Prove if a=b(mod n) then a^2=b^2(mod n)

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Homework Help Overview

This discussion revolves around a proof in modular arithmetic, specifically examining the relationship between integers a, b, and n, where n is greater than or equal to 2. The original poster seeks to prove that if a is congruent to b modulo n, then a squared is congruent to b squared modulo n.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the congruence a ≡ b (mod n) and explore whether n divides the difference a² - b². There are attempts to express a and b in terms of n and other integers, as well as considerations of factorization.

Discussion Status

The discussion has progressed with participants sharing hints and insights. Some have expressed uncertainty about their reasoning, while others have provided guidance on factorization and the implications of the initial congruence. There appears to be a productive exchange of ideas, with some participants feeling closer to understanding the proof.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the depth of exploration or the types of solutions discussed. There is an emphasis on understanding the reasoning behind the proof rather than simply arriving at a conclusion.

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Homework Statement



This is a question from the book I'm studying called 'Mathematical Proofs: A transition to advanced mathematics'

Homework Equations



Let a, b, n be integers, with n≥2. Prove that if a\equivb(mod n), then a2\equivb2(mod n).

The Attempt at a Solution



Following the examples I assumed that I'd start by stating that since a\equivb(mod n), then a-b=nx... but from this point I'm stuck.

Any help much appreciated.
 
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privyet said:

Homework Statement



This is a question from the book I'm studying called 'Mathematical Proofs: A transition to advanced mathematics'

Homework Equations



Let a, b, n be integers, with n≥2. Prove that if a\equivb(mod n), then a2\equivb2(mod n).

The Attempt at a Solution



Following the examples I assumed that I'd start by stating that since a\equivb(mod n), then a-b=nx...

In other words, n divides a-b. Does n divide a2-b2?
 
Yeah. I know its simple but I just don't know how to think about it.
 
privyet said:
Yeah. I know its simple but I just don't know how to think about it.

Are you saying you still don't know how to think about it after my hint? Can you answer my question in the hint?
 
privyet said:
Yeah. I know its simple but I just don't know how to think about it.
LCKurtz has probably provided the most direct way to think about it if you happen to know the well known factorization of the expression a2-b2, but alternatively you could think of it this way:

You have already stated that a-b=nx, or a=nx+b, for some integer x.

Can you find an integer, say y, such that a2=ny+b2 ?
 
I was saying that I could see that it looks pretty obvious that if n divides a-b, it will divide a2-b2, but that I didn't know how to prove it.

Using oay's hint about factorisation I've got a2-b2=(a+b)(a-b) and since n divides a-b, n therefore divides a2-b2 but this is looking very different to the examples in the book.

OK, after a pause to try again I think I've got it (with the help of oay's other hint):

Assume n divides a-b, then a=nx+b. Therefore a2=(nx+b)2=n2x2+2bnx+b2=n(nx2+2bx)+b2. Since a2=n(nx2+2bx)+b2 and nx2+2bx is an integer, a2-b2=n(nx2+2bx). Therefore a2\equivb2(mod n).

Is that right?
 
privyet said:
Is that right?
Spot on! :smile:
 
Great! Thank you both for your help.
 

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