Show that the congruence has solutions when

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- February 4th 2009, 02:20 PM #1

- February 4th 2009, 05:22 PM #2

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first see that satisfy the equation. we want to show that there are no other solutions. obviously such solution has got to be odd. so let where

and is an odd number. then: which we'll call it (1). now consider two cases:

__case 1__. : then and are both odd and thus (1) gives us: thus:

__case 2__: : in this case (1) gives us: but is odd. hence therefore: thus:

- February 4th 2009, 07:23 PM #3

- February 4th 2009, 08:47 PM #4

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Because and so .

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Here is another way to solve this problem. But it is overkill. It is a known result that is isomorphic to for .

Let be an isomorphism.

An element satisfies if and only if iff .

Thus, there is a 1-to-1 corresponded between solutions to in and solutions to in .

However, there are clearly 4 solutions to .