Approach of the value of a rent when non-central moments of the capitalization factor are known: an R application with interest rates following normal and beta distributions Ratio Mathematica Volume 37, 2019, pp. 39-48 39 Solving some specific tasks by Euler's and Fermat's Little theorem Viliam Ďuriš* Abstract Euler's and Fermat's Little theorems have a great use in number theory. Euler's theorem is currently widely used in computer science and cryptography, as one of the current encryption methods is an exponential cipher based on the knowledge of number theory, including the use of Euler's theorem. Therefore, knowing the theorem well and using it in specific mathematical applications is important. The aim of our paper is to show the validity of Euler's theorem by means of linear congruences and to present several specific tasks which are suitable to be solved using Euler's or Fermat's Little theorems and on which the principle of these theorems can be learned. Some tasks combine various knowledge from the field of number theory, and are specific by the fact that the inclusion of Euler's or Fermat's Little theorems to solve the task is not immediately apparent from their assignment. Keywords: Euler's theorem, coding, Fermat's Little theorem, linearcongruences, cryptology, primality testing, Matlab 2010 AMS subject classification: 11A07, 14G50† * Department of Mathematics, Faculty of Natural Sciences Constantine the Philosopher University in Nitra, Tr. A. Hlinku 1, 949 74 Nitra, Slovakia; vduris@ukf.sk. †Received on December 1st, 2019. Accepted on December 10th, 2019. Published on December 31th, 2019. doi: 10.23755/rm.v37i0.485. ISSN: 1592-7415. eISSN: 2282-8214. ©Viliam Ďuriš. This paper is published under the CC-BY licence agreement. Viliam Ďuriš 40 1 Introduction At present, mathematics provides apparatus for virtually all modern coding systems. The first coding system, with only two symbols - a dot and a comma, was Morse code which was used to send the first coding message by American inventor Samuel F. B. Morse in 1844 using an electric telegraph. Binary code encoding has become a better code for message encryption at a later time, in which each coded word consists of blocks of ones and zeroes, and this encoding is still used today [1]. Significant developments in coding occurred in the 20th century when Euler's theorem was used for coding and the coded text could be broadcasted publicly with the message kept secret. The principle of this coding is that the sender assigns a number to a coded word (e.g. 74) and encodes that word using two additional numbers (e.g. 247 and 5), which may be public in such a way that 745(mod 247) = 120 is calculated. This will give you a message “120” that will be sent to the recipient. Since numbers 247 and 5 are public keys, anyone can encode the message "74" to "120", but only the actual recipient can decode it correctly. The essence of the key to the cipher lies in the fact that only the recipient knows that number 247 was compiled as the product of primes 𝑝 = 13 and 𝑞 = 19 and using Euler's theorem searches for the value x for which the congruence is 5𝑥 ≡ 1(mod [(𝑝 − 1)(𝑞 − 1)]). The recipient can easily get the result 𝑥 = 173. Using this figure, the remainder by dividing 120173 by number 247 is found, thereby obtaining the original coded word 74 which can already be assigned to the message [2]. In practice, with this type of coding, the product of two very large primes is used, where the decomposition of the thus obtained number is very difficult, virtually impossible for someone who does not know the product of which two primes have been executed. Despite the fact that the principle of this coding was discovered and started to be used practically in the 20th century, it is actually derived from Euler's knowledge from the 18th century. Most of the results in mathematics in the 18th century stemmed from efforts to solve various separate problems discovered in the 17th century. In this period, the theory of numbers remained more or less in the background, and the only mathematician who dealt with the issues of number theory after 1730 to a greater extent was Euler. In 1736, he proved Fermat's Little theorem which claims that for any natural number a and prime p, 𝑎𝑝−1 ≡ 1(mod 𝑝). Later in 1760, after the introduction of Euler's totient function 𝜑(𝑛) he demonstrated the validity of congruence 𝑎𝜑(𝑚) ≡ 1(mod 𝑚) which is a generalization of Fermat's Little theorem. Euler also dealt with many other Fermat's claims. He also achieved several accomplishments related to the decomposition of certain expressions with the powers of natural numbers and to perfect and friendly numbers. He was also interested in the problem of integer roots of Pell's equation, about which he published several articles, and presented his own Solving some specific tasks by Euler's and Fermat's Little theorem 41 method of solution. Euler has introduced a number of concepts into number theory, such as the quadratic residue and the quadratic nonresidue in the law of quadratic reciprocity and his work and accomplishments, despite the lack of exact evidence in several areas, were generally accepted by respected mathematicians of the 18th and 19th centuries (e.g. Gauss or Legendre) [3]. We would like to mention there's also another principle of coding using Fibonacci numbers and can be seen in [4]. 2 Euler's theorem, Fermat's Little theorem Let us consider two natural numbers a, m where (𝑎,𝑚) = 1. Euler's theorem [5] then states that 𝑚|𝑎𝜑(𝑚) − 1, or that congruence 𝑎𝜑(𝑚) ≡ 1(mod 𝑚) applies. The symbol 𝜑(𝑛) denotes the number of natural numbers smaller than n and relatively prime to n and is called Euler's totient function [6]. To show the validity of Euler's theorem, we will use the basic properties of congruences and residue classes. Let's write all relatively prime numbers to 𝑚 less than 𝑚. These are 𝑥1,𝑥2,⋯,𝑥𝜑(𝑚). Let us further consider the sequence 𝑎𝑥1,𝑎𝑥2,⋯,𝑎𝑥𝜑(𝑚) and indirectly show that all its members are relatively prime to 𝑚. If ∃𝑖:(𝑎𝑥𝑖,𝑚) = 𝑑 > 1, then 𝑑|𝑎𝑥𝑖 ∧ 𝑑|𝑚. Then (𝑑,𝑎) = 1, because (𝑎,𝑚) = 1 ∧ 𝑑|𝑚. In that 𝑑|𝑥𝑖 and numbers 𝑚,𝑥𝑖 are commensurable which is a controversy. Furthermore, let us indirectly show that numbers 𝑎𝑥1,𝑎𝑥2,⋯,𝑎𝑥𝜑(𝑚) are non- congruent modulo 𝑚. ∃𝑖,𝑗:𝑎𝑥𝑖 ≡ 𝑎𝑥𝑗(mod 𝑚). Then 𝑚|𝑎𝑥𝑖 − 𝑎𝑥𝑗 = 𝑎(𝑥𝑖 − 𝑥𝑗) ∧ (𝑎,𝑚) = 1, of which 𝑚|𝑥𝑖 − 𝑥𝑗 and then 𝑥𝑖 ≡ 𝑥𝑗(mod 𝑚), which is a controversy, because 𝑥𝑖 are differently lower from each other than 𝑚, and therefore cannot give the same remainder after division by 𝑚. Before completing the evidence, we recall, that based on the basic properties of congruences, [7] we know that integers 𝑎 and 𝑏 belong to the same class 𝑅𝑖 modulo 𝑚 just when 𝑎 ≡ 𝑏 (mod 𝑚). If we first express the numbers a, b ∈ 𝑅𝑖 in the form 𝑎 = 𝑚 ∙ 𝑞 + 𝑖,𝑏 = 𝑚 ⋅ 𝑝 + 𝑖, then 𝑎 − 𝑏 = 𝑚(𝑞 − 𝑝), which means 𝑚|𝑎 − 𝑏, and thus 𝑎 ≡ 𝑏(mod 𝑚). On the other hand, let us assume that 𝑎 ≡ 𝑏 (mod 𝑚) and 𝑎 = 𝑚𝑞 + 𝑖,𝑏 = 𝑚𝑝 + 𝑗 (0 ≤ 𝑖, 𝑗 < 𝑚). For example, it is supposed that 𝑖 > 𝑗. Since 𝑎 ≡ 𝑏 (mod 𝑚), 𝑚|𝑎 − 𝑏. But then 𝑚 (𝑎 − 𝑏) = [𝑚(𝑞 − 𝑝) + (𝑖 − 𝑗)], of which 𝑚 (𝑖 − 𝑗). This would be a controversy though, because 0 < 𝑖 − 𝑗 < 𝑚. Similarly, a controversy arises even with the assumption 𝑖 < 𝑗. Therefore 𝑖 = 𝑗 must hold, hence the numbers 𝑎 and 𝑏 belong to the same residual class modulo 𝑚 with 𝑎 ≡ 𝑏 (mod 𝑚). As the class representative does not matter, we can write 𝑎𝑥1 ∙ 𝑎𝑥2 ∙ ⋯∙ 𝑎𝑥𝜑(𝑚) ≡ 𝑥1 ∙ 𝑥2 ∙ ⋯∙ 𝑥𝜑(𝑚)(mod 𝑚). Then 𝑚|𝑎𝑥1 ∙ 𝑎𝑥2 ∙ ⋯∙ 𝑎𝑥𝜑(𝑚) − 𝑥1 ∙ Viliam Ďuriš 42 𝑥2 ∙ ⋯∙ 𝑥𝜑(𝑚) = (𝑎 𝜑(𝑚) − 1)𝑥1 ∙ 𝑥2 ∙ ⋯∙ 𝑥𝜑(𝑚). Since (𝑚,𝑥1 ∙ 𝑥2 ∙ ⋯∙ 𝑥𝜑(𝑚)) = 1, then 𝑚|𝑎 𝜑(𝑚) − 1. If 𝑚 is a prime number and 𝑝 ∤ 𝑎, then 𝜑(𝑚) = 𝑚 − 1 and we get Fermat's Little theorem 𝑎𝑝−1 ≡ 1(mod 𝑝) directly from Euler's theorem. A variation of Fermat's Little theorem can be used to test primality [8]. If there exists 𝑎 ∈ {2,⋯,𝑛 − 1}, 𝑛 > 3, where 𝑎𝑛−1 ≢ 1(mod 𝑛), then n is a composite number and we call it Fermat's witness for the compositeness of number 𝑛 [9]. Fermat's primality test can be suitably algorithmically presented in a selected computational environment (e.g. Matlab). The algorithm consists of two steps: a) we randomly select number a for which 1 < 𝑎 < 𝑛 b) it is tested whether congruence 𝑎𝑛−1 ≡ 1(mod 𝑛) is satisfied If congruence 𝑎𝑛−1 ≡ 1(mod 𝑛) is satisfied, the number 𝑛 may or may not be a prime number. If congruence is not satisfied, the number 𝑛 is not a prime and number 𝑎 is the Fermat's witness for the compositeness of 𝑛. Fermat's primality test works well for numbers that are not products of prime numbers different from each other. It can be demonstrated that if we test the number 𝑛, which is not the product of different prime numbers, hence there is such a prime 𝑝 where 𝑝2|𝑛, then with a probability of at least 75% we can choose between numbers 2,⋯,𝑛 − 1 such a number which will be the Fermat's witness for the compositeness of 𝑛 [9]. First, in Matlab, we create a function that helps us test congruence 𝑎𝑛−1 ≡ 1(mod 𝑛) generally for two given numbers 𝑎 and 𝑛. The function will calculate the value 𝑎𝑛−1 mod 𝑛 which we will compare with 1 within the residue classes. function res = test_congruence(a, n) expn = n - 1; res = 1; while expn ~= 0 if rem(expn, 2) == 1 res = rem(res * a, n); end expn = floor(expn / 2); a = rem(a^2, n); end The second function randomly generates 𝑎 ∈ {2,⋯,𝑛 − 1} and we look for the Fermat's witness for the compositeness of 𝑛. Solving some specific tasks by Euler's and Fermat's Little theorem 43 function test_fermat(n, cnt) fo = false; ii = 1; while (ii <= cnt) && (~fo) a = 1 + unique(ceil((n - 2) * rand(1, 1))); tc = test_congruence(a, n); if(tc ~= 1) fermat_witness = a; fo = true; else ii = ii + 1; end end if fo disp(['Number ' num2str(n) ' is a composite number.']); disp(['Number ' num2str(fermat_witness) ' is a Witness for the compositeness of ' num2str(n) '.']); else disp(['Number ' num2str(n) ' can be a prime or a composite number.']); end The created test function is activated through the command line for any number 𝑛. >> test_fermat(223, 1) Number 223 can be a prime or a composite number. >> test_fermat(273, 1) Number 273 is a composite number. Number 220 is a Witness for the compositeness of 273. 3 Euler's, Fermat's Little theorem applications In this section, we have selected and compiled a number of specific tasks [10], [11] that guide on how to solve certain types of tasks using Euler's or Fermat's Viliam Ďuriš 44 Little theorem. We remark that for a natural number 𝑛 greater than 1 in canonical decomposition 𝑛 = 𝑝1 𝛼1 …𝑝𝑘 𝛼𝑘 it holds that 𝜑(𝑛) = 𝑛(1 − 1 𝑝1 )(1 − 1 𝑝2 )…(1 − 1 𝑝𝑘 ) [6] Example 3.1. First, we demonstrate that if we divide number 1724 by number 39, the remainder 1 is obtained. Solution. It is determined that 𝑎 = 17, 𝑚 = 39. (39,17) = 1 and Euler's theorem can be applied. Let us calculate 𝜑(𝑚) = 𝜑(39) = 39(1 − 1 3 )(1 − 1 13 ) = 24. Then according to Euler's theorem 39|1724 − 1, thus ∃𝑘 ∈ ℤ:1724 − 1 = 39𝑘. Then we can write 1724 = 39𝑘 + 1, and 1 is obtained as a remainder. Example 3.2. It is demonstrated that 𝑝 and 8𝑝2 + 1 are simultaneously prime just when 𝑝 = 3. Solution. 1. First, 𝑝 = 3. Then 8𝑝2 + 1 = 8 ∙ 9 + 1 = 73, which is a prime. 2. Now let 𝑝 and 8𝑝2 + 1 be prime numbers simultaneously. 8𝑝2 + 1 is adjusted as 8𝑝2 + 1 = 8𝑝2 − 8 + 9 = 8(𝑝2 − 1) + 9. Let 𝑝 be a prime number other than 3. Then (𝑝,3) = 1 a 3|𝑝𝜑(3) − 1 = 𝑝2 − 1 . Since 3|𝑝2 − 1, then 8(𝑝2 − 1) ∧ 3|9, then 3|8(𝑝2 − 1) + 9 = 8𝑝2 + 1 and 8𝑝2 + 1 would not be a prime number, which is a controversy, thus 𝑝 = 3. Example 3.3. We show if 𝑎 is not divisible by 5, then only one number from 𝑎2–1, 𝑎2 + 1 is divisible by 5. Solution. If 𝑎 is a multiple of 5, according to Euler's theorem 𝑎4 − 1 is a multiple of 5. Then only one of numbers 𝑎2 − 1 and 𝑎2 + 1 is a multiple of 5. They both concurrently cannot be, otherwise their difference would also be divisible by number 5, which is not, since (𝑎2 + 1) − (𝑎2 − 1) = 2. Example 3.4. We find all primes 𝑝 for which 5𝑝 2 + 1 ≡ 0(𝑚𝑜𝑑 𝑝2). Solution. The prime number 𝑝 = 5 does not satisfy the task and at the same time (𝑝,5) = 1. Then according to Euler's theorem 5𝑝−1 ≡ 1 (mod 𝑝). By exponentiation to 𝑝 + 1 we get 5𝑝 2−1 ≡ 1 (mod 𝑝), of which 5𝑝 2 ≡ 5 (mod 𝑝). Next, the task assignment states that 5𝑝 2 + 1 ≡ 0 (mod 𝑝2), that implies 5𝑝 2 ≡ −1 (mod 𝑝2) and also 5𝑝 2 ≡ −1 (mod 𝑝). Then congruences 5𝑝 2 ≡ Solving some specific tasks by Euler's and Fermat's Little theorem 45 5 (mod 𝑝) and 5𝑝 2 ≡ −1 (mod 𝑝) hold that 5 ≡ −1 (mod 𝑝). Then 𝑝|6. In that 𝑝 = 2 or 𝑝 = 3. For 𝑝 = 2 it holds that 54 + 1 ≡ 14 + 1 = 2 ≢ 0 (mod 4). For 𝑝 = 3 it holds that 59 + 1 = 56 ∙ 53 + 1 ≡ 53 + 1 = 126 ≡ 0 (mod 9). Then, the only prime number satisfying the task is 𝑝 = 3. Example 3.5. For the odd number 𝑚 > 1 we find the remainder after division of 2𝜑(𝑚)−1 by number 𝑚. Solution. Euler's theorem implies that 2𝜑(𝑚) ≡ 1 ≡ 1 + 𝑚 = 2 ∙ 1+𝑚 2 = 2𝑟(mod 𝑚) where 𝑟 is a natural number 0 ≤ 𝑟 < 𝑚. The basic properties of congruences [7] determine that if 𝑎 ≡ 𝑏(mod 𝑚) and 𝑑 is an integer with properties 𝑑|𝑎, 𝑑|𝑏, (𝑑,𝑚) = 1, then 𝑎 𝑑 ≡ 𝑏 𝑑 (mod 𝑚). Indeed 𝑎 = 𝑎1𝑑, 𝑏 = 𝑏1𝑑 and according to assumption 𝑚|(𝑎 − 𝑏), it holds that 𝑚|𝑑(𝑎1 − 𝑏1). Since (𝑑,𝑚) = 1, it holds that 𝑚|(𝑎1 − 𝑏1). Then 𝑎1 ≡ 𝑏1(mod 𝑚), thus 𝑎 𝑑 ≡ 𝑏 𝑑 (mod 𝑚). Then, however, we can divide both sides of the congruence 2𝜑(𝑚) ≡ 2𝑟(mod 𝑚) by their common divisor, number 2, which is relatively prime to the modulo. Then 2𝜑(𝑚)−1 ≡ 𝑟(mod 𝑚), and thus the remainder sought is 𝑟 = 1+𝑚 2 . Example 3.6. We find the last two digits of number 13742. Solution. The task leads to the search for the remainder when dividing number 13742 by number 100. Since (137,100) = 1, according to Euler's theorem it holds that 137𝜑(100) − 1 is a multiply of 100 (100|137𝜑(100) − 1). Next 𝜑(100) = 100(1 − 1 2 )(1 − 1 5 ) = 40. Then 13740 − 1 is a multiply of 100. Therefore 13742 = 137213740 − 1372 + 1372 = 1372(13740 − 1) + +1372 = 1372(13740 − 1) + (100 + 37)2 = 100𝑘 + (100 + 37)2. Next, we use the formula (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2. Then 13742 = 100𝑘 + 100𝑙 + 372 = 100𝑛 + 1369 = 100𝑛 + 1300 + 69 = 100𝑚 + 69. Thus, the remainder sought is 69. Example 3.7. We find the last 2 digits of number 𝑎 = 13747. Solution. The last 2 digits of number a are again obtained as the remainder after dividing the number 𝑎 by 100. (100,137) = 1 and Euler's theorem can be applied. Then 100|137𝜑(100) − 1, thus 100|13740 − 1. Then 13740 − 1 is a multiply of 100 and 13740 = 100𝑘 + 1. Viliam Ďuriš 46 Let us calculate 13747 = 13740 ∙ 1377 = (100𝑘 + 1) ∙ 1377 = 100𝑘 ∙ 1377 + 1377 Number 100𝑘 ∙ 1377 cannot specify last 2 digits (ending with 2 zeroes), and so just number 1377 has the last 2 digits of the given number 𝑎. Next, the binomial theorem is applied. 1377 = (130 + 7)7 = ( 7 0 )1307 + ( 7 1 )1306 ∙ 7 + ⋯+ ( 7 6 )130 ∙ 76 + ( 7 7 )77. In this summation only the members ( 7 6 )130 ∙ 76 a ( 7 7 )77 decide the last two digits (other contribute zeroes in last two digits). Their summation is calculated as ( 7 6 )130 ∙ 76 + ( 7 7 )77 = 130 ∙ 77 + 77 = 131 ∙ 77 = 107884133. Overall, we get the last 2 digits of the number 𝑎 = 13747 which are 33. Example 3.8. We find the remainder when dividing (8570 + 1932)16 by number 21. Solution. According to the binomial theorem 8570 = (84 + 1)70 = ( 70 0 )8470 + +( 70 1 )8469 ∙ 1 + ⋯+ ( 70 69 )84 ∙ 169 + ( 70 70 )170. We see that number 21 can be removed from every member except the last one. Then 8570 = (84 + 1)70 = 21𝑛 + 1. Because 𝜑(21) = 12, 1912 − 1 is a multiply of 21 (applying Euler's theorem), then 1932 = 198(1924 − 1) + 198 = 21𝑚 + 198. Therefore (8570 + 1932)16 = (21𝑛 + 1 + 21𝑚 + 198)16 = (21𝑘 + 1 + (21 − 2)8)16 = (21𝑞 + 1 + 28)16 = (21𝑟 + 5)16 = 21𝑡 + 516 = 21𝑡 + 54(512 − 1) + 54 = 21𝑡 + 21𝑟 + 625 = 21𝑢 + 16. The remainder sought is 16. Example 3.9. We demonstrate if 𝑥𝑝 + 𝑦𝑝 = 𝑧𝑝 where 𝑝 is a prime number, then 𝑥 + 𝑦 – 𝑧 is a multiply of 𝑝. Solution. According to Fermat's Little theorem, if 𝑝 is a prime and 𝑝 ∤ 𝑥, then 𝑥𝑝−1 ≡ 1(mod 𝑝), that means 𝑝|𝑥𝑝−1 − 1 and thus 𝑝|𝑥(𝑥𝑝−1 − 1) = 𝑥𝑝 − 𝑥. Similarly, 𝑝|𝑦𝑝 − 𝑦, 𝑝|𝑧𝑝 − 𝑧. Therefore we can write 𝑥𝑝 = 𝑝𝑡1 + 𝑥, 𝑦 𝑝 = 𝑝𝑡2 + 𝑦 a 𝑧 𝑝 = 𝑝𝑡3 + 𝑧. If we substitute in the equation 𝑥 𝑝 + 𝑦𝑝 = 𝑧𝑝, we get 𝑝(𝑡3 − 𝑡1 − 𝑡2) = 𝑥 + 𝑦 − 𝑧 after adjustment, thus 𝑥 + 𝑦 – 𝑧 is a multiply of 𝑝. These examples are the basis for understanding the principle of working with large numbers using congruences through Euler's and Fermat's Little theorem. Congruences are a modern and irreplaceable security tool for protecting data by a public key. It is important to realize that the public key uses such large numbers for which there is no effective method of decomposing to primes even in today's modern computer age. That is why Euler's theorem plays its role in encryption even today, when encryption uses keys of up to 256 bits in length Solving some specific tasks by Euler's and Fermat's Little theorem 47 and deciphering the word while trying out all the options would probably take more years than the age of the universe is. 4 Conclusion The paper points out some specific applications suitable for presenting and understanding the basic principle of Euler's and Fermat's Little theorems which are currently used in cryptography. Leonhard Paul Euler was such a great mathematician that many of the principles he had known almost 300 years ago were actually used by contemporary society. Euler, nicknamed as a "magician" in his time, had a great influence not only on number theory, but also on mathematical analysis or graph theory. 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