The article considers the following issues: – It’s of great interest for p and q primes to determine the number of those prime number divisors of a number 1 1 pq A p that are less than p. With this purpose we have considered: Theorem 1. Let’s p and q are odd prime numbers and p 2q 1. Then from various individual divisors of the 1 1 pq A p number, only one of them is less than p. A has at least two different simple divisors; Theorem 2. Let’s p and q are odd prime numbers and p 2q 1. Then all prime divisors of the number 1 1 pq A p are greater than p; Theorem 3. Let’s q is an odd prime number, and p N \{1}, p]1;q] [q 2; 2q] , then each of the different prime divisors of the number 1 1 pq A p taken separately is greater than p; Theorem 4. Let’s q is an odd prime number, and p{q1; 2q1}, then from different prime divisors of the number 1 1 pq A p taken separately, only one of them is less than p. A has at least two different simple divisors. Task 1. Solve the equation 1 2 1 z x y y in the natural numbers x , y, z. In addition, y must be a prime number. Task 2. Solve the equation 1 3 1 z x y y in the natural numbers x , y, z. In addition, y must be a prime number. Task 3. Solve the equation 1 1 z x y p y where p{6; 7; 11; 13;} are the prime numbers, x, y N and y is a prime number. There is a lema with which the problem class can be easily solved: Lemma ●. Let’s a, b, nN and (a,b) 1. Let’s prove that if an 0 (mod| ab|) , or bn 0 (mod| ab|) , then | ab|1. Let’s solve the equations ( – ) in natural x , y numbers: I. 2 z x y z z x y ; VI. (x y)xy x y ; II. (x y)z (2x)z yz ; VII. (x y)xy yx ; III. (x y)z (3x)z yz ; VIII. (x y) y (x y)x , (x y) ; IV. ( y x)x y x y , (y x) ; IX. (x y)x y xxy ; V. ( y x)x y yx , (y x) ; X. (x y)xy (x y)x , (y x) . Theorem . If a, bN (a,b) 1, then each of the divisors (a2 ab b2 ) will be similar. The concept of pseudofibonacci numbers is introduced and some of their properties are found.
On the order of the non-primes of type 6n+5 and 6n+1
