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

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The discussion centers on proving that if a ≡ b (mod n), then a² ≡ b² (mod n) for integers a, b, and n (n ≥ 2). The initial approach involves expressing a - b as a multiple of n, leading to the conclusion that n divides a² - b². The key insight comes from recognizing the factorization a² - b² = (a + b)(a - b), which confirms that since n divides a - b, it also divides a² - b². The final proof is validated by substituting a in terms of b and n, demonstrating that a² - b² is indeed a multiple of n. This confirms the original statement, effectively completing the proof.
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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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