Internat. J. Math. & Math. Sci.
Vol. 8 No. 4 (1985) 799-804
799
ON k-TRIAD SEQUENCES
HANSRAJ GUPTA
Panjab University, Chandigarh, India 160014
and
K. SlNGH
Department of Mathematics and Statistics
University of New Brunswick, Fredericton, Canada E3B 5A3
(Received May 29, 1984 and in revised June 29, 1984)
At the conference of the Indian Mathematical Society held at Allahabad
ABSTRACT.
in December 1981, S. P. Mohanty and A. M. S. Ramasamy pointed out that the three
numbers
I, 2, 7, have the following property:
increased by 2 is a perfect square.
the product of any two of them
They then showed that there is no fourth
integer which hsares this property with all of them.
They used Pell’s equation
and the theory of quadratic residues to prove their statement.
In this paper,
we show that their statement holds for a very large set of triads and our proof
of the statement is very simple.
KEY WORDS AND PHPASES.
Pell’s equation, congruence, Fibonacci, sequence.
1980 AMS SUBJECT CLASSIFICATION CODE. 10A10.
i.
INTRODUCTION.
DEFINITION.
Given any integer k, three numbers a I,
a2,
a
3
are said to form
a k-triad, if the numbers
ala 2 +
ala 3 +
k,
k
and
a2a 3 +
k
(I.I)
are all perfect squares.
An ascending sequence of integers
a
I, a 2, a 3,
a
n
is said to be a k-triad sequence if every three consecutive elements of the sequence
form a k-triad.
Evidently, if (I.i) is a k-triad sequence, then
a2, a3,
a
4
a
n
is also a k-triad sequence.
This elementary statement will prove useful to us in our work.
(1.2)
H. GUPTA AND K. SINGH
800
2.
CONSTRUCTION OF SEQUENCES (I.i).
In what follows, small letters denote integers and c.’s are positive.
i
Given any integer k, we can choose two integers a
such that
ala 2 +
I
and
a2,
a
2
>
al,
k is a perfect square:
+
ala 2
k
c
2
(2.1)
I
The problem of constructing the sequence (I.I) then reduces to finding
a number a
3
such that both
ala 3 +
ala 3 +
Let
Set
x
k and
k
x
2
a2a 3 + k will
a2a 3 + k
+ Cl,
a
y
a
2
be squares.
y
2
(2.2)
+ cI
Then from (2.2) we have
(a
(a
so that
a
x)(y + x)
(y
al)a 3
2
a
3
+ 2c I +
a2);
+ a 2 + 2c I
I
We assert that this value of a
-al)(a I
2
(2.3)
actually satisfies our requirements.
3
In
fact, we have
ala 3 +
+ 2c I + a 2) + k
2
a +
2alc I + (ala 2 + k)
al(a I
k
a
2
+
(a
+ c 2I
2alc
+
ci)2
+
Cl )2
Similarly
a2a 3 +
Writing
c
c
2
2
2
(a
k
a2a 3 +
a
2
2
(2.4)
k, we have
+ cI
(2.5)
Notice that (2.3) provides a formula for writing the third element of a
k-triad sequence interms of
a
and c
al, 2
I.
Applying this procedure to (1.2), we get
a
4
a
a
2
2
a
+
a
3
+ 2c 2
+ (a I + a 2 + 2c I) + 2(a 2 +
+ 4a 2 + 4c
c
I)
(2.6)
Treating this as a formula for the fourth element of a k-triad sequence and
applying it to (1.2), we get
a
5
+ 4a 3 + 4c 2
4a + 9a + 12c
I
2
I
a
2
The process can be repeated until the desired number of
been obtained.
(2.7)
elements of (I.I)
has
k-TRIAD SEQUENCES
801
Then assuming that
aj
uja I + vja 2 + wjc I
aj
uja 2 + vja 3 + wjc 2
we obtain
uja 2 +
+
vja
+ a 2 + 2c I) + wj(a 2 + c I)
+ vj + wj)a 2 + (2vj + wj)c
vj(a I
(uj
This gives us the following recurrence relations:
v
uj+ I
v
j,
+
u.
j+
+
v.
Wo,
wj +I
2v. +w.
Identically:
uj+ I
0
vj+
I
wj+
0
0
u.
I
I
v
2
I
wj
(2.9)
or
uj
1
-I
uj+ 1
vj
0
0
jv
LWj+l
i
wj
(2.10)
+I
0
Using the same technique, from (2.5), we obtain
c
3
a
3
(a
a
+
c
2
a
2c I) + (a +
I + 2 +
2
+ 2a 2 + 3c
c
I)
In general,
c. =a.
+c.
(2.11)
j-I
Assuming that
rja I + sja 2 + tjc I
cj
there is no difficulty in obtaining the recurrence relations:
r
s
t
+I
+I
+I
0
I
I
I
0
2
0
Irl
s
(2.12)
it_
and
rj
1
sj
tj
2
1
-I
0
0
sj+
0
I
tj+
rj+ 1
While relations (2.9) and (2.12) enable us to find expressions for ao’S and
c.’s
as
takes value in ascending order of magnitude, the relations (2.10)
and (2.13) enable us to extend them in the opposite direction.
(2.13)
802
H. GUPTA AND K. SINGH
Our k-triad sequence can thus be defined for all integral values of the
subscript; therefore, the sequence of
c.’s is defined
for all integral
values of the subscript.
IDENTIFICATION OF THE COEFFICIENTS u, v, w AND r, s, t.
3.
We hardly would expect that the Fibonacci sequence has anything to do with
the coefficients u, v, w and r, s, t introduced in the preceding section.
But
the unexpected happens.
Recall that the Fibonacci sequence is defined by the recurrence relations:
I,
fl
0
f0
f
f
m
m-2
+
m > 2
fm-i
(3.1)
This definiton can be extended to negative integral subscripts by noting
that
f
(-i)
-m
m+l
f
(3.2)
m
From results (2.3), (2.6) and (2.7), it will be seen that for the values
3, 4 and 5;
f2 2 al
aj
+ f2.j-1
a2 + 2fj -2
f
j-1
c
I
so that for these values of j,
f2.j-2
uj
v
f2.j-I
2fj_2 fj-I
wj
From (2.9), we will now have
2
and
f2.j-2 + f2. -i + 2fj -2 fj -I
(fj-2 + fj-i )2 f2.
Uj+l
fj-I
Vj+l
wj+
2f
2.
+ 2f j-2
(fj-i +
j-I
2f.j_l
fj_ i
fj-2
2fj_l f’j
We now leave it to the reader to complete the induction and show that
f2j-2
aj
al +
f2.]-I a2 + 2fj -2 fj -i
c
i
(3.3)
for all integral values of j.
It is no more difficult to prove that for all integral values of
fj_2fj_lal
cj
4.
S! OF CONSECUTIVE
+
fj_ifja 2
+
(f2_j fj-2fj-I )el
(3.4)
a’s.
From (2.11), we have
cj
a.
cj_ 1
Hence we have an interesting summation formula
n
a.
j=m+l
c
n
c
m
n > m
This provides a good check on the values of
a.’s and c.’s.
A GENERALIZATION OF THE STATEMENT OF MOHANTY
AND RAMASAMY.
Our generalization can be stated in the form
of the following.
THEOREM. If al,
a2, a3 is a k-triad and k 2 (mod 4), then there is no
integer a for which
5.
(4.1)
k-TRIAD SEQUENCES
ala +
a2a +
k,
k,
a3a
803
+ k
are all perfect squares.
We need two lemmas for the proof of our theorem.
LEMMA I.
PROOF.
Only two of the three numbers a I, a 2, a 3 are odd.
If a and a 2 are both even, let
2
ala 2 + k c I
This implies that c I is even.
Modulo 4, we have
0 (mod 4)
2
cannot be even.
and a
Hence both a
This is impossible.
2
I
and a are of opposite parity, then since
2
a + a + 2c
a
I
2
If a
3
a
must be odd, and our lemma holds.
3
If a
and a are both odd, then a 3 is even.
2
This completes the proof of our lemma.
LEMMA 2. The difference of the two odd elements of the given k-triad
is congruent to 2 (mod 4).
First let a I and a 2 be the odd elements of the k-triad. Then
0 or 2 (mod 4).
a H 0 (mod 4) and
a cannot be congruent to 0 (mod 4). Suppose a 2
I
PROOF.
a
a
I
2
But a
2
let
I
a
Since
ala 2
+
As
a
2
2
c
k
I
I + 2
c
a
even.
must be odd.
I
al(a I
a
2
and a
3
I
Again since
a
Hence
a
Now
a
Hence
a
2
+ 4d)
a
E
I (mod 4)
I (mod 4).
Hence
This is impossible.
Next let a
c
ala 2
we will have
for some integer d.
+ 4d
I
2 (mod 4)
be the two odd elements of the k-triad. Then a is necessarily
2
2
ala 2 + k c I c I must be even. We must, therefore, have
+ 2
0 (mod 4)
2 (mod 4)
2
+ a 2 + 2c
a
3
a
3
The case in which a
2
I
and
is even.
c
2 (mod 4).
and a
3
are the odd elements, can be dealt with in the
same manner and the lemma is proven.
PROOF OF THE THEOREM.
Since the existence or non-existence of the number
a does not depend on the order in which the three expressions are
can assume that a
k-triad.
I
is the even and a
2
and a
3
taken, we
the odd elements of the given
804
H. GUPTA AND K. SINGH
For some positive integers x, y, z, let
ala +
k
a2a +
a3a +
x
k
y
k
z
2
(i)
2
(ii)
2
(iii)
From (i) it is evident that x is even.
Replace x by 2g, a
(which is even)
by 2h, and k (which is congruent to 2 (mod 4)) by 2q where q is odd.
(i) takes the form
(2g) 2
+ q
ha
Then,
(iv)
Since q is odd and the right-hand side is even, h and a must both be odd.
This implies that y and z are both odd.
Now substracting (ii) and (iii),
we have
(a
3
a2)a
z
2
y
2
0 (mod 4).
(v)
a
2 (mod 4), (v) implies that a is even. This contradicts
3
2
the earlier statement that a is odd. Hence a does not exis and we are
Since a
through.
-4
-3
-2
-I
0
i
2
3
4
5
6
-67
-25
-i0
-3
-i
2
5
15
38
I01
263
690
c.
41
16
6
3
2
4
9
24
62
163
426
1116
f.
-3
2
-I
I
0
i
2
3
5
8
13
u
v
64
25
9
4
i
I
I
0
i
I
4
9
25
25
9
4
i
I
0
i
I
4
9
25
64
w.
-80
-30
-12
-4
-2
0
0
2
4
12
30
80
r.
-40
-15
-6
-2
-I
0
0
i
2
6
15
40
s.
-15
-6
-2
-i
0
0
I
2
6
15
40
104
t.
49
19
7
3
i
i
i
3
7
19
49
129
2,
a
a.
k
ACKNOWLEDGEMENT.
6,
a
I
2
5,
c
I
4
The second author was partially funded by a travel grant from the
University of New Brunswick.
REFERENCE
I.
ANDREWS, George E.
Number Theory
W. B. Saunders Company, Philadelphia, 1971.