Let m, n, r, e be pos. integers. I am trying to show that given m = n (mod e*r) that m^e=n^e (mod r*e^2). I have tried using the binomial theorem but keep running into problems.
Could someone help please?? thanks!
This is a little rough, but perhaps you can clean it up and make everything neat.
The first condition says
$\displaystyle m \equiv n \text{ (mod (er))}$
So we know that
m = a(er) + b
n = c(er) + d
where a, b, c, d are integers and 0 <= b < er and 0 <= d < er. Since m = n (mod er) we know that b = d.
Now, let's form m^e.
$\displaystyle m^e = (a(er) + b)^e = a^e (er)^e + e a^{e - 1}(er)^{e - 1}b + \text{ ... } + e a(er)b^{e-1} + b^e$
Note that every term but the last is proportional to e^2r. Thus m^e has the form:
$\displaystyle m^e = p(e^2r) + b^e$
where p is an integer. We don't need the usual 0 <= b^e < e^2 r condition as you will see.
Now do the same for n^e. Then use the fact that b = d from the initial condition.
-Dan