(rs) k(r2n+l-2k+s2n+I-2k ),
Now (1.12) implies
(2n+l)Bkv2
k
n+l-2k
v
If
2n+I
k
(2n+l_k)r s 2n+l-kl..
By (1.2) and (1.6),
LEMMA 2.5.
k=l.
k
k
Lemma 2.2 and (I.II) imply
LEMMA 2.4.
then
v
(1.4) implies
m
[
2n+l 2n+l-k k
)r
s +
k
k=l
k=l
V2n+l
0
B
2n+I
2n
(r+s) 2n+l
k=l
viv j
(2n+l
k=l
+ s 2n+l
A2n+l-{
V2n+l
too,
(rood 2); if p
2, then by
In either case, the conclusion now
0
k=
V2n+
P);
ab,
By (1.2) and the binomial theorem, we have
PROOF.
r
A2n+l
V2n+!
p-!
2-1 (rood 2).
follows from Lemma 2.1.
V2n+l
and p
n-m
(-I)P-I (aP b
(aPm)
bp
then
0
If
745
Y
pn-mlx,
v
P
so wy
m-I (rood
(Induction on k).
z+ip
pm ),
m,
and x
w
xz (rood
0 (rood
x+jp
n
pn-m),
so wy E
xz+ipmx
(rood p n ).
pn).
then v
m
kp
---v
kp
m-1
(mod
pm
Lemma 6 holds trivially for k=O, and by hypothesis
for
N. ROBBINS
746
m
m
v
(k+!)p
m
=v
kpm+p
=v
m
-BP
v
m m
kp p
v
=v
m tn
kp -p
kp
v
m
p
B
m
v
(k-l)p
m
by Lemma 2.4.
Now induction hypothesis and (1.13) imply that
v
(k+
)pro
(k+
)pro
v
=v
kp
v
m-
v
(k+
LEMMA 2 7.
p
-B p
m-1
v
(k-I )p
m-1
pro).
(rood
pm)
(mod
)pro-
m-1
If p is odd and
I, then
n
v
P
<
2.7 holds for all m
v
p
n-I
p
I/2 (pn-l-l)
n-I
I/2 (pn- I)
n
n
j=O
(P.)
then (I.15) implies
Therefore
v
p
m,
hp
ip
m
Ap
n
where
n
B
i
n
n
n-I
)Biv
(P)BJVpn_
>
n-
p
also
I-2i
2]
and m
<
pm),
(pp)BiPv p n-2 ip
Now
n.
by (1.13); also v
i
Bip v
Therefore
n-I
hPpm) -= (Phpm-l)
_=
P
B v
n-I
i
p
p
p n_ 2 ip
=- v p n-I -2i
(pn-m_2h)pm-
pn-2ip
p
p
i
V
(pn-m_2h)pm
and Lemma 2.6.
(p)
i
n-m
h
n-1
(rood
(rood
pro)
by induction hypothesis
pm
Also
by
Theorem
-2i
(rood p n)
implies
(P
h
n
Suppose Lemma
@ 0 (rood p
I/2 (pn-1-1)
(rood
V
pn_2hpm
J
j=O
ph
m-I
BiP BhP BhP
V
(rood p n)
n
p j,
Let
n-I
P
Then Lemma 2.2 implies
2.
n
i=O
Ap
v
If
n, where
Ap
-
v
Lemma 2.7 holds for n=1, by Lemma 2.3.
(Induction on n)
PROOF.
n
Now Lemma 2.4 implies
0 (rood p n-m)
(p)BiPvp n-2ip
Therefore Lemma 2 5 implies
n-I
(P i )Biv
p n-l_2 i
(mod p n)
Now (I 13) implles
2 I,
and
(I.15)
SOME CONGRUENCE PROPERTIES OF BINOMIAL COEFFICIENTS
v
p
v
P
Ap
n
.
1/2 (pn-l-l)
n-1
i
n(rood p n)
(p i )B v n-I
-2i
p
i=O
n
E
v
P
LEMMA 2.8.
If
21A,
then
v
2
(Induction on n)
PROOF.
2
2
=-
n-I
4 (mod 2
v
implies
2
v
n
n+l)
2
2
LEMMA 2.9.
n-1
If
v
-2B
2
2
(rood
2n-I
v
2
v
n
2
2
n-I
4-2(I)
2AB,
2n+l).-
_2B
LEMMA 2.10.
2 (rood
2n-I
then
2n-1 =-
v
2
2n),
so
B
so
-I (rood 2
A
v
=-I (rood
-I (rood
n+l)
2n),
B2
Again, B odd implies
I-2(I)
Now
(1.14) implies
B
2
n-I
(rood 2 n)
Now (1.9)
0.
for n
by (1.4) and hypothesis.
-I (rood 2)
v
Now
so (1.14) implies
n-I
-=
2n),
(rood
so (1.9) implies
2n+lj.
(rood 2
B, then v
If 2
=-
v20
2n-I
by (1.4) and hypothesis.
2n+l).
2 (rood
n
0.
for n
2 (rood 2
A
v
n-I
induction hypothesis implies
v
2n+l)
Hypothesis implies 2
(Induction on n)
PROOF.
2 (rood
n
v20
induction hypothesis implies
v
that is
pn).
(mod
n-I
747
n+l)-
for n
O.
2
PROOF.
v 0
(rood 2).
A
v
Hypothesis implies A is odd, so (1.4) implies
(Induction on n)
By hypothesis,
B
0
(rood 2), so
2
B
2n-1 -=
0 (rood 2
2
n-1
Since
induction hypothesis,
Now (1.9) implies
v
2
LEMMA 2.11.
PROOF.
where t
V2n_1
v
n
2
n-I
2
-2B
n-I
2
2
v
V2n 2n-I
2n),
(rood
Therefore
so (1.14) implies
v
1-2(0)
=-
(rood
2
2n_l
0 (rood 2 n)
(mod
2n+l).
2n+l)."
(mod 2 n)
Lemmas 2.8, 2.9, 2.10 imply
2 or +/-I.
B
n-I
E
2n-I
we have
n for n
V2n
t
v
2n-I
V2n_l
t (rood
(rood
2n).
2n),
V2n
E t
(rood
2n+l),
By
748
N. ROBBINS
THEOREM 2.2.
In
n
I, then v
P
PROOF.
PROOF.
PROOF.
2n
DI/2 Umn+l
D
and k
0, then v
n
kp
2n+
Dn
k=l
v
kp
pn)
n-I (rood
k
(-B)U2n+l_2 k
k
r
U2n+l
k
(_l)k(2n+l)r2n+l-k
s
k
.
2n
+
J=n+
2n+l-k in the second sum, we obtain
U2n+
I/2-
.
D
n+I/2_{
n
(-l)k(rs)k(2 I)
2n+l
k
.
k=l
n
Dn
k=l
LEMMA 2.13.
t
DI/2(pn)E
(t+ipn) p
2n+l-2k
k
(-B) U2n+l_2
k
2n+
s
.
2n+l-j sj
(-l)J( 2n+1)r
J
(-i)
k=
2n+l-2k
2n+l
2n-k( 2n+l-k
r
k 2n+l-k
s
by (1.12)
by (1.6) and (I.7), so
k
(-B) u
2n+ l-2k
If p is odd, then
u
P
=-
()
(rood p)
P
Follows from Lemma 2.12, (I.II), and Euler’s criterion.
LEMMA 2.14.
PROOF.
(2n+lk
(r
s
n
2n+l-k k
(2n+1)r
s
k
k
k=l
Dn/2-
PROOF.
(-I)
2n+
so (1.7) implies
k
I/2-{ n (-l)k(2n+l)r2n+l-k
s
k
n+
k=l
P
P
k=l
Setting j
()
n
U2n+l
k=l
U2n+l
If
(i.i) and (1.7) imply
2n+l
n+
Pn)-
(rood
v
Fol-los from Theorem 2.2 and Lemma 2.6.
LEMMA 2.12.
D
=-
Follows from Lemmas 2.7 and 2.11.
THEOREM 2.3.
I/9
n
If p is odd,
(Induction on n)
pD,
and
I, then DI/2 (pn)--
n
(D)
t (rood
t
pn),
pn).
Lemma 2.14 holds for n=l by Euler’s criterion.
Now induction hypothesis implies
11.
(mod
.
I/2(pn)
so D
p-I
p +
(ipn) +
pt
P
j--2
t+ip
n.
D
I/2(P n+ )=
(.P) tP-J(ipn) j.
3
(D
Now t p
i/2 (p n) )P
t
Let
SOME CONGRUENCE PROPERTIES OF BINOMIAL COEFFICIENTS
pn+llpnj
and
for j
LEMMA 2.15.
u
P
tu
m
P
k
If
2u
2U(k+ l)pm
kP
tu
kpm
u
kpm+ pm
kpm-
Theorem 2.2 implies
kp
pm ),
(rood
E tu
m
v
kP
2u( k+1)p m
tu
kp
m-I
p is odd, we have
LEMMA 2.16.
(),
pm).
(mod
kpm-
and
Lemma 2.15 is trivially true for k=O, and is true by
Now (I.I0) implies
hypothesis for k=l.
u
then u
(Induction on k)
PROOF.
pn+l).
(rood
0, m > I, p is odd,
)
pm ),
(rood
m-i
Dl/2(pn+l )--
2, so
)
v
p
p
p
kp
Upm --tUpm_l
v
(rood
kpm+ tu
p
-=
U(k+l)p m
m-I
tu
kP
If p is odd,
(rood
pm
pro);
O.
for k
9tu
v
kpm-I
m
By induction hypothesis,
m
and
m
m-I
+ u v
m
v
m m
kp
Therefore
pro).
(rood
m
(Induction on n)
Lemma 2.12 implies
u
p
pD,
n
and
)
-= (--D)u
p
I, then u n
Lemma 2.16 is true for n
DI/2 (pn-l_l) I/2 (p
n-I
Since
(mod p
P
PROOF.
749
n-I -I)
n-I
(mod
pn).
by Lemma 2.13.
n-I
(pff
i
P
(-B) i u
p n_l_ 2 i
n
(p)
u
p
Let
Now
-
Pn
(PJ) (-B)jupn-2 j.
j
0 (rood
pn).
I/2 (pn-I) I/2 (pn-I-I)
)-.
i=
m
hp
ip
ph
where
(-B) ip s (-B) hp
pj,
then (1.15) implies
Therefore
D
n
If
m
(-B
(pnp)
and m
(-B)
<
)hpm-
n.
iPu
p n-2 ip
Let
t
(-B) i (rood
(rood
pn).
().
pro)
by (1.13).
By induction
hypothesis and Lemma 2.15, we have
E
u
pn-2ip
u
pn-2hpm
E u
(pn-m-2h) pm
=-
tu
(pn-m-2h) p m-I
E tu
p
n-I
-2hp
m-I
N. ROBBINS
750
tu
(rood
p n_l_ 2 i
pro)
(-B)iPu
Therefore
t(-B)iu
pn-2ip
p n-l_2 i
(mod
pro).
As in the proof of Lemma 2.7, we have
t( p
n
(-B)iPu
(P)
pn-2ip
ip
u
pn
u
P
E
(-B)i u
p
(pn-l-l)
-t(D1/2
I/2 (pn-l)
D
n-!
-u
;
tu
n
P
(pn)
n
(rood
n-I
n-I
Lemma 2.14 implies
P -P
u
P
LEMMA 2.17.
PROOF.
If
v
pn).
pn)
that is
)u
n
P
2n-1
-=
THEOREM 2.4.
(rood
n-I
pn)
(-l)BU2n_l
u2 =n
(rood 2 n).
(rood
u2
2n).
If
(D)
P
(-I)
(-I)B
2n-I
(-l)Bu2n_l
n
-=
and p
n
If B
-I (rood 2n); if B is even then Lemma 2.10
n-I
2
(rood 2 n).
Therefore, in either case, v
v
(mod 2 n).
D, then u n
tu
P
P
n-l
(rood
pn),
where
if p is odd
B
2
if p
Follows from Lemmas 2.16 and 2.17.
THEOREM 2.5.
where
Since
By hypothesis and by the definitions of A, B, and D, A must be odd.
Now (1.8) implies
PROOF.
Therefore
pn)
(rood
and D is odd, then
n
is odd, then Lemma 2.9 implies
implies
(rood
DI/2 (pr-l_l) (DI/2 (pn_pn-l)_t)
+
n-I
n-I -2i
If
k
0, n > I, and
)
pD,
then u
n
kp
-=
tu
kp
n)
n-I (mod p
is defined as in Theorem 2.4.
PROOF.
Follows from Lemma 2.15 and Theorem 2.4.
Let
Concluding Remarks.
2.3 and 2 5, the sequences
v
kp
be defined as in Theorem 2.4 as a result of Theorems
n
0.
determine p-adic integers for each k
t u
n
n
kp
REFERENCE S
I.
LUCAS, E.
Theorie
Math.
2.
ROBBI NS, N.
des
fonctions
numeriques
simplement
periodiques,
Amer.
(187 7) 184-240; 289-321.
On the number of binomial coefficients which are divisible by their
row number, Canad. Math. Bull. 25 (3) 1982, 363-365.
J.