1. Secure Schemes for Secret Sharing and Key
Distribution
N. Chandramowliswaran, P. Muralikrishna and S. Srinivasan
School of Advanced Sciences,
Vellore Institute of Technology,
Vellore 632 014,
India.
e-mail: smrail@gmail.com
Abstract
In resent years the security of operations taking place over a computer
network become very important. It is necessary to protect such actions
against bad users who may try to misuse the system (e.g. steal credit
card numbers, read personal mail, or impersonate other users.) Many
protocols and schemes were designed to solve problem of this type. This
paper deals with two fundamental cryptographic tools that are useful in such
contexts: generalized secret sharing scheme and key distribution schemes.
Both secret sharing schemes and key distribution schemes are used in multi-
party systems. secret sharing schemes enables some predetermined sets of
parties to reconstruct a given secret. In this paper we have shown a novel
key pre distribution algorithm based on number theory which uses Chineese
Remainder Theorem (CRT), Continued fractions (CF) and pell’s equation.
1 Introduction
Secret sharing (also called secret splitting) refers to method for distributing a
secret amongst a group of participants, each of whom is allocated a share of the
1
2. 2
secret. The secret can be reconstructed only when a sufficient number, of possibly
different types, of shares are combined together; individual shares are of no use
on their own.
Secret sharing was invented independently by Adi Shamir [6] and George
Blakley [7] in 1979. Secret sharing schemes are ideal for storing information that is
highly sensitive and highly important. Examples include: encryption keys, missile
launch codes, and numbered bank accounts. Each of these pieces of information
must be kept highly confidential, as their exposure could be disastrous, however,
it is also critical that they not be lost. Traditional methods for encryption are
ill-suited for simultaneously achieving high levels of confidentiality and reliability.
This is because when storing the encryption key, one must choose between keeping
a single copy of the key in one location for maximum secrecy, or keeping multiple
copies of the key in different locations for greater reliability. Increasing reliability
of the key by storing multiple copies lowers confidentiality by creating additional
attack vectors; there are more opportunities for a copy to fall into the wrong
hands. Secret sharing schemes address this problem, and allow arbitrarily high
levels of confidentiality and reliability to be achieved.
A secure secret sharing scheme distributes shares so that anyone with fewer
than 𝑡 shares has no extra information about the secret than someone with 0
shares.
Consider for example the secret sharing scheme in which the secret phrase
security is divided into the shares 𝑠𝑒−−−−−−, −−𝑐𝑢−−−−, −−−−𝑟𝑖−−,
and − − − − − − 𝑡𝑦. A person with 0 shares knows only that the password
consists of eight letters. He would have to guess the password from 268
= 208
billion possible combinations. A person with one share, however, would have to
guess only the six letters, from 266
= 308 million combinations, and so on as
more persons collude. Consequently this system is not a secure secret sharing
scheme, because a player with fewer than t secret-shares is able to reduce the
problem of obtaining the inner secret without first needing to obtain all of the
necessary shares.
More generally, (𝑛; 𝑘) secret sharing is the problem of distributing a secret
3. 3
number 𝑠 among 𝑛 people so that no 𝑘 − 1 of them have any information
about 𝑠 but 𝑘 of them can determine 𝑠. Shamir’s secret sharing does this by
giving the 𝑛 -th party 𝑓(𝑛), where 𝑓 is an appropriately chosen polynomial. The
dealer Dan picks random field elements 𝑎1, 𝑎2, . . . , 𝑎𝑘−1, and uses the polynomial
𝑓(𝑡) = 𝑠 + 𝑎1𝑡 + 𝑎2𝑡2
+ ⋅ ⋅ ⋅ + 𝑎𝑘−1𝑡𝑘−1
. He gives the 𝑛 -th person 𝑓(𝑛). For
𝑘 people to recover the secret, they just pool their shares and use Lagrange
interpolation to find the unique degree 𝑘 − 1 polynomial passing through the 𝑘
points. (Lagrange interpolation works over any field. In practice, a large finite
field would probably be used.) The secret is just the constant term. Any 𝑘 − 1
people have no knowledge about 𝑠. They have 𝑘 − 1 points on the polynomial,
but there is a degree 𝑘 − 1 polynomial going through their 𝑘 − 1 points and
(0, 𝑡) for any integer 𝑡, so their combined knowledge reveals nothing about 𝑠.
The current trend towards cloud computing means that more and more data
is being processed and stored by online resources beyond physical and logical
control of the owner. It is a simple task for an adversary to intercept, copy, and
store any data sent across a public network and from this point on, confidentiality
is determined solely by the original encryption scheme. Hence re-encryption is
useless, and even deletion of data cannot be guaranteed.
Current cryptographic schemes in wide deployment today, such as RSA, Dife
Hellman, and AES, do not over long term confidentiality guarantees. This
is because concrete security is based on the current infeasibility of a specific
computational problem, such as factoring a 1024 -bit RSA modulus or computing
a 128 -bit AES key, and there are no known techniques to prove the hardness of
such problems.
We mention several related survey articles which overlap to some extent with
our exposition [1][2][3][4][5]
One of the standard topics in a first course in number theory is the Euler 𝜙
function, with 𝜙(𝑛) defined as the number of positive integers less than 𝑛 and
relatively prime to 𝑛. A famous theorem involving 𝜙 is that suppose 𝑎 and 𝑏
are any two positive integers with (𝑎, 𝑏) = 1 then 𝑎𝜙(𝑏)
+𝑏𝜙(𝑎)
≡ 1(𝑚𝑜𝑑 𝑎𝑏). The
Chinese Remainder Theorem (CRT) can also be used in secret sharing, there are
4. 4
two secret sharing schemes that make use of the Chinese Remainder Theorem,
Mignotte’s and Asmuth-Bloom’s Schemes. They are threshold secret sharing
schemes, in which the shares are generated by reduction modulo the integers
𝑚𝑖, and the secret is recovered by essentially solving the system of congruences
using the Chinese Remainder Theorem.
Theorem 1.0. (Chinese Remainder Theorem)
Suppose that 𝑚1, 𝑚2, . . . , 𝑚𝑟 are pairwise relatively prime positive integers, and
let 𝑎1, 𝑎2, . . . , 𝑎𝑟 be integers. Then the system of congruences, 𝑥 ≡ 𝑎𝑖(𝑚𝑜𝑑 𝑚𝑖)
for 1 ≤ 𝑖 ≤ 𝑟, has a unique solution modulo 𝑀 = 𝑚1 × 𝑚2 × . . . × 𝑚𝑟, which is
given by: 𝑥 ≡ 𝑎1𝑀1𝑦1 + 𝑎2𝑀2𝑦2 + . . . + 𝑎𝑟𝑀𝑟𝑦𝑟(𝑚𝑜𝑑 𝑀), where 𝑀𝑖 = 𝑀
𝑚𝑖
and
𝑦𝑖 ≡ (𝑀𝑖)−1
(𝑚𝑜𝑑 𝑚𝑖) for 1 ≤ 𝑖 ≤ 𝑟.
2 Main Results
Lemma 2.1. Let 𝑝, 𝑞 and 𝑟 be three given distinct odd primes. Then there exist
integers 𝑘1, 𝑘2 and 𝑘3 such that
𝑘1𝑝(𝑞𝑟−1
+ 𝑟𝑞−1
) + 𝑘2𝑞(𝑝𝑟−1
+ 𝑟𝑞−1
) + 𝑘3𝑟(𝑝𝑞−1
+ 𝑞𝑝−1
) + 2 ≡ 0 (𝑚𝑜𝑑 𝑝𝑞𝑟).
Proof:
Define: 𝑋 = (𝑝𝑞−1
+ 𝑞𝑝−1
) + (𝑝𝑟−1
+ 𝑟𝑝−1
) + (𝑞𝑟−1
+ 𝑟𝑞−1
) − 2. Then
𝑋 ≡ (𝑞𝑟−1
+ 𝑟𝑞−1
)(𝑚𝑜𝑑𝑝)
𝑋 ≡ (𝑝𝑟−1
+ 𝑟𝑝−1
)(𝑚𝑜𝑑𝑞) and
𝑋 ≡ (𝑝𝑞−1
+ 𝑞𝑝−1
)(𝑚𝑜𝑑𝑟).
By CRT, the above system of congruences has exactly one solution modulo the
product 𝑝𝑞𝑟.
Define 𝑀 = 𝑝𝑞𝑟
𝑀𝑝 = 𝑀
𝑝 = 𝑞𝑟, 𝑀𝑞 = 𝑀
𝑞 = 𝑝𝑟 and 𝑀𝑟 = 𝑀
𝑟 = 𝑝𝑞.
Since (𝑀𝑝, 𝑝) = 1, then there is a unique 𝑀′
𝑝 such that 𝑀𝑝 𝑀′
𝑝 ≡ 1 (𝑚𝑜𝑑 𝑝).
Similarly there are unique 𝑀′
𝑞 and 𝑀′
𝑟 such that
𝑀𝑞 𝑀′
𝑞 ≡ 1 (𝑚𝑜𝑑 𝑞) and 𝑀𝑟 𝑀′
𝑟 ≡ 1 (𝑚𝑜𝑑 𝑟).
Consider
𝑋 ≡ ((𝑝𝑞−1
+𝑞𝑝−1
)𝑀𝑟 𝑀′
𝑟+(𝑝𝑟−1
+𝑟𝑝−1
)𝑀𝑞 𝑀′
𝑞+(𝑞𝑟−1
+𝑟𝑞−1
)𝑀𝑝 𝑀′
𝑝) (𝑚𝑜𝑑 𝑝𝑞𝑟)
5. 5
𝑝𝑞−1
+ 𝑞𝑝−1
+ 𝑝𝑟−1
+ 𝑟𝑝−1
+ 𝑞𝑟−1
+ 𝑟𝑞−1
− 2
≡ ((𝑝𝑞−1
+𝑞𝑝−1
)𝑀𝑟 𝑀′
𝑟 +(𝑝𝑟−1
+𝑟𝑝−1
)𝑀𝑞 𝑀′
𝑞 +(𝑞𝑟−1
+𝑟𝑞−1
)𝑀𝑝 𝑀′
𝑝) (𝑚𝑜𝑑 𝑝𝑞𝑟)
−2 ≡ ((𝑝𝑞−1
+ 𝑞𝑝−1
)(𝑀𝑟 𝑀′
𝑟 − 1) + (𝑝𝑟−1
+ 𝑟𝑝−1
)(𝑀𝑞 𝑀′
𝑞 − 1)
+(𝑞𝑟−1
+ 𝑟𝑞−1
)(𝑀𝑝 𝑀′
𝑝 − 1)) (𝑚𝑜𝑑 𝑝𝑞𝑟)
Thus
𝑘1𝑝(𝑞𝑟−1
+ 𝑟𝑞−1
) + 𝑘2𝑞(𝑝𝑟−1
+ 𝑟𝑞−1
) + 𝑘3𝑟(𝑝𝑞−1
+ 𝑞𝑝−1
) + 2 ≡ 0 (𝑚𝑜𝑑 𝑝𝑞𝑟).
Theorem 2.0. Let 𝑆 be the given secret and 𝑁 = 𝑝𝑞𝑟 where 𝑝, 𝑞 and 𝑟 are
distinct large odd primes. Define three secret shareholders 𝑌1, 𝑌2, 𝑌3 as follows:
𝑌1 ≡ (−𝑆𝑘1𝑝(𝑞𝑟−1
+ 𝑟𝑞−1
)) (𝑚𝑜𝑑 𝑁), 𝑌2 ≡ (−𝑆𝑘2𝑞(𝑝𝑟−1
+ 𝑟𝑝−1
)) (𝑚𝑜𝑑 𝑁) and
𝑌3 ≡ (−𝑆(𝑘3𝑟(𝑝𝑞−1
+ 𝑞𝑝−1
) + 1)) (𝑚𝑜𝑑𝑁) then 𝑆 = 𝑌1 + 𝑌2 + 𝑌3(𝑚𝑜𝑑 𝑁)
Proof: By the above Lemma 2.1, we have
𝑘1𝑝(𝑞𝑟−1
+ 𝑟𝑞−1
) + 𝑘2𝑞(𝑝𝑟−1
+ 𝑟𝑞−1
) + 𝑘3𝑟(𝑝𝑞−1
+ 𝑞𝑝−1
) + 2 ≡ 0 (𝑚𝑜𝑑 𝑁).
1 ≡ (−(𝑘1𝑝(𝑞𝑟−1
+𝑟𝑞−1
))−(𝑘2𝑞(𝑝𝑟−1
+𝑟𝑞−1
))−(𝑘3𝑟(𝑝𝑞−1
+𝑞𝑝−1
)+1)) (𝑚𝑜𝑑 𝑁)
Thus 𝑆 = 𝑌1 + 𝑌2 + 𝑌3(𝑚𝑜𝑑 𝑁).
Algorithm 1.
∙ Choose two secret very large odd primes 𝑟1, 𝑟2 with 𝑟1 > 𝑟2
∙ Construct 𝑥2
+ 1 = (𝑟2
1 + 𝑟2
2)𝑦
∙ Select two large odd primes 𝑝 and 𝑞
∙ Define 𝑛 = 𝑝𝑞 then 𝜙(𝑛) = (𝑝−1)(𝑞−1) Where 𝜙(𝑛) is Euler phi function
∙ Select a random 𝑒 such that [ 1 < 𝑒 < 𝜙(𝑛) ] (𝑒, 𝜙(𝑛)) = 1
∙ For an 𝑒 there is a unique 𝑑 such that 𝑒𝑑 ≡ 1(𝑚𝑜𝑑 𝜙(𝑛))
∙ consider 𝑎 = (𝑟2
1 + 𝑟2
2)(𝑦 + 𝑑) − (𝑥 + 𝜙(𝑛))2
𝑎 = (𝑟2
1 + 𝑟2
2)𝑦 − 𝑥2
+ (𝑟2
1 + 𝑟2
2)𝑑 − [𝜙(𝑛)]2
− 2𝑥𝜙(𝑛)
= 1 + (𝑟2
1 + 𝑟2
2)𝑑 − [𝜙(𝑛)]2
− 2𝑥𝜙(𝑛)
𝑎 ≡ 1 + (𝑟2
1 + 𝑟2
2)𝑑(𝑚𝑜𝑑 𝜙(𝑛))
𝑎𝑒 ≡ 𝑒 + (𝑟2
1 + 𝑟2
2)(𝑚𝑜𝑑 𝜙(𝑛))
𝑠 ≡ 𝑒(𝑚𝑜𝑑 𝜙(𝑛)) where 𝑠 = 𝑎𝑒 − (𝑟2
1 + 𝑟2
2)
6. 6
∙ Public key: (𝑠, 𝑛)
∙ Represent the message 𝑚 in the interval [0, 𝑛 − 1] with (𝑚, 𝑛) = 1
∙ Encryption
𝐸 ≡ 𝑚𝑠
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑘𝜙(𝑛)+𝑒
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑘𝜙(𝑛)
𝑚𝑒
(𝑚𝑜𝑑 𝑛)
≡ [𝑚𝜙(𝑛)
]𝑘
𝑚𝑒
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑒
(𝑚𝑜𝑑 𝑛)
∙ Key distribution: Choose ℓ share holders then 𝑒 = 𝑘1 + 𝑘2 + ⋅ ⋅ ⋅ + 𝑘ℓ
𝐸 ≡ 𝑚𝑒
(𝑚𝑜𝑑 𝑛)
𝐸 ≡ 𝑚𝑘1+𝑘2+⋅⋅⋅+𝑘ℓ
(𝑚𝑜𝑑 𝑛)
𝐸 ≡ 𝑚𝑘1
𝑚𝑘2
. . . 𝑚𝑘ℓ
(𝑚𝑜𝑑 𝑛)
∙ For ℓ share holders we can distribute ℓ key’s such as 𝑚𝑘1
, 𝑚𝑘2
, . . . , 𝑚𝑘ℓ
.
Algorithm 2.
∙ Select a secret odd prime integer 𝑟
∙ Consider the Diophantine Equation:
𝑦2
− 𝑟𝑥2
= 1 (1)
∙ Let (𝑥0, 𝑦0) be the least positive integral solution of (1). Here 𝑥0, 𝑦0 are
kept secret
∙ Select two large odd primes 𝑝 and 𝑞
∙ Define 𝑛 = 𝑝𝑞 then 𝜙(𝑛) = (𝑝−1)(𝑞−1) Where 𝜙(𝑛) is Euler phi function
∙ Select a random 𝑒 such that [ 1 < 𝑒 < 𝜙(𝑛) ] such that (𝑒, 𝜙(𝑛)) = 1
∙ For an 𝑒 there is a unique 𝑑 such that 𝑒𝑑 ≡ 1(𝑚𝑜𝑑 𝜙(𝑛))
∙ consider 𝑎 = (𝑦0 + 𝜙(𝑛))2
− 𝑟(𝑥0 + 𝑒)2
(2)
∙ 𝑒3
is not congruent to 1(𝑚𝑜𝑑 𝜙(𝑛)) and 𝑑3
is not congruent to
1(𝑚𝑜𝑑 𝜙(𝑛))
7. 7
∙ From (2) 𝑎𝑑3
+ 𝑟𝑑 + 2𝑥0𝑑2
𝑟 ≡ 𝑑3
(𝑚𝑜𝑑 𝜙(𝑛))
∙ Let 𝑆 = 𝑎𝑑3
+ 2𝑥0𝑑2
𝑟 + 𝑟𝑑 then 𝑆 ≡ 𝑑3
(𝑚𝑜𝑑 𝜙(𝑛))
∙ Public key: (𝑠, 𝑛)
∙ Represent the message 𝑚 in the interval [0, 𝑛 − 1] with (𝑚, 𝑛) = 1
∙ Encryption
𝐸 ≡ 𝑚𝑠
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑘𝜙(𝑛)+𝑑3
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑘𝜙(𝑛)
𝑚𝑑3
(𝑚𝑜𝑑 𝑛)
≡ [𝑚𝜙(𝑛)
]𝑘
𝑚𝑑3
(𝑚𝑜𝑑 𝑛)
≡ 𝑚𝑑3
(𝑚𝑜𝑑 𝑛)
∙ Key distribution: Choose ℓ share holders then 𝑑3
= 𝑘1 + 𝑘2 + ⋅ ⋅ ⋅ + 𝑘ℓ
𝐸 ≡ 𝑚𝑑3
(𝑚𝑜𝑑 𝑛)
𝐸 ≡ 𝑚𝑘1+𝑘2+⋅⋅⋅+𝑘ℓ
(𝑚𝑜𝑑 𝑛)
𝐸 ≡ 𝑚𝑘1
𝑚𝑘2
. . . 𝑚𝑘ℓ
(𝑚𝑜𝑑 𝑛)
∙ For ℓ share holders we can distribute ℓ key’s such as 𝑚𝑘1
, 𝑚𝑘2
, . . . , 𝑚𝑘ℓ
.
Algorithm 3.
∙ Let 𝑝, 𝑞, 𝑟 and 𝑠 be the given distinct secrete odd primes.
∙ Define 𝑢 = 𝑝 𝑞 and 𝑣 = 𝑟 𝑠
∙ Select 𝑎, 𝑏 such that (𝑎, 𝑢) = 1 and (𝑏, 𝑣) = 1
∙ Select two positive integers 𝑒, 𝑓 such that
(𝑒, (𝑝 − 1)(𝑞 − 1)) = 1 and (𝑓, (𝑟 − 1)(𝑠 − 1)) = 1
∙ Select a common secret 𝑡 with 𝑝, 𝑞, 𝑟, 𝑠 should not divide 𝐻
∙ Define 𝑥1, 𝑥2 as follows:
𝑥1 ≡ 𝑎𝑡𝑒
(𝑚𝑜𝑑 𝑈)
𝑥2 ≡ 𝑏𝑡𝑓
(𝑚𝑜𝑑 𝑉 )
∙ Solve 𝑡 uniquely under (𝑚𝑜𝑑 𝑈𝑉 ) using Chineese Remainder Theorem
8. 8
∙ 𝑡 is the common secret shared by 𝑥1 and 𝑥2
BC code
∙ Let 𝑁 be a fixed positive integer
∙ Define 𝜙(𝑖,𝑁) =∣ {𝑥 ∣ 𝑖 ≤ 𝑥 ≤ 𝑁 𝑤𝑖𝑡ℎ (𝑥, 𝑁) = 1} ∣ where 𝑖 ∈ {1, 2, . . . , 𝑁}
∙ Define the 𝐵𝐶 code for 𝑁 ( 𝐵𝐶𝑁 ) as follows:
𝐵𝐶𝑁 = (𝜙(1,𝑁), 𝜙(2,𝑁), . . . , 𝜙(𝑁−1,𝑁), 𝜙(𝑁,𝑁))
Remark
For every positive integer 𝑁 we can write a unique 𝐵𝐶 code
Theorem 2.0. Let 𝑁 be any positive integer. Then 𝑁 is a prime if and only
if there exist a unique 𝐵𝐶 code such that 𝐵𝐶𝑁 = (𝑁 − 1, 𝑁 − 2, . . . , 2, 1, 0).
.............................
9. 9
∙ Let 𝑁 be an odd positive integer
∙ Let 𝑆𝑁 = {1, 2, 3, . . . , 𝑁 − 1, 𝑁}
∙ Define 𝐴 = {𝑥 ∈ 𝑆𝑁 ∣ 1 ≤ 𝑥 ≤ 𝑁, (𝑥, 𝑁) = 1} where 𝑔𝑐𝑑{𝑥, 𝑁} = 1 =
(𝑥, 𝑁)
∙ For each 𝑒 with (𝑒, 𝜙(𝑁)) = 1, the map 𝑥 −→ 𝑥𝑒
is a permutation on 𝐴
∙ ∣ 𝐴 ∣ = 𝜙(𝑁) = 𝑁
∏
𝑝∣𝑁
(
1 − 1
𝑝
)
, where the product is over the distinct
prime numbers dividing 𝑁
∙ Let 𝑓 : 𝐴 −→ 𝐴 with 𝑓(𝑥) = 𝑁 − 𝑥, ∀𝑥 ∈ 𝐴
Then 𝑓 is bijective on 𝐴
∙ Define 𝑆1 =
∑
𝑥∈𝐴
𝑥 and 𝑆1 =
∑
𝑥∈𝐴
𝑁 − 𝑥
Then 2𝑆1 =
∑
(𝑥,𝑁)=1
𝑁 = 𝑁𝜙(𝑁)
𝑆1 = 𝑁𝜙(𝑁)
2
∙ Define 𝐵 = {𝑥 ∈ 𝐴 ∣ (𝑥 + 1, 𝑁) = 1}
∙ ∣ 𝐵 ∣ = 𝜓(𝑁) = 𝑁
∏
𝑝∣𝑁
(
1 − 2
𝑝
)
, where the product is over the distinct
prime numbers dividing 𝑁 ( 𝐵 is non empty if and only if 𝑁 is odd)
∙ Let 𝑔 : 𝐵 −→ 𝐵 with 𝑔(𝑦) = 𝑁 − 𝑦 − 1, ∀𝑦 ∈ 𝐵
Then 𝑔 is bijective on 𝐵
∙ Define 𝑆2 =
∑
𝑦∈𝐵
𝑦 and 𝑆2 =
∑
𝑦∈𝐵
𝑁 − 𝑦 − 1
Then 2𝑆2 =
∑
(𝑥,𝑁)=(𝑥+1,𝑁)=1
𝑁 − 1 = (𝑁 − 1)𝜓(𝑁)
𝑆2 = 𝑁−1
2 𝜓(𝑁)
Problem 2.1 Let 𝑁 ≥ 3 be a given positive integer. Define a tree 𝑇𝑁 as
follows, for each 𝑥 > 1 such that (𝑥, 𝑁) = 1, then there is a unique vertex
𝑣𝑥 ∈ 𝑉 (𝑇𝑁 ) with deg 𝑣𝑥 = 𝑥 and remaining all leaves, then prove that the number
of vertices of 𝑇𝑁 is 𝑛 = 𝜙(𝑁)
2 (𝑁 − 2) + 2.
Proof:
Let 𝑁 ≥ 3 be a positive integer then there exists 𝑥𝑖 such that gcd (𝑥𝑖, 𝑁) =
10. 10
1, ∀𝑖 = {2, 3, . . . , 𝜙(𝑁)}. Construct a tree 𝑇𝑁 such that for each 𝑥𝑖 there exists
a unique vertex 𝑣𝑥𝑖
of degree 𝑥𝑖.
Clearly,
𝑛∑
𝑖=2
𝑑(𝑣𝑥𝑖
) = 2(𝑛 − 1)
(𝑁𝜙(𝑁)
2 − 1) + 𝑛 − (𝜙(𝑁) − 1) = 2𝑛 − 2 where 𝑉 (𝑇𝑁 ) = 𝑛.
Problem 2.2 Let 𝑁 ≥ 3 be a positive integer. Construct a tree 𝑇𝑁 such that
for each 𝑥 > 1 with (𝑥, 𝑁) = (𝑥 + 1, 𝑁) = 1 then there is a unique vertex 𝑣𝑥
of degree 𝑥. Prove that the number of vertices of 𝑇𝑁 is 𝑛 = 𝜓(𝑁)
2 (𝑁 − 3) + 2.
Proof:
Let 𝑁 ≥ 3 be a positive integer then there exists 𝑥𝑖 such that (𝑥𝑖, 𝑁) = 1, ∀𝑖 =
{2, 3, . . . , 𝜓(𝑁)}. Construct a tree 𝑇𝑁 such that for each 𝑥𝑖 there exists a unique
vertex 𝑣𝑥𝑖
of degree 𝑥𝑖.
Clearly,
𝑛∑
𝑖=2
𝑑(𝑣𝑥𝑖
) = 2(𝑛 − 1)
(𝑁−1
2 )𝜓(𝑁) − 1 + 𝑛 − (𝜓(𝑁) − 1) = 2𝑛 − 2 where 𝑉 (𝑇𝑁 ) = 𝑛.
Problem 2.3 Let 𝑁 ≥ 3 be an odd positive integer. Construct a tree 𝑇𝑁 such
that for each 𝑥 > 1 with (𝑥, 𝑁) = (𝑥 + 1, 𝑁) = (𝑥 + 2, 𝑁) = 1 then there
is a unique vertex 𝑣𝑥 of degree 𝑥. Prove that the number of vertices of 𝑇𝑁 is
𝑛 = ?.
Proof:
Let 𝑁 ≥ 3 be an odd positive integer then there exists 𝑥𝑖 such that (𝑥𝑖, 𝑁) = 1
(𝑥𝑖 + 1, 𝑁) = 1 and (𝑥𝑖 + 2, 𝑁) = 1, ∀𝑖 = {2, 3, . . . , 𝜂(𝑁)}. Construct a
tree 𝑇𝑁 such that for each 𝑥𝑖 there exists a unique vertex 𝑣𝑥𝑖
of degree 𝑥𝑖.
.......................................
11. 11
Problem 2.4 Let 𝑛 be a composite positive integer and let 𝑝 be the smallest
prime divisor of 𝑛 with 𝑛
𝑝 = 𝑛1. Prove that if 𝑞 > 𝑛
1
3
1 then 𝑛1
𝑞 is prime where
𝑞 be the smallest prime divisor of 𝑛1.
Solution: Suppose 𝑛1
𝑞 = 𝑎𝑏 where 1 < 𝑎, 𝑏 < 𝑛1
𝑞
Let 𝑟 and 𝑠 be the prime divisors of 𝑎 and 𝑏 respectively, then 𝑟 and 𝑠 are
also prime divisors of 𝑛1, so that
𝑟 ≥ 𝑞 and 𝑠 ≥ 𝑞.
This implies that
𝑞3
= 𝑞.𝑞.𝑞 ≤ 𝑞.𝑟.𝑠 ≤ 𝑞.𝑎.𝑏,
that is, 𝑞3
≤ 𝑛1 which is a contradiction. Therefore, 𝑛1
𝑞 is prime.
Problem 2.5 Let 𝑛 be a composite positive integer and let 𝑝 be the smallest
prime divisor of 𝑛 such that 𝑝2
∣ 𝑛. Prove that if 𝑝2
> 𝑛
𝑝2 then 𝑛
𝑝2 is prime.
Solution: Suppose 𝑛
𝑝2 = 𝑎𝑏 where 1 < 𝑎, 𝑏 < 𝑛
𝑝2
Let 𝑟 and 𝑠 be the prime divisors of 𝑎 and 𝑏 respectively, then 𝑟 and 𝑠 are
also prime divisors of 𝑛, so that
𝑟 ≥ 𝑝 and 𝑠 ≥ 𝑝.
This implies that
𝑝4
= 𝑝2
.𝑝.𝑝 ≤ 𝑝2
.𝑟.𝑠 ≤ 𝑝2
.𝑎.𝑏,
that is, 𝑝4
≤ 𝑛 which is a contradiction. Therefore, 𝑛
𝑝2 is prime.
............................
12. 12
∙ Let 𝑁 = 𝑝𝑞 where 𝑝 and 𝑞 are distinct odd primes
with 3 ∤ 𝑝 and 3 ∤ 𝑞
∙ 𝜂(𝑁) = (𝑝 − 3)(𝑞 − 3)
𝜂(𝑁) = 𝑁 − 3𝑝 − 3𝑞 + 9
3(𝑝 + 𝑞) = 𝑁 + 9 − 𝜂(𝑁)
𝑝 + 𝑞 = 𝑁+9−𝜂(𝑁)
3
∙ Consider the quadratic equation 𝑥2
− (𝑝 + 𝑞)𝑥 + 𝑝𝑞 = 0
∙ Solve the equation and find their roots 𝛼 and 𝛽
∙ Now give 𝑝 and 𝑞
........................................
13. 13
∙ Let 𝑇 be the given graceful tree on 𝑞 edges ( 𝑞 is very large)
∙ Let 𝑓 : 𝑉 (𝐺) −→ {0, 1, 2, . . . , 𝑞} be the graceful labeling of 𝑇 (which is
kept secret)
∙ Choose: {𝑚1, 𝑚2, . . . , 𝑚𝑘} ⊆ 𝐸(𝐺)
such that (𝑚𝑖, 𝑚𝑗) = 1, 𝑖 ∕= 𝑗 where 𝑘 is largest in size
∙ FACT 1: 𝑥 ≡ 𝑑𝑒𝑔(𝑚𝑖) (𝑚𝑜𝑑 𝑚𝑖), 1 ≤ 𝑖 ≤ 𝑘
∙ FACT 2: 𝑥 ≡
∑2
𝑖=1(𝛾(𝐺𝑖) − 𝛼(𝐺𝑖) (𝑚𝑜𝑑 𝑚𝑖), 1 ≤ 𝑖 ≤ 𝑘
......................
3 Conclusion
This paper dealt with two fundamental cryptographic tools that are useful in
such contexts: generalized secret sharing scheme and key distribution schemes.
Both secret sharing schemes and key distribution schemes are used in multi-
party systems. secret sharing schemes enables some predetermined sets of parties
to reconstruct a given secret. These schemes make it possible to store secret
information in a network, such that only good subsets can reconstruct the
information. Furthermore, by using these schemes we can allow only better
subsets to perform action in a system.
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