Proof of infinitely many solutions

[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.

 

[haruspex]

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.

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).

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

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

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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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, ...

 

[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

 

[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.

 

[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.

 

[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.

 

[Klungo]

Once you fabricate a proof, Go step by step with the intention of finding a flaw. If you can't 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)

 

[nicnicman]

I'm just not seeing it. Anymore advice?

 

[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.

 

[nicnicman]

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

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[Klungo]

Here's my final hint. You got from the last post:

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

Based on your construction of z, what is in the blank in terms of n and m? By construction, i mean what you assign z to be by assumption. Look at z in terms of m and n.

 

[nicnicman]

Here's my final hint. You got from the last post:

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

Based on your construction of z, what is in the blank in terms of n and m? By construction, i mean what you assign z to be by assumption. Look at z in terms of m and n.

Well, I assigned z to equal m^2 + n^2. Then, z^2 = (m^2 + n^2)^2.

 

[Klungo]

That's right. So what would you write to complete the equality. Look at each side of the equation where _____ is.

 

[Klungo]

Thats it! Now write the revised proof.

 

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

 

[Klungo]

Correct.

Be sure to understand this structure as many proofs are similar. I.e. assumptions first, then constructions, then the magic. I would recommend you work out a few similar problems until you get the hang of it.

 

[nicnicman]

Yeah, I'll be practicing these a lot. Thanks for the help.

 

[haruspex]

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

Almost! The one thing missing is a detail many would overlook. You can choose infinitely many different combinations of m and n, but does this generate infinitely many different solutions of a²+b²=c²?

 

[nicnicman]

Since m and n dictate the values of a, b, and c doesn't it follow if there are infinite values for m and n there would also be infinite values for a, b, c and also infinite solutions.

 

[haruspex]

Since m and n dictate the values of a, b, and c doesn't it follow if there are infinite values for m and n there would also be infinite values for a, b, c and also infinite solutions.

The proof is trivial in this case, but it's not automatically true in general. One can imagine a more complicated relationship between m and n, on the one hand, and a, b, c on the other, such that many combinations of m and n lead to the same triple (a, b, c).
All you need to say is something like "keeping n fixed, no two values of m lead to the same triple (a, b, c), therefore infinitely many distinct triples are generated". It's no big deal in this case, but being rigorous is a good habit to get into.

 

[Klungo]

Im surprised I forgot about the uniqueness part.