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• May 1, 2002
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--- "S.R.Sudarshan Iyengar" <gayathrisr@...> wrote:
> >> Prove that for each prime number (that is not 2 or 5) and each
> >> natural number N ,there is a power of P that ends with
> 000...001,
> >> where the number of zeroes is N-1
>
> >Look at the finite fields F_(2^N) and F_(5^N). What are the
> orders,
> >of the element p in the multiplicative groups of this field?
> >Call these orders o5 and o2. What is p^lcm(o2,o5) in this field?

> Well Mr. Phil

Just 'Phil', no need for 'Mr.' :-)

> what do you mean by F_(2^N) do you mean that it is
> the field
> containing all elements >=1 which are relatively prime to 2^N?

Nope, I mean the finite field with p^N elements, where in this case
p=2 and p=5.

The statement, seen a few posts back, that p \in {1,3,7,9} (mod 10)
implies p^4 == 1 (mod 10) is a statement that Order(p) divides 4 for
such p in (Z/10Z)* . The statement that if q == 1 (mod p^n), then
q^10 == 1 (mod p^(n+1)) for p=2,5 effectively says that the order of
q in F_(2^(n+1)) divides 10 times the order of q in F_(2^n).

I think my hints get to the root of the problem, but the
demonstrative proof we saw was probably simpler as it didn't care
what the order actually was, it didn't matter - as long as you raise
a number by a multiple of its order you get the result you need.
(e.g. raising 10n+1 to the power 4 in order to get a number of the
form 10n+1 is a tad unnecessary, but it works.)

Phil

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