Number theory Proof Question

[CalleighMay]

Hey guys,

 

I'm in the class "Number Theory" at my college. It's WEEK 1 and we are already going over these weird proofs and i am COMPLETELY lost.

 

The question asks:

 

"Show that if n is any odd positive integer, and m = (n^2-1)/2, then m^2 + n^2 = (m+1)^2."

 

My professor as awful and has given us absolutely NO background.

He's given us two proof examples in class, but i don't understand either of them, so they don't help much.

 

From what i can gather, odd numbers are "2k+1" and even numbers are "2k". I guess this makes sense, but i don't understand how to use it.

 

So for the problem on hand, i understand i need to show how you get from point A to point B- but the question is how.

 

We can assume that n is odd. So n=2k+1.

We can also assume that m= (n^2-1)/2

And we need to get these two equations to look like: m^2 + n^2 = (m+1)^2 in the end

 

Working backwards:

 

m^2 + n^2 = (m+1)^2

.............= m^2 + 2m + 1 (expanding)

m^2 + n^2 = m^2 + 2m + 1

........n^2 = ........2m + 1 (subtraction)

so n^2 = 2m+1

 

Working from the given that m = (n^2 - 1) / 2

 

m = (n^2 - 1) / 2

2m = n^2 - 1 (multiplication)

2m + 1 = n^2 (addition)

n^2 = 2m + 1 (switch things around)

 

So this gets the "n^2 = 2m + 1" in the final step. But how do i get the "+ m^2" ??

 

My biggest problem is that i don't know how this proof should "look". I understand the concept, but i am having difficulty getting it into proof form.

 

So it should look like:

 

A: Assume n is positive and odd, and assume m = (n^2 - 1) / 2

A1:

A2:

A3:

A4:

A5:

A6:

A7:

A8:

A9:

B: Show that m^2 + n^2 = (m+1)^2

 

Then we have to write it in paragraph form....

 

Can anyone help me out with this problem? I just don't get it...

 

This problem is from the text "Elementary Number Theory" 2/e by Charles Vanden Eynden.

It's Chapter 0 page 9 number 11.

 

Thanks for any help!!

[battletoad]

Hey guys,

 

I'm in the class "Number Theory" at my college. It's WEEK 1 and we are already going over these weird proofs and i am COMPLETELY lost.

 

The question asks:

 

"Show that if n is any odd positive integer, and m = (n^2-1)/2, then m^2 + n^2 = (m+1)^2."

 

 

Your derivations are right for the most part so you definitely on the right track when it comes to these AWFUL (fun for me tho;)) number theory exercises

 

My A1-A9 would look like this...(im going to draw it out so you can get the full understanding behind it all)

.

Given n is odd. Thus n=2k+1 for some integer k.

It follows that (n^2)=4(k^2)+4k+1.

After factorisation we get n^2=2([2k^2]+2k)+1. <---Just to show that n^2 can be written in the form of 2t+1 for some integer t (in this case our t=([2k^2]+2k), which is our definition of an odd integer

Clearly n^2 is odd as well.

 

Since n^2 is odd, n^2-1 is even, i.e. divisible by 2. As such, (n^2-1)/2 is an integer.

We have now shown that m=(n^2-1)/2 is an integer <---AFAIS, not needed for your proof but lets be precise shall we

 

(the part that follows is your derivation of earlier with an added 'twist'...)

 

Given m=(n^2 - 1)/2

2m=n^2-1

2m + 1=n^2

m^2 + 2m + 1=m^2 + n^2. ....Added m^2 both sides, permissible since its an integer

(m+1)^2 = m^2 + 2m + 1 = m^2 + n^2

=> (m+1)^2 = m^2 + n^2

 

And there you go. As with all maths, doing plenty of exercises will give you the experience needed to get to know the tricks of the trade and to know when to use them. I advise you to do as much of the exercises at the end of the section and all this will be a breeze at the end of the day

 

Good luck with your course

[triclino]

Hey guys,

 

I'm in the class "Number Theory" at my college. It's WEEK 1 and we are already going over these weird proofs and i am COMPLETELY lost.

 

The question asks:

 

"Show that if n is any odd positive integer, and m = (n^2-1)/2, then m^2 + n^2 = (m+1)^2."

 

My professor as awful and has given us absolutely NO background.

He's given us two proof examples in class, but i don't understand either of them, so they don't help much.

 

From what i can gather, odd numbers are "2k+1" and even numbers are "2k". I guess this makes sense, but i don't understand how to use it.

 

So for the problem on hand, i understand i need to show how you get from point A to point B- but the question is how.

 

We can assume that n is odd. So n=2k+1.

We can also assume that m= (n^2-1)/2

And we need to get these two equations to look like: m^2 + n^2 = (m+1)^2 in the end

 

Working backwards:

 

m^2 + n^2 = (m+1)^2

.............= m^2 + 2m + 1 (expanding)

m^2 + n^2 = m^2 + 2m + 1

........n^2 = ........2m + 1 (subtraction)

so n^2 = 2m+1

 

Working from the given that m = (n^2 - 1) / 2

 

m = (n^2 - 1) / 2

2m = n^2 - 1 (multiplication)

2m + 1 = n^2 (addition)

n^2 = 2m + 1 (switch things around)

 

So this gets the "n^2 = 2m + 1" in the final step. But how do i get the "+ m^2" ??

 

My biggest problem is that i don't know how this proof should "look". I understand the concept, but i am having difficulty getting it into proof form.

 

So it should look like:

 

A: Assume n is positive and odd, and assume m = (n^2 - 1) / 2

A1:

A2:

A3:

A4:

A5:

A6:

A7:

A8:

A9:

B: Show that m^2 + n^2 = (m+1)^2

 

Then we have to write it in paragraph form....

 

Can anyone help me out with this problem? I just don't get it...

 

This problem is from the text "Elementary Number Theory" 2/e by Charles Vanden Eynden.

It's Chapter 0 page 9 number 11.

 

Thanks for any help!!

 

I think the right question is :

 

If n is odd then there exists an integer m such that :

 

[math] n^2 + m^2 = (m+1)^2[/math].

 

Proof:

 

If n is odd ,then [math]n^2[/math] is odd ,hence there exists an integer m and [math]n^2 = 2m+1[/math].

 

Thus [math] n^2 + m^2 = 2m+1+m^2 = (m+1)^2[/math].

 

Now the book or your lecturer to make it easier for you they added the extra formula [math] m=\frac{n^2-1}{2}[/math] ,or solving for [math]n^2[/math] is: [math]n^2 = 2m+1[/math],which is the conclusion of the theorem:

 

If n is odd ,then [math]n^2[/math] is odd.

 

Actually one of the facts : "n is odd" or " [math]m=\frac{n^2-1}{2}[/math]" is redundant because either of the two can lead to the conclusion:

 

[math]n^2+m^2 = (m+1)^2[/math]

 

 

 

Now in your opening post you ask for a step wise proof :

 

[math]A_{1},A_{2},A_{3}....[/math].

 

Stepwise proofs can be quite difficult sometimes>

 

Anyway here is a step wise proof:

 

[math]A_{1}[/math] : n is odd......................................................................................................................................................assumption

 

[math]A_{2}[/math] : [math]m=\frac{n^2-1}{2}[/math]..................................................................................................................................assumption ,or derived from [math]A_{1}[/math]

 

 

[math]A_{3}[/math]: [math]n^2 = 2m+1[/math]....................................................................................................by multiplying [math]A_{2}[/math] by 2 and then adding 1 to both sides

 

 

[math]A_{4}[/math]: [math]n^2+m^2 = 2m+1+m^2[/math] ...............................................................................by adding [math]m^2[/math] to both sides of [math]A_{3}[/math]

 

 

[math]A_{5}[/math]: [math] (m+1)^2= 2m+1+m^2[/math].................................................................................................................expanding [math](m+1)^2[/math].

 

 

 

[math]A_{6}[/math] :[math] n^2+m^2 = (m+1)^2[/math]................................................................................by substituting [math]A_{5}[/math] into [math]A_{4}[/math]

 

A more detailed stepwise proof can be done ,where the axioms and the laws of logic involved in the proof can be explicitly mentioned.

[CalleighMay]

can you simply just "add m^2 to both sides in A4 ?

 

I had a little help last night and came up with this method:

 

Let n = 2k + 1 (since n is odd)

 

plugging it into the m equation given, you get m = ( (2k+1)^2-1) )/2

 

1: so m = 2k^2 + 2k

 

2: so m+1 = 2k^2 + 2k + 1

 

3: so (m+1)^2 = 4k^4 + 8k^3 + 8k^2 + 4k + 1

 

4: and m^2 = 4k^4 +8k^3 + 4k^2

 

And using n = 2k + 1,

 

5: n^2 = 4k^2 + 4k +1

 

So we have that (4k^4 +8k^3 + 4k^2) + (4k^2 + 4k +1) = 4k^4 + 8k^3 + 8k^2 + 4k + 1

 

which is m^2 + n^2 = (m+1)^2

[triclino]

Yes you can add [math]m^2[/math] to both sides of [math]A_{3}[/math] by using the following axiom,concerning equality:

 

for all ,x,y,z : [math] x=y\Longrightarrow x+z = y+z[/math] and if we put:

 

x=[math]n^2[/math] , y= 2m+1 ,z = [math]m^2[/math] the above becomes:

 

[math]n^2 = 2m+1\Longrightarrow n^2+m^2 = 2m+1+m^2[/math].

 

By the way the law that allow us to do that is called Universal Elimination.

 

Now coming back to your exercise ,the right state of affairs is:

 

If n is odd ,then n= 2k+1 and thus [math]n^2 =4k^2 +4k+1 =2(2k^2+2k)+1[/math],hence there exists a natural No m such that [math]n^2=2m+1[/math] ,or [math] m=\frac{n^2-1}{2}[/math] and from here onwards we follow the proof as shown in my post

 

But, by making the substitution n = 2k+1 into [math] m=\frac{n^2-1}{2}[/math]

 

you simply make the proof longer .

 

To give you another simple example.

 

Suppose you want to prove : [math](x+1)^2 =x^2+2x+1[/math].

 

You can do that the simple way by simple multiplying (x+1) and(x+1) and using the law of distribution obtain the desired result, or you can do it the long way as follows:

 

Let x=y+1 ,then [math](x+1)^2 =[(y+1)+1]^2 = [(y+1)+1][(y+1)+1][/math]= (Y+1)[(y+1)+1] +1[(y+1)+1] =[math]y^2+4y+4[/math] =[math](x-1)^2 +4(x-1) +4 = x^2-2x+1+4x-4+4 =x^2+2x+1[/math]

 

You can make the above simple problem as long as you like.

 

So if there is a direct fly from L.A to New-York,would you fly thru S.Africa??