Math Challenge - April 2021

[fresh_42]

"Question 12"
First we prove by contradiction that for any primitive Pythagorean triplet $(x, y, z)$ ($z^2$ being the sum on RHS), exactly one of either $x, y$ must be odd, while the other must be even.

• If both $x, y$ are even, then the RHS $z^2$ would also be even, in turn implying that $z$ is also even. This in turn implies that 2 is a common divisor of $x, y, z$, contradicting the requirement of primitiveness.
• If both of $x, y$ are odd, they can be written as $x=(4a + r_1), y=(4b + r_2)$ for some positive integers $a, b$ such that $r_1 = x \mod 4, r_2 = y \mod 4$ and $r_1, r_2 \in \{1, 3\}$. Then the Pythagorean sum becomes $z^2 = 4a^2 + 8ar_1 + r_{1}^{2} + 4b^2 + 8br_2 + r_{2}^{2} \\ \Rightarrow z^2 \mod{4} \equiv (r_1^2 + r_2^2) \mod{4} \equiv 2 \mod{4}$. This implies that $z^2$ is even but not a multiple of 4. But this would mean that there can be no integer solution for $z$ since an odd $z$ cannot given an even $^2$ and an even $z$ will mean $z^2 \equiv 0 \mod{4}$, again a contradiction.

Hence the only possibility is that one of $x,y$ is odd and the other is even. This implies $z^ = x^2 + y^2$ is odd and hence $z$ too must be odd.

(Note to self:) This was only shown for primitive triplets. $x\equiv y\equiv 0 \;(2)$ isn't ruled out, yet.

Proof for the if part:
Given $u, v$ are coprime with $u > v$ and exactly one of them is odd. Let $x = u^2 - v^2, y = 2uv, z = u^2 + v^2$. $x^2 + y^2 = (u^4 + v^4 - 2u^2v^2) + 4u^2v^2 = (u^4 + v^4 + 2u^2v^2) = (u^2 + v^2)^2$
$\Rightarrow x^2 + y^2 = z^2$. Thus $x, y, z$ form a Pythagorean triplet. To prove that it is a primitive triplet, let us assume that on the contrary $x, y$ have some common prime divisor $p$.
$p \mid x \Rightarrow p \mid x^2 \Rightarrow p \mid (u - v)(u + v)$. (Eq.1)
Note that $p$ cannot be 2 since $x$ is odd.

If I got you right, then you want to show that the triplet is primitive. But $x\equiv y\equiv 0 \;(2)$ isn't ruled out, yet. Only for primitive triplets, which we don't know it is, yet. Hence why is $x$ odd?

Thus $p \mid y \Rightarrow p \mid 2uv \Rightarrow uv$
$\Rightarrow p \mid u$ or $p \mid v$ (Eq. 2)
Since $u, v$ are coprime, we may assume without loss of generality in (Eq. 2) that $p \mid u$. But this contradicts (Eq. 1) since both $(u-v)$ and $(u+v)$ must be coprime w.r.t. $u$ and hence neither of them can have $p$ as their divisor. Thus the assumption of there being a common prime divisor for $x,y$ must be wrong and hence the triplet must be primitive.

Proof for the only if part:
Suppose $(x, y, z)$ is a primitive Pythagorean triplet with $x^2 + y^2 = z^2$. We already proved that any such triplet must have an odd $z$ and exactly one odd number on LHS. Without loss of generality, let us assume that $x$ is the odd number on LHS (and so $y$ must be even). Since both $x$ and $z$ are odd and $z > x$, we may rewrite them as $x = (a - b), z = (a+b)$ where $a, b$ are positive integers given by $a = \dfrac {x + z} {2}$ and $b = \dfrac {z - x} {2}$. (Eq. 3)

We only have $z^2>x^2$. If you write $z>x$ then you already made an assumption. It can be made since we can restrict ourselves to natural numbers, but why?

Thus $y^2 = z^2 - x^2 \Rightarrow (a+b)^2 - (a-b)^2 = 4ab \Rightarrow y = 2 \sqrt {ab}$. (Eq. 4)
Since $y$ is an integer, $ab$ must be a perfect square. (Eq. 5)

$a, b$ must be coprime since otherwise $ab$, $(a-b)$ and $(a+b)$ will have some common prime divisor $p$ which would contradict the initial assumption that $(x, y, z)$ have no common divisor. This condition together with (Eq. 5) implies that $a$ and $b$ must themselves be perfect squares coprime w.r.t. each other. Thus we may write $a=u^2, b=v^2$ for some coprime integers $u, v$ with $u > v$. Substituting for $a, b$ in (Eq. 3) and (Eq. 4) we get $x = u^2 - v^2, z = u^2 + v^2$ and $y = 2 \sqrt {u^2 v^2} = 2uv$. Hence proved.

Why can we write all triplets this way? What about those which are not primitive?

 

[mathwonk]

SPOILER: #9:In reference to fresh42's post #88, yes, I had in mind looking at a map f from an interval [0,1] to X that sends both end points to the same base point p lying in UmeetV. Then subdivide the interval until each subinterval goes either entirely into U or entirely into V, (or both). I will assume there are only two subintervals, [0, 1/2] and [1/2, 1], with the first subinterval mapping entirely into U and the second one mapping entirely into V, (since you just repeat the argument if there are more). Then since the point f(1/2) = q lies in both U and V, we can find a path connecting q to p and lying in UmeetV. By adjoining this path to the map f on [0,1/2], we get a loop based at p, and lying entirely in U. Then by also adjoining this path in the other direction (from p to q) to the map f on [1/2,1] we get another loop based at p and lying entirely in V. Moreover if we adjoin these two extended loops end to end, the two path components joining q to p and then back again, cancel out, leaving us with our original f. Thus the new loop, which is a product of one loop in U and one loop in V, is homotopic to the original loop f, rel base points. Since both U and V have trivial fundamental group, both factors of the new product loop are homotopic to a constant, hence so is their product and so is the original f. This is what I called the surjectivity part of Seifert Van Kampen, and thanks to you I now realize it is rather elementary!

 

[Infrared]

Right, the surjectivity in Van Kampen isn't so mysterious! Even easier than surjectivity is the fact that the subgroup identifying classes in $\pi_1(U)$ and $\pi_1(V)$ coming from the same element of $\pi_1(U\cap V)$ is contained in the kernel. The meat of the theorem is that nothing else lies in the kernel. Hatcher's argument looks pretty direct but also un-fun to read (though I guess I should sit myself down and go through it since I don't remember it exactly...)

 

[mathwonk]

I didn't read Hatcher's argument either, but was enlightended by his statement, emphasizing the surjectivity as the first step. I just visualized this argument in my head afterwards. nice point about the inclusion of the natural candidate subgroup in the kernel.

 

[Office_Shredder]

I allegedly took a class on algebraic topology in 2011, but this question and reading material about your posts was very educational.

 

[graphking]

No one try Q5？ I believe it requires the Gauss Law of Quaduatic Reciprocity, which is a very beautiful formula in general number theory, related to the equation with variable x: x^2=a(modb).

 

[Not anonymous]

But $x \equiv y \equiv 0\ (2)$ isn't ruled out, yet. Only for primitive triplets, which we don't know it is, yet. Hence why is $x$ odd?

In the if-part of the proof the initial conditions stated not only that are $u, v$ are coprime but also that exactly one of them is odd and the other one is even. Since the square of an odd number is odd and the square of an even number is even, it follows that exactly one of $u^{2}, v^{2}$ must be odd and the other must be even. $x = u^{2} - v^{2}$ in that section of the proof and we know that the difference between an odd number and an even number (regardless of whether odd is subtracted from even or vice-versa) must be odd, hence $x$ must be odd.

We only have $z^{2} > x^{2}$. If you write $z > x$ then you already made an assumption. It can be made since we can restrict ourselves to natural numbers, but why?

Sorry, I don't know what assumption I have made other than $x, y, z$ being natural numbers, and even that assumption follows directly from the definition (as I know of it) of Pythagorean triples, since unless otherwise stated, such triples must consist only of positive integers. Should it be proven that the squares of natural numbers also follow the same ordering as the natural numbers (i.e. $a > b \Leftrightarrow a^{2} > b^{2}$ for natural numbers $a,b$), or are you referring to some other assumption that I am unable to identify?

Why can we write all triplets this way? What about those which are not primitive?

If $(x, y, z)$ form a non-primitive Pythagorean triple, then by definition they have some common divisor greater than 1. Let $p > 1$ be the greatest common divisor of $x, y$. Then we can write $x = px_{0}, y = py_{0}$ for some natural numbers $x_0, y_0$ that would be coprime. $x^2 + y^2 = z^2 \Rightarrow p^2 x_{0}^{2} + p^2 y_{0}^{2} = z^{2} \Rightarrow p^2 (x_{0}^{2} + y_{0}^{2}) = z^2$. Since $p^2 | z^2$ and $z^2$ is a perfect square, $z^2$ can be written as $p^2 z_{0}^{2}$ for some natural number $z_{0}$. Dividing the Pythagorean triple equation by $p^{2}$ gives a different Pythagorean triple $x_{0}^{2} + y_{0}^{2} = z_{0}^{2}$ which must be primitive since $x_{0}, y_{0}, z_{0}$ cannot have a common divisor greater than one since $x_{0}$ and $y_{0}$ are coprime. Thus any non-primitive Pythagorean triple can be reduced to a primitive Pythagorean triple and can be generated by multiplying a primitive Pythagorean triple by some positive integer greater than 1.

 

[fresh_42]

The only-if part looks correct.

By assuming $u>v$ you lost your advantage to interchange the two. Without any need, by the way, because the assumption that we haven't all integers is artificial as it is unnecessary. We only consider squares, so $x,y,z$ can as well be negative.

For the if-part: Why is $x$ odd?
($x^2=u^2-v^2=(2k)^2-(2l+1)^2\equiv 1\vee 3 \mod 4\Longrightarrow x\not\equiv 0\mod 2\quad \checkmark$.)

Ok. Now $p\,|\,u$ or $p\,|\,v$ and $p\neq 2.$ As mentioned, you cannot assume $p\,|\,u$ because you already demanded $u>v.$ However, I assume that neither assumption has to be made. Let's see.
($p\,|\,u \wedge p\,|\,x\Longrightarrow x^2=p^2k^2=(u^2-v^2)=(p^2m^2-v^2)\Longrightarrow p\,|\,p^2\,|\,v^2 \Longrightarrow p\,|\,v.$ The same argument can be used if $p\,|\,v.$)

Thus $x$ and $y$ are coprime.

Ok. I'm convinced now. That leaves only the part to be proven, namely the statement that all Pythagorean triples can be found that way, primitive or not.

 

[Not anonymous]

By assuming $u > v$ you lost your advantage to interchange the two

$u > v$ is a condition already mentioned in the question, so it is not an assumption introduced in the proof.

That leaves only the part to be proven, namely the statement that all Pythagorean triples can be found that way, primitive or not

Sorry, I am confused again. The original question refers to only primitive Pythagorean triples and says "and which are primitive (no common divisor of $x, y, z$)", so the proof I attempted to provide was also specific to primitive triples. The proof does not say that non-primitive Pythagorean triples too can be expressed as $(u^2 - v^2, 2uv, u^2+v^2)$ for coprime $u, v$. However, since your previous reply asked about non-primitive triples, my reply to that said that any non-primitive triple can be derived from a primitive triple by multiplying every element of the the primitive triple by the same positive integer. So if $(x, y, z)$ is a primitive Pythagorean triple, then $(px, py, pz)$ will also be a Pythagorean triple for any natural number $p$, but it will be non-primitive when $p > 1$.

As an example of a non-primitive Pythagorean triple, we take $(9, 12, 15)$. It is derived from the primitive Pythagorean triple $(3, 4, 5)$ using the multiplier 3. And we cannot express $9 = u^ - v^2, 6 = 2uv$ for any coprime $u, v$ since if they were coprime, then $u^2 - v^2$ and $2uv$ will also be coprime whereas 9 and 6 are not coprime.

 

[fresh_42]

statement one:

Show that all Pythagorean triples $x^2+y^2=z^2$ can be found by
$(x,y,z)=(u^2−v^2,2uv,u^2+v^2)$ with $u,v\in \mathbb{N}\, , \,u>v$

statement two:

and which are primitive (no common divisor of $x,y,z$) if and only if $u\, , \,v$ are coprime and one is odd and the other one even.

$u>v$ is an add-on at the end, it is not really required. If we find such a representation, then it can be assumed. Anyway, the signs are not of importance.

 

[Not anonymous]

statement one:
Show that all Pythagorean triples $x^2 + y^2 = z^2$ can be found by
$(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N} , u > v$

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

 

[fresh_42]

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

You are right. I made a mistake (sloppy translation). It should have been ... of the form $d\cdot (u^2-v^2,2uv,u^2+v^2)$ ...

I apologize for that negligence, @Not anonymous.

 

[Not anonymous]

I apologize for that negligence, @Not anonymous.

No problem (problem ≠ math problem in this context ). Thanks to you as always for promptly reviewing my answers and patiently explaining what I may have missed or got wrong.