Wierd Mathematical Proof

1. Feb 8, 2005

CollectiveRocker

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.

2. Feb 8, 2005

robert Ihnot

Just look at the four cases of integers: 4k, 4k+1, 4k+2, 4k+3.

3. Feb 8, 2005

CollectiveRocker

Ok, so it is even, odd, even, odd. What are you getting at?

4. Feb 8, 2005

dextercioby

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.

5. Feb 9, 2005

vsage

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!

Last edited by a moderator: Feb 9, 2005
6. Feb 9, 2005

dextercioby

It's definitely in the integers.Positive integers,even.

Daniel.

7. Feb 10, 2005

robert Ihnot

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?

8. Feb 10, 2005

Zurtex

3*3 = 1 mod 4

Woo! I love modulo arithmetic.

9. Feb 10, 2005

mathwonk

i wish to point out that "weird" is spelled "weird".

10. Feb 10, 2005

vincentchan

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 "if" statement can never be true

11. Feb 10, 2005

mathwonk

in that case i guess if m^2 = 4k+3, then weird is spelled wierd, and m = 4n.

12. Feb 10, 2005

vsage

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. Feb 10, 2005

Zurtex

*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

$$a,b,p,n \in \mathbb{N}$$

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?

Last edited: Feb 10, 2005