p-adic numbers

LaTeX4Web 1.4 OUTPUT 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 Zp and Qp 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 = åi=0 aipi
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 x0 and y0 (both smaller than m) we can express the general solutions as
x = x0 + kb/gcd(a,p)
y = y0 + 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
x2 º 5 mod p.
This has no rational solution in Z, but there are solutions in Zp. We choose a prime in which there will exist a x0 that will satisfy the congruence. For instance, if p equals 7 there exists no x0, but it does exist if p equals 11, and the two solutions are x0 = 4 and x0 = 7. Note that 7 is the inverse of 4 modulo 11, since 7+4 = 11. Then we start working in Z11 and we take x0 to be 4. We denote the p-adic expansion as a. Recall a is defined to be of the form
a = a0 + a1p + a2p2 + a3p3 + ...
Then we want a2 to be 5 and hence
5 = a2 = (a0 + a1p + a2p2 + a3p3 + ...)(a0 + a1p + a2p2 + a3p3 + ...)
If we multiply we obtain the following expression
a02 + (2a0a1)p + (2a0a2+a02 )p2 + (20a3)p3 + (2a0a4+2a1a3+a22 )p4 + ...
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 a0 equals 4 and then
a_02 = 16 = 5 + 11.
We carry one to the next parenthesis, which needs to be congruent to 0 modulo 11.
2a0a1 + 1 º 0 mod 11.
By substituting a0 we find a1.
8a1 + 1 º 0 mod 11
8a1 º -1 º 10 mod 11.
By inspection we see that a1 equals 4. If we keep repeating this process we find the first terms of the p-adic expansion, which is
a = 4 + 4p + 10p2 + 4p3 + 0p4 + ...
With this method any quadratic equation can be solved.