# Proof of infinitely many solutions

1. Oct 29, 2012

### nicnicman

Hello all,

I'm practicing proofs and I'm stuck. Here it is:

Prove that there are infinitely many solutions in positive integers x, y, and z to the equation x^2 + y^2 = z^2. Evidently I'm supposed to start by setting x, y, and z like this:

x = m^2 - n^2
y = 2mn
z = m^2 + n^2

So then we have:

(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2

Now I'm sort of at a standstill. I understand that I can plug any integer into m and n and x^2 + y^2 = z^2 will be true, but I'm not sure how to prove it.

Also, another way to show the proof would be to let x be:

x = 3m
y = 4m
z = 5m

Since any number can be plugged into m then there are infinite solutions.

However, I would like to understand how to derive the proof from my first method.

I'm really trying to understand this stuff so any help would be appreciated.

2. Oct 30, 2012

### haruspex

You have proved that you can plug and m and n in and they will generate a solution. What you have not proved is that this procedure will generate infinitely many different solutions. But for that, all you need to show is that it will generate infinitely many different x values (say).
Yes, but that feels like cheating. I suspect the problem was supposed to say "where x, y and z have no common factor". If so, that also complicates the 'proper' proof.

3. Oct 30, 2012

### bossman27

You have almost proved it. Everything you have left to do is trivial.

To prove that an equation (or rather, a system of equations) has infinitely many solutions, you need only reduce it to an absolute truth. All this means is that you want a statement that doesn't involve variables, that is true.

For example, if we had a system of equations:
$y = 2x + 3$
$2x^{2} = y^{2} - 12x - 9$

Then solving for x gives:
$2x + 3 = 2x + 3 \Rightarrow 3 = 3$

Just take your last step, and reduce it to an absolute truth to show that there are infinitely many solutions.

Conversely, if, in some other situation, you end up with an absolutely false statement, i.e. $2 = 3$ you would know that there are no solutions.

Last edited: Oct 30, 2012
4. Oct 30, 2012

### nicnicman

Okay so to finish it off I would do this:

(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2

2(mn)^2 = 2(mn)^2 . . . Subtract m^4 and n^4 from both sides.
2 = 2 . . . Divide both sides by (mn)^2

Since 2 always equals 2 there are infinite solutions.

5. Oct 30, 2012

### Klungo

That doesn't work. Rather, factor the end terms of the equation. What do you get?

Edit: m and n are vague. Be sure to include their purpose (we want to have infinitely many solutions).

Edit 2: Well it could work actually but then you'd have to go "On the other hand, z^2=…". You're better off doing a clean argument.

6. Oct 30, 2012

### nicnicman

Thanks for the reply Klungo, but I'm not sure I know what you mean by end terms.

Also, the purpose of m and n are to represent arbitrary integers.

7. Oct 30, 2012

### Klungo

Right.

What i mean is factor m^4+n^4+2(mn)^2.

Let's clean up.

For arbitrary m and n. Let x,y,z be given by (what you assigned). Then

x^2+y^2= plug in for x and y (you did that) = manipulated terms (you did that) = m^4+n^4+2(mn)^2 = _______ = z^2.

Thus, ...

8. Oct 30, 2012

### nicnicman

Ok I'm at a loss.

So factor
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2
mmmm + nnnn + 2mmnn = mmmm + nnnn + 2mmnn

Still not sure what I'm looking for, but thanks for your help

Last edited: Oct 30, 2012
9. Oct 30, 2012

### nicnicman

Maybe factor like this:

m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2
(m^2 + n^2)^2 = (m^2 + n^2)^2
m^2 + n^2 = m^2 + n^2 . . . square root of both sides

Again though, I'm not sure what I'm looking for.

10. Oct 30, 2012

### Klungo

You've got it (kinda).

m^4+n^4+(mn)^2 = (m^2+n^2)^2.

Your proof goes:

Let
x = m^2 - n^2
y = 2mn
z = m^2 + n^2
for arbitrary m and n.

So then we have:

x^2+y^2= (m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2 = _____= ____. Since m and n were arbitrary, we have an infinite number of solutions to x^2+y^2=z^2.

Do you see the logic and flow? You are really just missing that last part in the long equation.

11. Oct 30, 2012

### nicnicman

Well I'm still a little fuzzy but I'll take a shot. Here it is:

Let
x = m^2 - n^2
y = 2mn
z = m^2 + n^2
for arbitrary m and n.

So then we have:

x^2+y^2= (m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2 = m^2 + n^2 = z^2. Since m and n were arbitrary, we have an infinite number of solutions to x^2+y^2=z^2.

12. Oct 30, 2012

### Klungo

Once you fabricate a proof, Go step by step with the intention of finding a flaw. If you cant find one, continue to the next step and try to find a flaw. The point of a proof is to convince the truth to a proposition. Convince yourself.

Hint (I'm not)

13. Oct 30, 2012

### nicnicman

I'm just not seeing it. Anymore advice?

14. Oct 30, 2012

### nicnicman

Let
x = m^2 - n^2
y = 2mn
z = m^2 + n^2
for arbitrary m and n.

So then we have:

x^2+y^2= (m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2 = x^2+y^2= z^2. Since m and n were arbitrary, we have an infinite number of solutions to x^2+y^2=z^2.

15. Oct 30, 2012

### nicnicman

Am I trying to factor out the variables? Thereby, reducing the equation to an absolute truth, as Bossman said.

16. Oct 30, 2012

### Klungo

Thats pointless though. The goal is to prove equality. That is x^2+y^2 = *magic* = z^2.

You have one error in the magic. Just read through your work. It is possible you missed something out that you typed on here vs what you may have (probably) done correctly on paper. Look at you previous post (just to be sure) and do the revision from there.

Just so you know. Its a bit pointless reducing some expressions to something like 1=1. Just think about that. Ex: sin(pi)=0 implies 0=0. Thats not useful. The only useful part is sin(pi)=0.

17. Oct 30, 2012

### nicnicman

Well I've been poring over this for hours and I'm still not sure what I'm missing, but I'll look again.

Thanks again for all your help.

18. Oct 30, 2012

### Klungo

No problem. This is a huge part of learning to write proofs. Once you get a few of these down, almost all (similar) 'theorems' of this type can be proven in seconds. One suggestion is to cover up what you did and rewrite the proof. Compare the two proofs and look for any math errors (this doesn't clear logic errors).

Again. I feel you got the proof correct on paper but typed it up incorrectly on here.

19. Oct 30, 2012

### nicnicman

Well, I looked and I have it written the same way. It's missing the magic. I keep trying but I'm still not sure what I'm missing.

20. Oct 30, 2012

### nicnicman

I know that m must be greater than n so that x is positive, but I don't think this is what I'm missing.

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