Direct proof by using if then technique

[woundedtiger4]

Hi all,

I am trying to proof the following question.

If a is an integer, divisible by 4, then a is the difference of two perfect squares

now by the definition of divisibility if 4 divides a then there is a natural number k such that

a = 4k

Can someone how should I do it with direct proof by using if then technique?

 

[micromass]

You want to find $x$ and $y$ such that

$$a = x^2 - y^2$$

Now, what if you write

$$a = (x-y)(x+y)$$

Does that give you any ideas?

 

[woundedtiger4]

You want to find $x$ and $y$ such that

$$a = x^2 - y^2$$

Now, what if you write

$$a = (x-y)(x+y)$$

Does that give you any ideas?

ahhh got it

proof: if a is divisible by 4 then

4a = (a+1)^2 - (a-1)^2

QED

PS: I have followed the proof method given at http://zimmer.csufresno.edu/~larryc/proofs/proofs.direct.html
Theorem. Every odd integer is the difference of two perfect squares.

 

[micromass]

NO :(

isn't a=a/4 then a = (a/2)^2 - x ?

Why would a = a/4 ??

Anyway, you can always write $a = 4k$. So we have

$$4k = (x-y) (x+y)$$

Does that help? You need to identify some $x$ and $y$ that do that.

 

[micromass]

ahhh got it

proof: if a is divisible by 4 then

4a = (a+1)^2 - (a-1)^2

QED

PS: I have followed the proof method given at http://zimmer.csufresno.edu/~larryc/proofs/proofs.direct.html
Theorem. Every odd integer is the difference of two perfect squares.

Something is not right. You say that $a$ is divisible by $4$ and then you say something about $4a$. Shouldn't you be giving a decomposition of $a$?

 

[woundedtiger4]

Something is not right. You say that $a$ is divisible by $4$ and then you say something about $4a$. Shouldn't you be giving a decomposition of $a$?

Proof:

now by the assumption and definition of divisibility if 4 divides a then there is a natural number k such that

a = 4k =(k+1)^2 - (k-1)^2

QED

PS. is it correct now?

 

[micromass]

Yes, it's right now! Congratulations!

 

[woundedtiger4]

Yes, it's right now! Congratulations!

hehehe thanks a lot

 

[micromass]

Anyway, it might beneficial to show my solution.

I want to find integers $x$ and $y$ such that

$$4k = (x-y)(x+y)$$

So if I find $x$ and $y$ such that

$$2 = x-y~\text{and}~2k = x+y$$

then I'm done. So I get

$$x = 2 + y~\text{and}~x = 2k -y$$

and thus $2 + y = 2k - y$, which gives us $y = k-1$. Then I set $x=(k-1) + 2 = k+1$.

So this gets me that $4k = (k+1)^2 - (k-1)^2$. This is the same answer you found, but it might be good to see how you can find it.

 

[woundedtiger4]

Anyway, it might beneficial to show my solution.

I want to find integers $x$ and $y$ such that

$$4k = (x-y)(x+y)$$

So if I find $x$ and $y$ such that

$$2 = x-y~\text{and}~2k = x+y$$

then I'm done. So I get

$$x = 2 + y~\text{and}~x = 2k -y$$

and thus $2 + y = 2k - y$, which gives us $y = k-1$. Then I set $x=(k-1) + 2 = k+1$.

So this gets me that $4k = (k+1)^2 - (k-1)^2$. This is the same answer you found, but it might be good to see how you can find it.

How did you set 2=x-y and 2k=x+y ?

 

[micromass]

How did you set 2=x-y and 2k=x+y ?

It's not that $4k = (x+y)(x-y)$ implies $2=x-y$ and $2k = x+y$. It might be that $x$ and $y$ are different numbers.

But I said: if I can find $x$ and $y$ such that $2 = x-y$ and $2k = x+y$, then $4k = (x+y)(x-y)$ will be satisfied. So I took a guess about what x and y looked like. It might have happened that the guess didn't give anything useful.