Prime numbers and their variations are extremely useful in applied research areas such as cryptography, feedback and control in engineering. In this paper we discuss about prime numbers, perfect numbers, even perfect and odd perfect numbers, amicable numbers, semiprimes, mersenne prime numbers, triangular numbers, distribution of primes, relation between \(\pi\) and prime numbers. In the process we also obtain interesting properties of some of them and raise a set of open problems for further exploration.
It is widely acknowledged that mathematics is a base for all science and engineering concepts and number theory is the base for mathematics. A vital and challenging task of answering questions in number theory depends heavily on finding an integer’s unique factor decomposition. When researchers show involvement, new research approaches emerge. Number theory is largely classified as elementary number theory dealing with divisibility and congruence; analytic number theory supported by complex analysis; algebraic number theory due to the evolution of the study on ring of integers; geometric number theory depending on the perspectives of geometry to explain the pattern of distribution and computational number theory that uses computer algorithms to solve certain mind boggling problems. The universe now witness the rapid use of the concepts of number theory in applied fields such as physics, biology, chemistry, communication, acoustics, electronics, cryptography, computing etc, [1].
Proposition 1. Suppose that \(p\) is a prime. Then \(p^2 + p + 1 \neq r^3\) for some \(r \in Z^+\).
Proof. Suppose that \(p^2 + p + 1 \neq r^3\) for some \(r \in Z^+\) then \(r^3-1=p(p+1)\) implies \(p(p+1)=(r-1)(r^2+r-1)\). As \(p\) is a prime, by one of the properties of primes, we have either \(p | r -1\) or \(p | r2 + r + 1\). . If \(p | r -1\) then clearly \(p \leq r -1\). So \(p + 1\) \(r \rightarrow r^3\) \((p + 1)^3 > p^2 + p + 1 = r^3\), a contradiction. Next if \(p | r^2 + r + 1\) implies \(p = 3\) and \(r = 1\), else, \(p = 3\mathcal{L} + 1\) for some \(\mathcal{L} \in Z^+\). Hence \(p^2 + p + 1 \equiv 3 (mod 9)\). But it can be easily seen that a perfect cube cannot be written as \(9t -3\) for some \(t \in Z^+\).
Proposition 2. Let \(p_1\), \(p_2\), \(p_3\) be odd primes and \(p_{1}^2 + p_1+ 1 = (p_{2}^2 + p_2+ 1)( p_{3}^2 + p_3+ 1)\) then either \(p_{2}^2 + p_2+ 1\) or \(p_{3}^2 + p_3+ 1\) is not a prime.
Proof. Suppose that both \(p_{2}^2 + p_2+ 1\) and \(p_{3}^2 + p_3+ 1\) are primes. Then one of (a) \(p_1\equiv p_2(mod p_{2}^2 + p_2+ 1 )\) or \(p_1\equiv p_{2}^2(mod p_{2}^2 + p_2+ 1 )\) or (b) \(p_1\equiv p_3(mod p_{3}^2 + p_3+ 1 )\) or \(p_1\equiv p_{3}^2(mod p_{3}^2 + p_3+ 1 )\) hold good. Without loss of generality assume that a) holds good. As \(p_1 > p_2\) and \(p_1\) is a prime it readily follows that \(p_1 \neq p_{2}^2\). Further, the other choice for \(p_1\) is \(p_{2}^2+p_{2}+1+p_{2}=(p_2+1)^2\). As \(p_2\) is an odd prime \((p_2 + 1)\) is even and hence \((p_2 + 1)^2\) is also even, and contradiction. Finally, \(p_1\) also cannot be equal \(p_{2}^2+p_{2}+1+p_{2}^2=2p_{2}^2+p_2+1\) as \(2p_{2}^2+p_2+1\) is even. From this it follows that \(p_1\geq 2(p_{2}^2+p_{2}+1)+p_2 > 2(p_{2}^2+p_{2}+1)\). On similar lines we can also get \(p_2 \geq 2(p_{3}^2+p_{3}+1). So 3(p_{2}^2+p_{2}+1)(p_{3}^2+p_{3}+1)=p_{1}^2+p_1+1 >p_{1}^2 > 4(p_{2}^2+p_{2}+1)(p_{3}^2+p_{3}+1)\). This is again absurd. Hence one of the two \(p_{2}^2+p_{2}+1\), \(p_{3}^2+p_{3}+1\) is not a prime.
Proposition 3. Let \(p_1\), \(p_2\) be two distinct odd primes and \(p_{1}^2+p_{1}+1\) is also a prime. If \(p_{1}^2+p_{1}+1|p_{2}^2+p_{2}+1\) then \(p_{1}^2+p_{1}+1 < p_{2}/2\).
Proof. It is easy to see that \(p_2=k(p_{1}^2+p_{1}+1)-p_{1}\) or \(p_2=k(p_{1}^2+p_{1}+1)-p_{1}^2\). Also \(p_{2}\) cannot be equal to \(p_{1}^2\). Moreover \(P_{2}\) canot be equal to \((p_{1}^2+p_{1}+1)-p_{1}\) and \((p_{1}^2+p_{1}+1)-p_{1}^2\) as both these terms are even. So we have \(p_{2}\geq p_{1}+2(p_{1}^2+p_{1}+1) > p_{1}^2+p_{1}+1\).
Proposition 4. It is not possible to produce three odd primes \(p_1\), \(p_2\), and \(p_3\) with \(p_{1}^2+p_{1}+1\) and \(p_{2}^2+p_{2}+1\) primes such that \(p_{3}^2+p_{3}+1=3(p_{1}^2+p_{1}+1)(p_{2}^2+p_{2}+1)\).
Proof. Suppose we assume the contrary then by invoking Proposition 3 one can deduce \((p_{1}^2+p_{1}+1)< p_{3}/2\) and \((p_{2}^2+p_{2}+1)< p_{3}/2\). So, \(p_{3}^2+p_{3}+1=3(p_{1}^2+p_{1}+1)(p_{2}^2+p_{2}+1) < 3(p_{3}/2)(p_{3}/2)=\frac{3p_{3}^2}{4}< p_{3}^2+p_{3}+1\) yields a contradiction to conclude the proof.
Proposition 5. If \(d^{*}|n\) then \(\frac{\sigma(d^{*})}{d^{*}}\leq \frac{\sigma(n)}{n}\) and equality holds good only if \(d^{*}=n\).
Proof. If \(d|n\) then \(n=\mathcal{L}d\) for some \(\mathcal{L}\), so \(\mathcal{L}=(n/d)|n\). Hence \(\sigma (n)=\sum _{d|n}d=\sum _{d|n}\frac{n}{d}=n\sum_{d|n}\frac{1}{d}\). If \(d^{*}\) is proper factor of \(n\) then \(\sigma(n)/n=\sum _{d|n}(1/d)>\sum_{d’|d^{*}}\frac{1}{d’}=\frac{\sigma (d^{*})}{d^{*}}\).
Proposition 5 implies that \(\sigma (n)=\sum _{d|n}d=\sum _{d|n}\frac{n}{d}=n\sum_{d|n}\frac{1}{d}=2n\). So \(\sum_{d|n}1/d=2\) is \(n\) is perfect. Euclid established that if \(2^n-1\) is a prime then \(2^{n-1}(2^n-1)\) is a perfect number. This is because, the only prime divisors of \(2^{n-1}(2^n-1)\) are \(2^n-1\) and \(2\). So \(\sigma[2^{n-1}(2^n-1)\sigma(2^{n-1})\sigma(2^n-1)=\left(\frac{2^n-1}{2-1}\right)2^n=2\{2^{n-1}(2^n-1)\}\). Similarly, \(2^n-1\) is a prime then \(n\) itself is a prime. This is because, \(y^n-1=(y-1)(y^{n-1}+y^{n-2}+…+y+1)\). If \(n=\mathcal{L}n\) then \(2^n-1=(2^{\mathcal{L}})^{m}-1=(2^{\mathcal{L}}-1)(1+2^{\mathcal{L}}+…+(2^{\mathcal{L}})^{m-1})\). Hence \((2^{\mathcal{L}}-1)|2^n-1\), which is a prime, a contradiction. Its converse is not true. For example, 11 is a prime but \(2^{11}-1=2047=23 \times 89\) is not a prime. Perfect numbers have some interesting properties. We call a number a triangular number if it can be arranged as a triangular lattice.
Proposition 6. If \(n\) is an even perfect number then it is triangular.
Proof. We deem \(\mathcal{L}\) as a triangular number if \(\mathcal{L}=\sum_{j=1}^{t-1}j=1+2+…+(t-1)t/2\) for some t. However note here that \(n^{*}=2^{n-1}(2^n-1)=\frac{1}{2}(2^n)(2^n-1)\). So an even perfect number is triangular.
Proposition 7. \(n^{*}=2^{n-1}(2^n-1)\) is perfect then \(n=1^3+3^3+…+(2^{(n-1)/2}-1)^3\).
Proof. We know that \(\sum_{j=1}^{n}=n^{2}(n+1)^2/4\). Let \(\mathcal{L}=2^{(n-1)/2}\). Then \(n=1^3+3^3+…+(2\mathcal{L}-1)^3=(1^3+2^3+…+(2\mathcal{L})^3)-(2^3+4^3+…+(2\mathcal{L})^3) =[(2\mathcal{L})^2(2\mathcal{L}+1)^2]/4-2^3[(\mathcal{L})^2(\mathcal{L}^2+1)^2]/4 (\mathcal{L})^2(\mathcal{L}^2-1)\). Put \(\mathcal{L}=2^{(n-1)/2}\) to complete the proof.
Proposition 8. All even perfect numbers have its unit digit as 6 or 8.
Proof. Note that if \(p\) is an odd prime then it is either congruent to 1 (mod 4) or congruent to 3 (mod 4). Let it be first congruent to 1 (mod 4). Then \(2^{n-1}(2^n-1)=2^{4\mathcal{L}}(2^{4\mathcal{L}+1}-1)=16^{\mathcal{L}}(2\times 16^{\mathcal{L}}-1)\equiv 6^{\mathcal{L}}(2\times 6^{\mathcal{L}}-1)\equiv 6(12-1) \equiv 6(mod 10) \). In a similar way, \(2^{n-1}(2^n-1)=4\times 16^{\mathcal{L}}(8\times 16^{\mathcal{L}}-1)\equiv 4 \times 6 (8\times 6-1)\equiv 4(8-1)\equiv 8(mod10)\). In both instances we made use of the fact that \(6^{\mathcal{L}}\equiv 6(mod 10)\).
A pair of integers \((r, s)\) with \(r, s \in Z^+\) and \(r < s\) is called amicable if each of \(r\) and \(s\) is the sum of the proper divisors of the other. Euler was the first mathematician to investigate amicable pairs in a systematic manner. He looked at amicable pairs of the form \((rm, rn)\) where \(r\) is a given known factor and m, n are unknowns with \(gcd(m, n) = 1\). By setting \(r = 2^s\), \(s \in Z^+\), \(m = pq\), \(n = t\) with \(p, q, t\) are distinct primes, one can obtain the forms of Thabit period. By taking \(r = 3^2 \times 7 \times 13, m = pq\) and \(n = t,\) Euler got the first amicable pair whose elements are odd, \((3^2 \times 7 \times 13 \times 5 \times 17 = 69615, 3^2 \times 7 \times 13 \times 107 = 87633)\). A pair of positive numbers \((r_1, r_2)\) is called a breeder if \(r_1 + r_2x = \sigma (r_1) = \sigma (r_2) (x + 1)\) have a positive integer solution x. Borho suggested a rule to find an amicable pair by using breeders. Suppose that \((rd, r)\) be a breeder, with integer solution x. If a pair of distinct prime numbers \(p_1, p_2\) exist with \(gcd(r, p_1p_2) = 1\), fulfilling the bilinear equation \((p_1 -x) (p_2 -x) = (x + 1) (x + d)\) and if a third prime \(p_3\) exists with \(gcd(rd, p_3) = 1\) such that \(p_3 = p_1 + p_2 + d\) then \((rdp_3, rp_1p_2)\) is an amicable pair. Another great mathematician Erdos suggested: for a given \(s \in Z^+\) if \(x_j, j = 1, 2, … \)are solutions of \(\sigma(x) = s\) then any pair \((x_i, x_k), i \neq k\) for which \(x_i + x_k = s\) is an amicable pair.
We call a set of three integers \(m_1, m_2, m_3\) form an amicable triple if \(\sigma(m_1) = m_1 + m_2\), \(\sigma(m_2) = m_2 + m_2\) and \(\sigma(m_3) = m_3 + m_1\), and a set of four integers \(m_1, m_2, m_3\) and \(m_4\) is said to form an amicable quadruple if \(\sigma(m1) = m2, \sigma(m2) = m3, \sigma(m3) = m4\) and \(\sigma(m4) = m1\). Fig. 1 shows a python program to find perfect numbers, amicable pairs, amicable triples and amicable quadruples. Accordingly we record that \(6, 28, 496, 8128\) are perfect numbers, \((220, 284), (1184, 1210), (2620, 2924), (5020, 5564)\) are amicable pairs, \((1980, 2016, 2556), (9180, 9504, 11556)\), are amicable triples and \((1236402232, 1369801928, 1603118392, 1412336648)\) is an amicable quadruple.
It is worth to re-publicize the following open questions. 1) Are there an infinite number of amicable pairs 2) Is there an amicable pair whose elements are of different parity 3) Is there an amicable pair whose elements are relatively prime 4) Are there amicable pairs whose elements have smallest but different prime divisors Some of the amicable pairs are (\(1184, 1210), (2620, 2924), (5020, 5564)\) etc. These numbers have the feature that one represents the other. This stands for love, harmony, friendship etc. There numbers have lot of applications in astrology in casting horoscopes and magic.
When analyzing the logic gates, one can analyze both AND gate and OR gate together. This is because both generate almost identical information. For any given state of AND or OR there is a 22% probability that the state could be an error. This occurs when at least two I/O’s are defined at the same time. If one of the I/O is defined there is 50% probability that at least one more I/O will be deduced. Given a gate with random inputs and no error there is 62% chance that the gate is fully define. It is wonderful to observe that the probability of error do not get altered while changing from the Full-Adder to Array Multiplier Cell. As the I/O count increases, the ability to generate new information decreases. This is crucial to factorize the semiprime numbers, because as the semiprimes grows, the complexity and I/O count of the array multiplier will grow and result in lesser quantity of information generation. With a fully operating reversible array multiplier, one can test a huge quanta of possibilities for the semiprime number and perform data analysis to quantity the information generated. Python code is involved to generate prime numbers for various binary lengths, from 2 bits to 512 bits. The prime numbers are then multiplied and fed into the output of the reversible array multiplier. After deducing all information, it can be saved as a CSV file. Then one can depict it as a graph to show the information as a function of the size of the semiprime number.