Congruences between a square and itself

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Homework Statement



What are the congruences between a square and itself?

Homework Equations



none

The Attempt at a Solution



Let A, B, C, and D be the sides of the square and e, f, g and h the four right angles.

All four sides are congruent to themselves--same with the angles.

Side A is congruent to Side B
Side A is congruent to side C
Side A is congruent to side D
Side B is congruent to side C
Side B is congruent to side D
Side C is congruent to side D

And we can symmetrically list the congruences between the angles.

And the whole square is congruent to itself...

It's a simple problem but there's no answer key in the text and I'm wondering if I'm right to count all the reflexive congruences...
 
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Saying "side A is congruent to side B" does not say the square is conguent to itself.

Since you can "label" a square by listing its vertices in order, ABCD, how many different ways of ordering ABCD give the same square? BCDA does but BACD does not. Do you see why?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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