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- Jun 19th 2008, 04:10 PM #1

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## 2^a 2^b == 2^b+3 (mod 9)

I did a case analysis on b mod 6 and found the congruences a has to satisfy, but I don't think my answer is right... Any good resources on modular arithmetic are appreciated also!

EDIT: Sorry, clarification. I am looking for necessary and sufficient conditions that a and b satisfy the congruence. Like I said, I broke into cases on a and b mod 6 and got as my answer

but the process is ugly and I don't think I'm right - -;

- Jun 19th 2008, 06:24 PM #2

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- Jun 20th 2008, 03:30 AM #3

- Jun 20th 2008, 08:36 AM #4
Hi

-->

So one of the conditions will be that is a multiple of 3.

- If a is even, that is to say a=2k, then . But .

Therefore << OK.

- If a is odd, that is to say a=2k+1, << NOT OK.

------------------------------------------------

Now, .

denotes Euler's totient function, and is used for Euler's theorem :

if x and n are coprime, then :

, therefore,

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Now, you know that a is even. So the only possibilities are :

And from here, I don't know any other way than trial and errors , and being helped by the property .

I hope this helps you...

- Jun 20th 2008, 08:37 AM #5

Thus:

In fact (we can also go from bottom to top in the above argument) supposing

If but (one of those options) for all naturals , thus it is absurd to assume that b is even

If (1) which is possible and in fact there are infinitely many s that satisfy that congruence.

Note that: ,since it follows that satisfies (1) for all natural numbers (in fact it must be of that form)

In conclusion must be odd and for some natural number

- Jun 20th 2008, 09:52 AM #6
## new release

If ,

-->

--------------------------

If

-->

We want .

.

We want b such that . Dividing by 3, we get

We can use Euler's theorem (or Fermat's little theorem, since 3 is prime).

2 & 3 are coprime.

So , with *

Therefore

is solution.

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If

-->

We want b such that , that is to say .

By dividing by 3 :

Similarly to above, we get this :

, with *

Therefore

Therefore, b has to be odd.

is solution.