A Theorem on Squares and Congruences

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    Squares Theorem
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Discussion Overview

The discussion revolves around a theorem related to congruences in abstract algebra, specifically examining the relationship between the squares of two numbers that sum to a natural number. The scope includes theoretical exploration and potential novelty of the theorem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes a theorem stating that for any natural number n, if a + b = n, then a² ≡ b² (mod n), and provides a proof based on algebraic manipulation.
  • Another participant suggests that the theorem is trivial and provides an alternative proof using the relationship (a - b)(a + b) = 0 (mod n).
  • A third participant expresses curiosity about whether the theorem has been previously established, acknowledging their limited experience in the field.
  • Some participants affirm that the theorem has indeed been proven before, while also encouraging the exploration of such results.

Areas of Agreement / Disagreement

Participants generally agree that the theorem has been previously established, but there is a recognition of its value in the context of learning and discovery.

Contextual Notes

There is an acknowledgment of the theorem's triviality, which may affect its documentation in formal texts. The discussion reflects varying levels of familiarity with the topic among participants.

Who May Find This Useful

Undergraduate students in mathematics or those interested in abstract algebra and congruences may find this discussion relevant for understanding foundational concepts.

middleCmusic
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Hey guys --

I was wondering whether the following theorem was already established, or if it was something novel. I couldn't find it in my abstract algebra textbook and a Google search was not productive ("a list of congruence theorems" just turns up elementary geometry theorems on triangle congruences.)

Thm For any n in ℕ, if a+b=n, then a2 ≡ b2 (mod n).

Proof: Let a+b=n. Then a=b-n.

a2 = (b-n)2 = b2 - 2bn + n2 = b2 - n*(2b)+n) ≡ b2 (mod n)
 
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middleCmusic said:
Hey guys --

I was wondering whether the following theorem was already established, or if it was something novel. I couldn't find it in my abstract algebra textbook and a Google search was not productive ("a list of congruence theorems" just turns up elementary geometry theorems on triangle congruences.)

Thm For any n in ℕ, if a+b=n, then a2 ≡ b2 (mod n).

Proof: Let a+b=n. Then a=b-n.

a2 = (b-n)2 = b2 - 2bn + n2 = b2 - n*(2b)+n) ≡ b2 (mod n)



Well, this is a trivial statement. Observe (and this also serves as a proof) that

$$a^2=b^2\pmod n\Longleftrightarrow (a-b)(a+b)=0\pmod n$$

so if [itex]\,a+b=0\pmod n\,[/itex] then we can go leftwards in the above and get the result.

DonAntonio
 
Thanks DonAntonio for your explanation.

I agree that it is a rather trivial result, but I thought it was "nice" in a way. Just curious if it's already been proven/noticed. (Perhaps no one thought to write it down because it is so trivial and it's relegated instead to the Exercises section of some introductory textbook.) Please forgive me if I'm writing the equivalent of "Hey guys, I discovered that the sky is blue!" but I'm only an undergraduate and my breadth of knowledge is still pretty thin.
 
Yes, this has certainly been proven before. However, it's great that you're discovering little results like this own your own--keep it up!
 
A. Bahat said:
Yes, this has certainly been proven before. However, it's great that you're discovering little results like this own your own--keep it up!

Thanks for the encouragement!
 

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