let be n,a and b 3 positive integers..my question is if n>2 ┐how could i solve the congruence:
?..if possible please indicate an "step by step" demonstration... thanks.
It is not always possible. For example, x^(b-1)=1 (mod b) where b is not a prime or a power of a prime. This is based on the integers under multiplication and addition modulo b form a field. Since we are dealing with a field b is either prime or a power of a prime.
Furthermore, if (b-1) does not divide n then there is no solution. This is based the the order of an element most divide the order of the group. Which in this case is not satisfied.
Thus, you made your problem too general. And let me also add that there is no general method of finding:
But there is a badder side of number theory:
Initially number theory was done elementary by basic arguments and properties of the integers and divisibility from the time of Fermat (credited as the father of number theory) to the time of Gauss/Cauchy. That was the end period when "Elementary" number theory was practiced among mathematicians.
During this end period mathematicians made an important discovery of a group (in French groupe was given by Galois) and field. It turned out that many properties of the set of integers can be converted to the study of finite fields and groups. Important figures were Krouncker and Kummer this gave rise to "algebraic" number theory.
But some poeple, Riemann for example, studied functions that work only with integers as countinous/analytic functions. Hence number theory began to use concepts from analysis (analysis is the formal side of calculus) and hence became known as "analytic" number theory. This is what most mathematcians in the number theory field study today (I think). This type of number theory can get real messy and very complicated sometimes. This is the bad side of number theory I was refferening to.
In the future people might study algebraic geometry for number theory because that is the connection Wiles made with his prove on the Last theorem.