G RINSTEAD ’ S M ETHOD F OR S OLVING
S IMULTANEOUS D IOPHANTINE E QUATIONS
S ANDA B UJA ČI Ć , D EPARTMENT OF M ATHEMATICS , U NIVERSITY OF Z AGREB , C ROATIA
I NTRODUCTION
A PPLICATION OF THE T HEOREM (B AKER , D AVENPORT )
We define triangular, square and centered hexagonal numbers as the numbers given by the formulas:
Also, P = Q + Q−1 + 12 P −1 < Q + 32 Q−1 < Q + 32 , so Q > P − 32 . We may suppose that l ≥ 3, so
1
2
xn = n(n + 1), yn = n , zn = 1 + 3n(n − 1),
2
respectively. The centered hexagonal numbers
arise geometrically by starting with one point,
putting six points around the first point to form
a hexagon and continuing to build outward with
12 points, 18 points,. . . .
P ≥
√ 3
(3+2 2)
√
2
> 125. Therefore,
P
61 −1
P , 0 < log
= − log
P −Q <
40
Q
P −Q
1−
P
2
2 61 −2
61 −2
61 −2
61
1
−2
<
P +
P
<
P
1+
< 1.526P .
40
40
40
40 125
If we substitute for P, Q in the last inequality, we obtain:
√
√
√
√ −2l
−2
0 < log P − log Q = l log(3 + 2 2) − log 2 − (2r − 1) log(2 + 3) < 1.526P
< 3.052(3 + 2 2) .
Theorem 1 Suppose that k ≥ 2 and α1 , . . . , αk are non-zero algebraic numbers, whose degrees do not
exceed d and whose heights do not exceed A, where d ≥ 4, A ≥ 4. If the rational integers b1 , . . . , bk
satisfy
−δH
0 < |b1 log α1 + · · · + bk log αk | < e
,
where 0 < δ ≤ 1 and H = max{|b1 |, . . . |bk |}, then
k2 −1 2k
H < ((4 δ
Are there any integers, besides 1, which are
of all three of the above types?
We will show that if any other solution exists,
10455
it must be less than 4
. To do this, we follow the proof of theorem by Baker and Davenport.
We wish to solve the following system of
Diophantine equations:
d
2
2
√
√ 4r+2
√ 2l
3 2r+1
2.27
1
(3 + 2 2)2l
4r+2
2r+1
2r+1
> (2 + 3)
⇒ (3 + 2 2) > 3e
< 2r+1 .
>e
> e
> 2.27e
⇒
√
2l
2
2
e
(3 + 2 2)
The equations satisfied by α1 , α2 , α3 are α12 − 6α1 + 1 = 0, α22 − 4α2 + 1 = 0, α32 − 2 = 0. Hence, the
maximum height of α1 , α2 , α3 is A = 6 and d = 4. The theorem gives:
2r + 1 = H < (4 4 log 6)
k + k = 2m , 3n − 3n + 1 = m .
So, m <
The set of solutions of the first equation form a
second order linear recurrent sequence with
√ l
√ l
(3 + 2 2) − (3 − 2 2)
√
ml =
, l ≥ 1.
4 2
The corresponding recurrent sequence is given by:
m1 = 1, m2 = 6, ml = 6ml−1 − ml−2 , l ≥ 3.
Similarly, the solutions of the second equation are
given by:
√ 2r−1
√ 2r−1
(2 + 3)
+ (2 − 3)
0
mr =
, r ≥ 1,
4
and:
0
m1
= 1,
0
m2
= 13,
0
mr
=
0
14mr−1
−
0
mr−2 ,
r ≥ 3.
We wish to find l, r such that ml = m0r . Letting
this common value be m, we have:
√ l
√ −l
(3 + 2 2)
(3 + 2 2)
√
√
4m =
−
=
2
2
√ 2r−1
√ −(2r−1)
= (2 + 3)
+ (2 + 3)
.
(1)
Let
√ l
√ 2r−1
(3 + 2 2)
√
P =
, Q = (2 + 3)
.
2
Now, the last equation (1) becomes
1 −1
−1
P− P
=Q+Q .
2
Since P −1 > 0, Q−1 > 0, it follows P > Q.
).
√
√
√
We see from inequality above the theorem that k = 3, α1 = 3 + 2 2, α2 = 2 + 3, α3 = 2, H =
max{l, 2r − 1} = 2r − 1. Since P > Q, we have P 2 > Q2 , so
2
S OLUTIONS
log A)
2
9 6
2
(2k+1)
Q
4
+ 1 < (2 +
√
10 455
3)
49
<4
735
49
(1.7918)
455
< 10
.
.
G RINSTEAD ’ S M ETHOD 1/2
0
mr .
We wish to find l i r such that ml =
With the
0
aid of computer we are able to show that ml = mr
implies l ≡ 1 (mod p), Q
for all primes p < 1095.
Therefore, l ≡ 1 (mod p<1095 p). Thus, either
Q
460
l = 1 or l >
. But if l = 1,
p<1095 p > 10
460
10460
then ml = 1 and if l > 10 , then ml > 5
>
10455
4
, so ml is outside the range of solutions.
Given a prime p < 1095, we will assume that the
statement ml = m0r implies l ≡ 1 (mod q) for all
primes q < p. We assume that l ≡ 1 (mod 24 ·
32 · 52 ). We define L(q) as the lenght of the period
of the linear reccurence defining ml (mod q). We
generate a list of primes q such that
1. p|L(q), p2 does not divide L(q)
2. no prime larger than p divides L(q)
3. L(q) has no multiple prime factors except
possibly 22 , 23 , 24 , 32 , 52
0
the two sequences {ml } and {mr } are gene(mod q). We know that ml ≡ m0r (mod q)
Now,
rated
for any solution ml and l ≡ 1 (mod L(q)/p). If we
write the sequence {ml } (mod q), we only have p
possible positions for solutions, namely those positions ≡ 1 (mod L(q)/p). If any number in these
positions does not occur in the sequence {m0r }
(mod q), that position may be ruled out. If we
have not ruled out all remaining positions, except
the first position (which will never be ruled out
because ml = 1 is a solution), then we record
which positions have not been ruled out (mod p)
and move to the next q in our list.
G RINSTEAD ’ S M ETHOD 2/2
When this was done by a computer, in every case,
l was shown to be 1 (mod q) by using at most
eight q 0 s.
We will show that l ≡ 1 (mod 5). We assume that
l ≡ 1 (mod 24 32 ). Two values of q which we use
are q = 19, 29. We have L(19) = 20, L(29) = 10.
The sequence {ml } (mod 19) is:
{1, 6, 16, 14, 11, 14, 16, 6, 1, 0, 18, 13, 3, 5, 8, 5, 3, 13, 18, 0}.
Since we know that l ≡ 1 (mod 4), we only look
at every fourth term, starting with the first one.
The sequence {m0r } (mod 19) is {1, 13, 10, 13}.
By comparing sequences, we see that l ≡ 1, 9
(mod 20) are the only possible positions for solutions, so l ≡ 1, 4 (mod 5). We rule out the possibility that l ≡ 4 (mod 5). The sequence {ml }
(mod 29) is:
{1, 6, 6, 8, 0, 28, 23, 28, 0, 1}.
Sequence
0
{mr }
(mod 29) is
{1, 13, 7, 27, 23, 5, 18, 15, 18, 5, 23, 27, 7, 13, 1}.
Since the number 0 is in the position 9 (mod 10)
in the first sequence, and it does not occur in the
second sequence, we have l ≡ 1 (mod 5).
R EFERENCES
[1] Charles M. Grinstead: Solving A Class Of Diophantine Equations, Mathematics Of Computation, Vol
32, Num.143, 936-940
[2] Ezra Brown: Sets in Which xy + k is Always a Square, Mathematics Of Computation, Vol 45, Num.172,
613-620