Abstract.We say that a regular graph G of order n and degree r ≥ 1 (which
is not the complete graph) is strongly regular if there exists non-negative
integers τ and θ such that |Si ∩ Sj| = τ for any two adjacent vertices i
and j, and |Si ∩ Sj| = θ for any two distinct non-adjacent vertices i and j,
where Sk denotes the neighbourhood of the vertex k. We describe here the
parameters n,r,τ and θ for strongly regular graphs of order 7(2p+1), where
2p + 1 is a prime number.
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JOURNAL OF SCIENCE OF HNUE
Natural Sci., 2010, Vol. 55, No. 6, pp. 113-127
ON STRONGLY REGULAR GRAPHS OF ORDER n = 7(2p+ 1)
WHERE 2p+ 1 IS A PRIME NUMBER
Vu Dinh Hoa(∗)
Hanoi National University of Education
Do Minh Tuan
Nam Dinh Teacher Training College
(∗)E-mail: hoavd@fpt.com.vn
Abstract.We say that a regular graph G of order n and degree r ≥ 1 (which
is not the complete graph) is strongly regular if there exists non-negative
integers τ and θ such that |Si ∩ Sj| = τ for any two adjacent vertices i
and j, and |Si ∩ Sj| = θ for any two distinct non-adjacent vertices i and j,
where Sk denotes the neighbourhood of the vertex k. We describe here the
parameters n, r, τ and θ for strongly regular graphs of order 7(2p+1), where
2p+ 1 is a prime number.
Keywords: Strongly regular graph, conference graph, integral graph.
1. Introduction
1.1. Some results of strongly regular graphs
In recent years, determining the parameter of strongly regular graph is a im-
portant problem of the theory graph. However, when the number of vertices of a
strongly regular graph (called n) is any number, determining its parameter is very
difficult and nobody can solve this problem. Therefore, we have some results in
special cases of n:
Due to Theorem 1.1 we have recently obtained the following result [2]:
(i) there is no strongly regular graph of order 4p+ 3 if 4p+ 3 is a prime number.
(ii) the only strongly regular graph of order 4p+ 1 are conference graphs if 4p+ 1
is a prime number. Besides [2-5], we have decribed the parameters n, r, τ and
θ for strongly regular graphs of order 2(2p+ 1), 3(2p+ 1), 4(2p+ 1), 5(2p+ 1)
and 6(2p+ 1), where 2p+ 1 is the prime number.
In this article, We now proceed to establish the parameters of strongly regular
graphs of order 7(2p+ 1), where 2p+ 1 is a prime number, as follows.
In order to solve the problem. First, we have some results about relations of
the parameters of strongly regular graphs:
113
Vu Dinh Hoa and Do Minh Tuan
(i) A strongly regular graph G of order 4n+ 1 and degree r = 2n with τ = n− 1
and θ = n is called the conference graph.
(ii) A strongly regular graph is the conference graph if and only if m2 = m3 and
(iii) If m2 6= m3 then G is an integral graph.
(iv) If G is disconnected strongly regular graph of degree r then G = mKr+1, where
mH denotes the m−fold union of the graph H .
(v) G is a disconnected strongly regular graph if only if θ = 0.
1.2. Definitions
Definition 1.1. A regular graph is a graph where each vertex has the same number
of neighbours, i.e. every vertex has the same degree or valency. A regular graph with
vertices of degree k is called a k-regular graph or regular graph of degree k.
Definition 1.2. A regular graph G of order n and degree r ≥ 1 (which is not the
complete graph) is strongly regular if there exist non-negative integers τ and θ such
that |Si ∩ Sj| = τ for any two adjacent vertices i and j, and |Si ∩ Sj| = θ for any
two distinct non-adjacent vertices i and j, where Sk denotes the neighbourhood of
the vertex k. Set of strongly regular graph with parameters (n, r, τ, θ) is denoted
SRG(n, r, τ, θ)
1.3. Property of strongly regular graph
Let G be a simple graph of order n. The spectrum of G consists of the
eigenvalues λ1 ≥ λ2 ≥ · · · ≥ λn of its (0, 1) adjacent matrix A (0, 1) and denoted
by σ(G). We say that a regular connected graph G is strongly regular if and only if
it has exactly three distinct eigenvalues [1]. Let λ1 = r, λ2, λ3 denote the distinct
eigenvalues of G. Let m1 = 1, m2, m3 denote the multiplicity of r, λ2 and λ3,
respectively.
Theorem 1.1 (Lepovic [2]). Let G be a connected strongly regular graph of order n
and degree r. Then m2m3δ
2 = n.r.r where δ = λ2 − λ3 and r = (n− 1)− r.
Remark 1: Let r = (n−1)− r, λ2 = −λ3−1 and λ3 = −λ2−1 denote the distinct
eigenvalues of the strongly regular graph G, where G denotes the complement of G.
Then τ = n− 2r − 2 + θ and θ = n− 2r + τ where τ = τ(G) and θ = θ(G).
If G ∈ SRG(n, r, τ, θ) then G ∈ SRG(n, n− 1− r, n− 2r− 2 + θ, n− 2r+ τ).
Proposition 1.1 (Elzinga [1]). Let G is a connected or disconnected strongly regular
graph of order n and degree r. Then
r2 − (τ − θ + 1)r − (n− 1)θ = 0 (1.1)
114
On strongly regular graphs of order n = 7(2p + 1) where 2p+ 1 is a prime number
Proposition 1.2 (Elzinga [1]). Let G be a connected strongly regular graph of order
n and degree r. Then
2r + (τ − θ)(m2 +m3) + δ(m2 −m3) = 0, (1.2)
δ = λ2 − λ3 (1.3)
Remark 2: In [1], we have some formula:
τ − θ = λ2 + λ3 (1.4)
δ2 = (τ − θ)2 + 4(r − θ) (1.5)
In order to solve the problem when n = 7(2p+1), we use these formulas to con-
truct an equation which contains parameters of strongly regular graph (Diophantine
Equation). Then, we can solve this equation by tools of number theory.
2. Main result
Theorem 2.1. Let G be a connected strongly regular graph of order 7(2p+1) and de-
gree r, where 2p+1 is a prime number. Then G in one of 144 classes in Parameter
Table (Section 4.)
Lemma 2.1. Let G be a connected strongly regular graph of order 7(2p + 1) and
degree r, where 2p+ 1 is a prime number. If 2p+ 1|δ then G belongs to the class 1
in Parameter Table (Section 4.).
Lemma 2.2. Let G be a connected strongly regular graph of order 7(2p + 1) and
degree r, where 2p+ 1 is a prime number. If 2p+ 1|m2 then G belongs to class 2 to
class 72 represented in Parameter Table (Section 4.).
Lemma 2.3. Let G be a connected strongly regular graph of order 7(2p + 1) and
degree r, where 2p + 1 is a prime number. If 2p + 1|m2 then G belongs to class 73
to class 144 in Parameter Table (Section 4.).
3. Proof
3.1. Proof of Lemma 2.1
2p+ 1|δ then δ = k.(2p+ 1), where 1 ≤ k ≤ 6. Using Theorem 1.1 we have:
m2.m3.k
2.(2p+ 1)2 = 7.(2p+ 1).r.r ⇔ m2.m3.k2.(2p+ 1) = 7.r.r
which means that 2p+ 1|r or 2p+ 1|r or 2p+ 1|7
+) Case 1. 2p+1|7 then 2p+1 = 7. n = 21 = 3.7 we have sovled this problem
with n = 3(2p+ 1) in [3]
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Vu Dinh Hoa and Do Minh Tuan
+) Case 2. 2p + 1|r or 2p + 1|r. Without loss of generality we may consider
only the case when 2p+ 1|r.
Therefore r = l.(2p + 1) with 1 ≤ l ≤ 6 ⇒ r = 14p + 6 − l.(2p + 1) =
(14− 2l)p+ 6− l
⇒ m2.m3 =
7l(2p+ 1).[(14− 2l)p+ 6− l]
k2(2p+ 1)
=
7l(14− 2l)p + 7l(6− l)
k2
.
We have (m2 − 1).(m3 − 1) + 1 > 0 hence m2m3 > m2 +m3 − 2
⇔ 7l(14− 2l)p + 7l(6− l)
k2
> 14p+ 4
⇔ 14[(l − 7
2
)2 + k2 − 49
4
].p + 7(l − 3)2 + (4k2 − 63) < 0 (∗) If k ≥ 4 then it
doesn’t satisfy the condition (∗). So that k = 1; 2; 3.
we have m2 +m3 = 14p+ 6.
1) k = 1. We have m2.m3 = 7l(14− 2l)p+ 7l(6− l).
Let ∆2 =
(
m2 +m3
2
)2
−m2m3
a) l = 1⇒ δ = 2p+ 1, r = 2p+ 1. m2.m3 = 84p+ 35
m2, m3 is two roots of the equation: X
2 − 2(7p+ 3)X + 84p+ 35 = 0.
∆2 = (7p+ 3)2 − (84p+ 35) = 49p2 − 42p− 26. We can easily see that ∆2 is
a perfect square only for p = 3 and ∆ = 17. Otherwise, m2,3 = 7p + 3 ± ∆,
so we have two cases:
+) m2 = 41, m3 = 7. Using (1.2), we obtain 48(τ − θ) + 828 = 0 ⇒ τ − θ =
−69
4
. A contradiction because τ, θ is integers.
+)m2 = 7, m3 = 41. Using (1.2) , we obtain: 48(τ−θ)−736 = 0⇒ τ−θ =
46
3
.
A contradiction because τ, θ is integers.
b) l = 2⇒ δ = 2p+ 1, r = 2(2p+ 1).
m2.m3 = 140p+ 56. We have ∆
2 = 49p2 − 98p− 47.
Because ∆ is an integer, p = 3 and ∆ = 10. Therefore δ = 7, r = 14. We
have two cases:
+) m2 = 34, m3 = 14⇒ 168+ 48(τ − θ) = 0⇒ τ − θ = −
7
2
. A contradiction.
+) m2 = 14, m3 = 34⇒ −112+ 48(τ − θ) = 0⇒ τ − θ =
7
3
. A contradiction.
c) l = 3⇒ δ = 2p+ 1, r = 3(2p+ 1).
m2.m3 = 168p+ 63. ∆
2 = 49p2 − 126p− 54.
116
On strongly regular graphs of order n = 7(2p + 1) where 2p+ 1 is a prime number
⇒ p = 3; 11,
p = 3 ⇒ ∆ = 3, δ = 7, r = 21.
+) m2 = 27, m3 = 21⇒ 48(τ − θ) + 84 = 0⇒ τ − θ = −
7
4
.
+) m2 = 21, m3 = 27 ⇒ 48(τ − θ) = 0 ⇒ τ − θ = 0. Using (8) we obtain
49 = 02 + 4(21− θ)⇒ θ = 35
4
. A contradiction because θ is an integer.
p = 11⇒ δ = 23, r = 69, ∆ = 67 .
Sovle m2, m3, we have two cases:.
+) m2 = 147, m3 = 13 ⇒ 48(τ − θ) + 3220 = 0 ⇒ τ − θ = −
161
8
. A
contradiction .
+)m2 = 13, m3 = 147⇒ 160(τ−θ)−2944 = 0⇒ τ−θ =
92
5
. A contradiction.
d) l = 4⇒ m2.m3 = 168p+ 56. δ = 2p+ 1, r = 4(2p+ 1).
∆2 = 49p2 − 126p− 47⇒ p = 3; 6
p = 3⇒ ∆ = 4, δ = 7, r = 28.
Sovle m2, m3, we have two cases:
+) m2 = 28, m3 = 20⇒ 48(τ − θ) + 112 = 0⇒ τ − θ = −
7
3
. A contradiction
.
+) m2 = 28, m3 = 20⇒ 48(τ − θ) = 0⇒ τ − θ = 0. Theo (8) ta c θ =
63
4
. A
contradiction
p = 6⇒ ∆ = 31, δ = 13, r = 52.
Sovle m2, m3, we have two cases:
+) m2 = 76, m3 = 14⇒ 90(τ −θ)+910 = 0⇒ τ −θ = −
91
9
. A contradiction
+) m2 = 14, m3 = 76⇒ 90(τ − θ)− 702 = 0⇒ τ − θ =
39
5
. A contradiction
e) l = 5⇒ δ = 2p+ 1 v r = 5(2p+ 1).
m2.m3 = 140p+ 35⇒ ∆2 = 49p2 − 98p− 26.
Because ∆ is an integer, p = 3.
p = 3⇒ ∆ = 11, δ = 7, r = 35.
Solve equations, we have two cases:
+) m2 = 35, m3 = 13⇒ 48(τ −θ)+224 = 0⇒ τ −θ = −
14
3
. A contradiction
+) m2 = 13, m3 = 35⇒ 48(τ − θ)− 84 = 0⇒ τ − θ =
7
4
. A contradiction
117
Vu Dinh Hoa and Do Minh Tuan
f) l = 6⇒ δ = 2p+ 1, r = 6(2p+ 1).
m2.m3 = 84p, ∆
2 = 49p2 − 42p+ 9 = (7p− 3)2.
Solve equations, we have two cases:
+) m2 = 14p,m3 = 6. Then τ − θ = −2p− 1.
Using (8), we obtain θ = 12p+ 6 and τ = 10p+ 5.
λ2 =
τ − θ + δ
2
= 0, λ3 =
τ − θ − δ
2
= −2p− 1.
G ∈ SRG(14p+ 7, 12p+ 6, 10p+ 5, 12p+ 6).
⇒ G ∈ SRG(14p+ 7, 2p, 2p− 1, 0). But, θ = 0⇒ G is disconnected ⇒ G =
7K2p+1 and G = 7K2p+1.
+) m2 = 6, m3 = 14p. Then τ − θ =
(7p− 9)(2p+ 1)
7p+ 3
.
UCLN(2p+1, 7p+3) = UCLN(2p+1, 7p+3−3(2p+1)) = UCLN(2p+1, p) =
UCLN(2p + 1− 2.p, p) = UCLN(1, p) = 1.
Do 7p+3|(7p−9)(2p+1)⇒ 7p+3|7p−9. A contradiction, because 7p−9 <
7p+ 3.
2) k = 2⇒ δ = 2(2p+ 1). And m2.m3 =
7l(7− l)
2
p+
7l(6− l)
4
.
If l is odd then
7l(7− l)
2
p is an integer. But 7l(6− l) is odd; therefore m2.m3 is
not an interger, a contradiction. So that l is even, and l = 2; 4; 6.
a) l = 2, r = 2(2p+ 1), δ = 2(2p+ 1).
m2.m3 = 35p+ 14. ∆
2 = 49p2 + 7p− 5. Because ∆2 is a perfect square, @p.
b) l = 4, r = 4(2p+ 1), δ = 2(2p+ 1).
m2.m3 = 42p + 14, ∆
2 = 49p2 − 5. We have (7p − 1)2 < ∆2 < (7p)2. A
contradiction .
c) l = 6, r = 6(2p+ 1), δ = 2(2p+ 1),
m2.m3 = 21p, ∆
2 = 49p2 + 21p+ 9.
Do (7p+ 1)2 < ∆2 < (7p+ 2)2
3) k = 3⇒ δ = 3(2p+ 1) and
m2.m3 =
14l(7− l)
9
p+
7l(6− l)
9
.
a) If l = 1 then m2.m3 =
84p+ 35
9
. A contradiction because 84p + 35 ≡ 2
mod 3.
118
On strongly regular graphs of order n = 7(2p + 1) where 2p+ 1 is a prime number
b) If l = 2 then m2.m3 =
140p+ 56
9
⇒ p ≡ −4 mod 9, and p = 9h − 4 v
m2.m3 = 140h− 56, m2 +m3 = 126h− 50.
∆2 = 3969h2 − 3290h+ 681 = (63h− 26)2 − (14h+ 5)
We obtain (63h− 27)2 < ∆2 < (63h− 26)2. A contradiction
c) If l = 3 then m2m3 =
56p
3
+ 7. Therefore p = 3h and m2.m3 = 56h + 7,
m2 +m3 = 42h+ 6.
Hence ∆2 = 441h2 + 70h+ 2 = (21h+ 1)2 + 28h+ 1.
We have (21h+ 1)2 < ∆2 < (21h+ 2)2. A contradiction
d) If l = 4 then m2m3 =
168p− 56
9
. A contradiction because 168p − 56 ≡ 1
mod 3.
e) If l = 5 then m2.m3 =
140p+ 35
9
. Hence p ≡ 2 mod 9 v p = 9h + 2 and
m2.m3 = 140h+ 35, m2 +m3 = 126h+ 34.
∆2 = 3969h2 + 2002h+ 254 = (63h+ 15)2 + 112h+ 29.
We obtain (63h+ 15)2 < ∆2 < (63h+ 16)2. A contradiction
f) If l = 6 then m2.m3 =
28p
3
. Hence p = 3h and
m2m3 = 28h, m2 +m3 = 42h+ 6.
∆2 = 441h2 + 98h+ 9 = (21h+ 2)2 + (14h+ 5). Therefore, we have
(21h+ 2)2 < ∆2 < (21h+ 3)2. A contradiction
3.2. Proof of Lemma 2.2
m2 = k(2p+ 1) (1 ≤ k ≤ 6), m3 = n− 1−m2 = (7− k).2p+ 6− k.
Using (1.2), we obtain : 2r + (m2 +m3)(τ − θ) + δ.(m2 −m3) = 0.
Using (1.3),(1.4), we have:
2r + (m2 +m3).(λ2 + λ3) + (λ2 − λ3).(m2 −m3) = 0
r + kλ2 + (6− k)λ3 = 2p(−kλ2 + (k − 7)λ3).
Let t = −kλ2 + (k − 7)λ3. Hence r = (2p+ 1)t+ λ3
1) If k = 1 then :
m2 = 2p+ 1, m3 = 12p+ 5, λ2 = −t+ 6a, λ3 = −a,
r = (2p+ 1)t− a,
Using (1.3), we have δ = λ2 − λ3 = −t+ 7a.
Using (1.4), we have τ − θ = λ2 + λ3 = −t+ 5a.
Otherwise, Using (1.5), we obtain: θ = (2p+ 1 + a)t− 6a2 − a.
119
Vu Dinh Hoa and Do Minh Tuan
Therefore, τ = (2p+ a)t− 6a2 + 4a.
Replace them into Equation (1.1) and transform it, we have:
2(p+ 1)t2 − (14p+ 7 + 14a)t+ 42a2 + 7a = 0 (3.1)
a) If t = 0 then 42a2 + 7a = 0⇒ a = 0. Hence θ = 0 and G is disconnected. A
contradiction.
b) If t = 1 then 12p = 5− 7a+42a2 thus a = 12h− 5⇒ p = 504h2− 427h+90.
G ∈ SRG(7(1008h2−854h+181), 1008h2−866h+186, 144h2−74h+5, 144h2−
134h+ 31)
+) m2 = 1008h
2 − 854h+ 181, m3 = 6048h2 − 5124h+ 1085
λ2 = 72h− 31, λ3 = −12h+ 5.
c) If t = 2 then 20p = −6−21a+42a2. Hence, a = 20h+2⇒ p = 840h2+14h+6.
G ∈ SRG(7(1680h2+ 294h+ 13), 3360h2+ 568h+ 24, 12 + 228h+ 960h2, 4 +
128h+ 960h2)
+) m2 = 1680h
2 + 294h+ 13, m3 = 10080h
2 + 1764h+ 77
λ2 = 10 + 120h, λ3 = −20h− 2.
d) If t = 3 then 24p = −3− 35a+ 42a2.
⇒ a = 24h− 3⇒ p = 1008h2 − 287h+ 20.
G ∈ SRG(7(2016h2−574h+41), 6048h2−1746h+126, 45−690h+2592h2, 63−
810h+ 2592h2)
and m2 = 2016h
2 − 574h+ 41, m3 = 12096h2 − 3444h+ 245
λ2 = −21 + 144h, λ3 = −24h + 3.
e) If t = 4 then 24p = 4− 49a+ 42a2.
⇒ a = 24h+ 4⇒ p = 1008h2 + 287h+ 20.
G ∈ SRG(7(2016h2+574h+41), 8064h2+2272h+160, 96+1336h+4608h2, 80+
1216h+ 4608h2)
and m2 = 2016h
2 + 574h+ 41, m3 = 12096h
2 + 3444h+ 245
λ2 = 20 + 144h, λ3 = −24h− 4.
f) If t = 5 then 20p = 15− 63a+ 42a2.
⇒ a = 20h− 1⇒ p = 840h2 − 147h+ 6.
G ∈ SRG(7(1680h2−294h+13), 8400h2−1490h+66, 45−1050h+6000h2, 55−
1150h+ 6000h2)
and m2 = 1680h
2 − 294h+ 13, m3 = 10080h2 − 1764h+ 77
λ2 = −11 + 120h, λ3 = −20h + 1.
120
On strongly regular graphs of order n = 7(2p + 1) where 2p+ 1 is a prime number
g) If t = 6 then 12p = 30− 77a+ 42a2.
⇒ a = 12h+ 6⇒ p = 504h2 + 427h+ 90.
G ∈ SRG(7(1008h2 + 854h + 181), 6048h2 + 5112h + 1080, 924 + 4380h +
5184h2, 900 + 4320h+ 5184h2)
and m2 = 1008h
2 + 854h+ 181, m3 = 6048h
2 + 5124h+ 1085
λ2 = 30 + 72h, λ3 = −12h− 6.
h) If t = 7 then 42a2 − 91a+ 49 = 0⇔ a = 1 or a = 7
6
(a contradiction)
If a = 1 then λ2 = −1 < 0 (a contradiction).
i) If t 7 then t2 − 7t > 0.
Using 3.1, we have 2p(t2 − 7t) = −2t2 + (14a+ 7)t− 42a2 − 7a.
Thus −2t2 + (14a+ 7)t− 42a2 − 7a > 0⇔ 2t2 − (14a+ 7)t+ 42a2 + 7a < 0.
This inequation have roots if and only if (14a+ 7)2 − 4.2(42a2 + 7a) > 0
⇔ −140a2 + 140a+ 49 > 0⇔ 1
2
−
√
15
5
< a <
1
2
+
√
15
5
thus a = 0; 1.
Case 1:a = 0. We have 2t2 − 7t < 0⇔ 0 < t < 7
2
. A contradiction with t < 0
or t > 7.
Case 2:a = 1. We have 2t2 − 21t+ 49 < 0⇔ 7
2
< t < 7. A contradiction.
2) We consider cases: k = 2, 3, 4, 5, 6, we have all of results in Lemma 2.2
3.3. Proof of Lemma 2.3
m3 = k(2p+ 1) (1 ≤ k ≤ 6), m2 = n− 1−m2 = (7− k).2p+ 6− k.
Using (1.2), we have : 2r + (m2 +m3)(τ − θ) + δ.(m2 −m3) = 0.
Using (1.3),(1.4), we have:
2r + (m2 +m3).(λ2 + λ3) + (λ2 − λ3).(m2 −m3) = 0
r + (6− k)λ2 + kλ3 = 2p((k − 7)λ2 − kλ3).
Let t = (k − 7)λ2 − kλ3. Hence, r = (2p+ 1)t+ λ2
1) k = 1,, we have :
m2 = 12p+ 5, m3 = 2p+ 1, λ2 = a, λ3 = −t− 6a,
r = (2p+ 1)t+ a,
Using (1.3), we have δ = λ2 − λ3 = t+ 7a.
Using (1.4), we have τ − θ = λ2 + λ3 = −t− 5a.
Using (1.5), we have: θ = (2p+ 1 + a)t− 6a2 − a.
121
Vu Dinh Hoa and Do Minh Tuan
It follows τ = (2p+ a)t− 6a2 + 4a.
Replacing these variable in (1.1) and simplifying, we have:
2(p+ 1)t2 − (14p+ 7− 14a)t+ 42a2 − 7a = 0 (3.2)
a) If t = 0 then 42a2 − 7a = 0 hay a = 0. Thus θ = 0 and G is disconnected. A
contradiction .
b) If t = 1 then 12p = 42a2+7a−5. Thus a = 12h+5⇒ p = 504h2+427h+90.
G ∈ SRG(7(1008h2+854h+181), 1008h2+866h+186, 74h+5+144h2, 144h2+
134h+ 31)
and m2 = 6048h
2 + 5124h+ 1085, m3 = 1008h
2 + 854h+ 181
λ2 = 12h+ 5, λ3 = −31− 72h.
c) If t = 2 then 20p = −6 + 21a+ 42a2. Thus a ≡ −6;−2 mod 20.
+) Case 1: a = 20h− 6⇒ p = 840h2 − 483h+ 69.
G ∈ SRG(7(1680h2 − 966h + 139), 3360h2 − 1912h + 272,−612h + 96 +
960h2, 960h2 − 512h+ 68)
and m2 = 10080h
2 − 5796h+ 833, m3 = 1680h2 − 966h+ 139
λ2 = 20h− 6, λ3 = 34− 120h.
+) Case 2: a = 20h− 2⇒ p = 840h2 − 147h+ 6.
G ∈ SRG(7(1680h2−294h+13), 3360h2−568h+24,−228h+12+960h2, 960h2−
128h+ 4)
and m2 = 10080h
2 − 1764h+ 77, m3 = 1680h2 − 294h+ 13
λ2 = 20h− 2, λ3 = 10− 120h.
d) If t = 3 then 24p = −3 + 35a+ 42a2.
⇒ a = 24h+ 3⇒ p = 1008h2 + 287h+ 20.
G ∈ SRG(7(2016h2+574h+41), 6048h2+1746h+126, 690h+45+2592h2, 2592h2+
810h+ 63)
and m2 = 12096h
2 + 3444h+ 245, m3 = 2016h
2 + 574h+ 41
λ2 = 24h+ 3, λ3 = −21− 144h.
e) If t = 4 then 24p = 4 + 49a+ 42a2.
⇒ a = 24h− 4⇒ p = 1008h2 − 287h+ 20.
G ∈ SRG(7(2016h2 − 574h + 41), 8064h2 − 2272h + 160,−1336h + 96 +
4608h2, 4608h2 − 1216h+ 80)
and m2 = 12096h
2 − 3444h+ 245, m3 = 2016h2 − 574h+ 41
λ2 = 24h− 4, λ3 = 20− 144h.
f) If t = 5 then 20p = 15 + 63a+ 42a2.
⇒ a ≡ 1; 5 mod 20.
122
On strongly regular graphs of order n = 7(2p + 1) where 2p+ 1 is a prime number
+) Case 1: a = 20h+ 1⇒ p = 840h2 + 147h+ 6.
G ∈ SRG(7(1680h2+294h+13), 8400h2+1490h+66, 1050h+45+6000h2, 6000h2+
1150h+ 55)
and m2 = 10080h
2 + 1764h+ 77, m3 = 1680h
2 + 294h+ 13
λ2 = 20h+ 1, λ3 = −11− 120h.
+) Case 2: a = 20h+ 5⇒ p = 840h2 + 483h+ 69.
G ∈ SRG(7(1680h2 + 966h + 139), 8400h2 + 4850h + 700, 3450h + 495 +
6000h2, 6000h2 + 3550h+ 525)
and m2 = 10080h
2 + 5796h+ 833, m3 = 1680h
2 + 966h+ 139
λ2 = 20h+ 5, λ3 = −35− 120h.
g) If t = 6 then 12p = 30 + 77a+ 42a2.
⇒ a = 12h+ 6⇒ p = 504h2 + 581h+ 167.
G ∈ SRG(7(1008h2 + 1162h + 335), 6048h2 + 6984h + 2016, 5988h+ 1728 +
5184h2, 5184h2 + 6048h+ 1764)
and m2 = 6048h
2 + 6972h+ 2009, m3 = 1008h
2 + 1162h+ 335
λ2 = 12h+ 6, λ3 = −42− 72h.
h) If t = 7 then 42a2 + 91a+ 49 = 0⇔ a = −1 or a = 7
6
(Loi)
If a = −1 then λ2 = −1 < 0 (V l).
i) If t 7 then t2 − 7t > 0.
Using (3.2), we have 2p(t2 − 7t) = −2t2 + (−14a+ 7)t− 42a2 + 7a.
It follows −2t2+(−14a+7)t−42a2+7a > 0⇔ 2t2−(−14a+7)t+42a2−7a < 0.
This inequation have roots if and only if (−14a + 7)2 − 4.2(42a2 − 7a) > 0
⇔ −140a2 − 140a+ 49 > 0⇔ −1
2
−
√
15
5
< a < −1
2
+
√
15
5
thus a = 0;−1.
Case 1: If a = 0 then 2t2 − 7t < 0 ⇔ 0 < t < 7
2
. A contradiction with t < 0
or t > 7.
Case 2: If a = −1 then 2t2 − 21t+ 49 < 0⇔ 7
2
< t < 7. A contradiction .
2) We consider cases: k = 2, 3, 4, 5, 6, we have all of results in Lemma 2.3
3.4. Proof of Theorem 2.1
Using Theorem 1.1 we have m2.m3.δ
2 = 7(2p+ 1).r.r. We shall now consider
the following three cases:
Case 1 (2p+ 1|m2): Since Lemma 2.2, we prove that G in class 2 to class 72.
Case 2 (2p+ 1|m3): Since Lemma 2.3, we prove that G in class 73 to class 144.
Case 3 (2p+ 1|δ2): Since Lemma 2.1, we prove that G in class 1.
123
Vu Dinh Hoa and Do Minh Tuan
4. Table of parameters on strongly regular graph when n =
7(2p+ 1)
Parameter Table
Class n = 7(A.h
2 +B.h + C) r = 7(A.h2 + B.h +C) τ = 7(A.h2 + B.h+ C) θ = 7(A.h2 + B.h +C)
A B C A B C A B C A B C
1 n = 7 (2p + 1) r = 12p + 6 τ = 10p + 5 θ = 12p + 6
2 1008 −854 181 1008 −886 186 144 −74 5 144 −134 31
3 1680 294 13 3360 568 24 960 228 12 960 128 4
4 2016 −574 41 6048 −1746 126 2592 −690 45 2592 −810 63
5 2016 574 41 8064 2272 160 4608 1336 96 4608 1216 80
6 1680 −294 13 8400 −1490 66 6000 −1050 45 6000 −1150 55
7 1008 854 181 6048 5112 1080 5184 4380 924 5184 4320 900
8 420 −350 73 420 −362 78 60 −38 5 60 −56 13
9 420 −70 3 420 −82 4 60 2 −1 60 −16 1
10 420 70 3 420 58 2 60 22 1 60 4 0
11 420 350 73 420 338 68 60 62 15 60 44 8
12 700 −294 31 1400 −608 66 400 −158 15 400 −188 22
13 700 406 59 1400 792 112 400 342 36 400 212 28
14 840 −658 129 2520 −1998 396 1080 −864 165 1080 −882 180
15 840 −238 17 2520 −738 54 1080 −306 21 1080 −342 27
16 840 −98 3 2520 −318 10 1080 −126 3 1080 −162 6
17 840 322 31 2520 942 88 1080 414 39 1080 378 33
18 840 −1022 311 3360 −4112 1258 1920 −2348 717 1920 −2384 740
19 840 −322 31 3360 −1312 128 1920 −748 72 1920 −784 80
20 840 98 3 3360 368 10 1920 212 5 1920 176 4
21 840 238 17 3360 928 64 1920 532 36 1920 496 32
22 700 −406 59 3500 −2050 300 2500 −14