Show that the congruence has solutions when
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:
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 .