Prove If m Not Form 4k+3, m^2 Is Not Form 4k+3

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Discussion Overview

The discussion revolves around the statement that if \( m^2 \) is of the form \( 4k+3 \), then \( m \) must also be of the form \( 4k+3 \). Participants explore various approaches to proving or analyzing this statement, including contraposition, case analysis, and modular arithmetic. The scope includes mathematical reasoning and debate over the validity of the initial claim.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests starting the proof by contraposition, indicating uncertainty about how to begin.
  • Another participant proposes examining the four cases of integers: \( 4k \), \( 4k+1 \), \( 4k+2 \), and \( 4k+3 \), but later refines this to focus only on odd integers.
  • A participant questions whether \( k \) remains an integer if \( k' \) is, expressing curiosity rather than directly addressing the original post.
  • There is a calculation presented involving the product of odd integers \( (4k+1)(4k+3) \) and its form, leading to a discussion about modulo arithmetic.
  • One participant humorously critiques the validity of the "if" statement by providing absurd examples, suggesting that the statement can never be true.
  • Another participant questions the entire proof's relevance, implying that the conditions may not hold true for integers.
  • A later reply attempts to clarify the situation by defining a product of two numbers of the form \( 4k+3 \) and demonstrating that the product cannot be of that form, suggesting a conclusion about the original claim.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the original statement, with some suggesting it is moot while others attempt to provide proofs or counterexamples. The discussion remains unresolved, with multiple competing perspectives present.

Contextual Notes

There are limitations regarding the assumptions made about the integers involved, particularly concerning whether \( k \) and \( k' \) can both be integers. The discussion also reflects varying levels of mathematical rigor among participants.

CollectiveRocker
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I'm given the statement: if m^2 is of the form 4k+3, then m is of the form 4k+3. I don't even know how to begin proving this. I'm guessing by contraposition.
 
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Just look at the four cases of integers: 4k, 4k+1, 4k+2, 4k+3.
 
Ok, so it is even, odd, even, odd. What are you getting at?
 
Yes,Robert presented you with 2 many numbers:4 instead of 2.So you're left only with the odd numbers.
Can u do the analysis in this case...?

Daniel.

P.S.Relabel k and k' not to create confusion when putting them in the same equation.
 
In response to the above:

m^2 = 4k + 3 = 4(4k'^2 + 6k' + 3/2) + 3 = [16k'^2 + 24k' + 6] + 3 = (4k'+3)^2

but is k an integer is k' is? I'm not really trying to answer the OP's post just sort of curious, so pardon my hijacking!
 
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It's definitely in the integers.Positive integers,even.

Daniel.
 
Well, O.K., we need only consider the odd numbers: 4k+1 and 4k+3. Suppose we multiply them together: (4k+1)(4k+3) = 16k^2+16k+3. What form is that?
 
3*3 = 1 mod 4

Woo! I love modulo arithmetic.
 
i wish to point out that "weird" is spelled "weird".
 
  • #10
if m and k are positive integer... then your "THEN" statement is always true, no matter what is following..., here are some examples
if m^2 is of the form 4k+3, then the world is peace
if m^2 is of the form 4k+3, then I have 3 hands and 4 legs
if m^2 is of the form 4k+3, then US government will give us $10billions dollars to build a particle accelatator

can you see the reason?
the answer is in white:

the "if" statement can never be true [/color]
 
  • #11
in that case i guess if m^2 = 4k+3, then weird is spelled weird, and m = 4n.
 
  • #12
I don't think too mathematically, but is this entire proof moot since the condition is never true? (alluding to my previous post where k and k' can't both be integers)
 
  • #13
*sigh*

This thread is not coherent at all and hardly a help to anyone.

This is quite simple, take any 2 numbers:

a=4p + 3
b=4n + 3

[tex]a,b,p,n \in \mathbb{N}[/tex]

Define c such that:

c = ab

We know that:

a ≡ b ≡ 3 (mod 4)

Therefore:

c ≡3*3 ≡ 9 ≡ 1 (mod 4)

Therefore there exists some t in N such that:

c = 4t + 1

And looking at the original post we can safely say that m2 is not of the form 4k + 3.

Is this not valid in some way?
 
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