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The p-adic numbers were introduced by Kurt Hensel in 1897. They were originally used in order to introduce power series in Number Theory. Their applications are various and wide, and they extend some properties of Z and Q to Z_{p} and Q_{p} respectively. The p-adic system for any given prime p extends the metric of rational numbers. Given a natural number n, if we take a fixed primer p then n can be expressed in the form n = å_{i=0} a_{i}p^{i}
In the traditional arithmetic, this can be understood as the expression in base p of a natural number n. For instance, the p-adic expansion of 52 if p is 3 would be 1212, which is 52 in base 3. When it comes to linear equations modulo p, we can always find a solution and it will be unique.

LaTeX4Web 1.4 OUTPUT
** Proof: ** The existence and uniqueness of the solution can be shown with diophantine equations. Take the equation ax + b º 0 mod p. It can be rewritten as ax - py = b, which is indeed a diophantine equation since it is linear and there are two variables. Since p is prime, gcd(a,p) = 1. The necessary condition for a diophantine to have infinite solutions is that b needs to be a multiple gcd(a,p). Clearly, 1 is a divisor of b, and hence this equation has infinite solutions. However, they can be no bigger than m, since we are working with congruences. Once we find a base solution x_{0} and y_{0} (both smaller than m) we can express the general solutions as x = x_{0} + kb/gcd(a,p) y = y_{0} + ka/gcd(a,p). If we add kp to the base solutions, they will become larger than p except if k = 0. It follows that the solution is unique.

LaTeX4Web 1.4 OUTPUT
Once linear equations can be solvable, the next step is to study quadratic equations and find its p-adic expansion. Take the equation x^{2}
º 5 mod p. This has no rational solution in Z, but there are solutions in Z_{p}. We choose a prime in which there will exist a x_{0} that will satisfy the congruence. For instance, if p equals 7 there exists no x_{0}, but it does exist if p equals 11, and the two solutions are x_{0} = 4 and x_{0} = 7. Note that 7 is the inverse of 4 modulo 11, since 7+4 = 11. Then we start working in Z_{11} and we take x_{0} to be 4. We denote the p-adic expansion as a. Recall a is defined to be of the form a = a_{0} + a_{1}p + a_{2}p^{2}
+ a_{3}p^{3}
+ ... Then we want a^{2}
to be 5 and hence 5 = a^{2}
= (a_{0} + a_{1}p + a_{2}p^{2}
+ a_{3}p^{3}
+ ...)(a_{0} + a_{1}p + a_{2}p^{2}
+ a_{3}p^{3}
+ ...) If we multiply we obtain the following expression a_{0}^{2}
+ (2a_{0}a_{1})p + (2a_{0}a_{2}+a_{0}^{2}
)p^{2}
+ (2_{0}a_{3})p^{3}
+ (2a_{0}a_{4}+2a_{1}a_{3}+a_{2}^{2}
)p^{4}
+ ... All parentheses need to be congruent to 0 modulo 11. Also, for each 11 that we encounter in one parenthesis we will take one carry (such as when we add in base 11). We already know that a_{0} equals 4 and then a_0^{2}
= 16 = 5 + 11. We carry one to the next parenthesis, which needs to be congruent to 0 modulo 11. 2a_{0}a_{1} + 1 º 0 mod 11. By substituting a_{0} we find a_{1}. 8a_{1} + 1 º 0 mod 11 8a_{1} º -1 º 10 mod 11. By inspection we see that a_{1} equals 4. If we keep repeating this process we find the first terms of the p-adic expansion, which is a = 4 + 4p + 10p^{2}
+ 4p^{3}
+ 0p^{4}
+ ... With this method any quadratic equation can be solved.