Thread: congruence question

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!

Originally Posted by arlingtonbassett

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


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.


Note that every term but the last is proportional to e^2r. Thus m^e has the form:


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

thanks for that. it all makes sense except two of the beginning lines...how did you get

m=aer+b and n= cer+d?

everything else makes sense to me.

thanks!

Originally Posted by arlingtonbassett

thanks for that. it all makes sense except two of the beginning lines...how did you get

m=aer+b and n= cer+d?

everything else makes sense to me.

thanks!

Sorry, I wasn't very clear about that was I? We can get both equations by the division algorithm.

-Dan

nm got it thanks again for the help