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- Mar 22, 2013

- 573

You have to verify what your intended equation was. Your title says \(\displaystyle x^2 + y^2 = 5z^3\) whereas your thread material says \(\displaystyle x^2 + y^2 = 5z^2\).

For the later one, invoke a rational transformation to convert it into \(\displaystyle X^2 + Y^2 = 5\) over \(\displaystyle \mathbb{Q}[X, Y]\). Now try to seek a nontrivial solution to this and then use transformations \(\displaystyle P = x - x_0\) and \(\displaystyle Q = y - y_0\) for the initial solution \(\displaystyle (x_0, y_0)\) -- this is the basic line of thought to approach all the Pythagoras-like forms over \(\displaystyle \mathbb{Z}[x, y, z]\)

Balarka

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- Mar 5, 2012

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First find 1 solution.Hello!

Could anyone give me a hint to solve this diophantine equation?

Show how to construct an infinite amount of triples x, y and z e N with (x,y)=1 and x^2 + y^2=5z^2?

Let's call it $(x_0, y_0, z_0)$.

Did you already find one?

Then pick $x$ and $y$ by multiplying $x_0$ and $y_0$ by some $a^k$ or something like that, such that it pans out.

That is, pick $x = a^k \cdot x_0$ and $y = a^k \cdot y_0$.

Can you find an $a$ and corresponding $z$ such that the equation is satisfied?

- Aug 30, 2012

- 1,143

Coolerino! As I've never seen the method I'll ask the obvious question: Does this generate all possible solutions?First find 1 solution.

Let's call it $(x_0, y_0, z_0)$.

Did you already find one?

Then pick $x$ and $y$ by multiplying $x_0$ and $y_0$ by some $a^k$ or something like that, such that it pans out.

That is, pick $x = a^k \cdot x_0$ and $y = a^k \cdot y_0$.

Can you find an $a$ and corresponding $z$ such that the equation is satisfied?

-Dan

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- Mar 22, 2013

- 573

I didn't look much at I like Serena's solution, but an $(x, y, z)$ can easily be found. Try $(1, 2, 1)$.

What I worked out last night was a heuristic analysis, i.e., Note that there are heuristically $\frac{N}{\log N}$ numbers which can be written as $x^2 + y^2$ for $(x, y) \leq (N, N)$. Hence, the n-th such number should be heuristically $\gg \frac{N}{(\log N)^\epsilon}$, implying that the total number of such integers, upto a correct error, is $\gg \frac{N}{(\log N)^\epsilon}$ for some $\epsilon \geq 1$, therefore, there should be asymptotically more or less $\frac{\log(N)}{N^{2/3}}$ such triple satisfying the given Diophantine equation.

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- Mar 5, 2012

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I'm afraid not.Coolerino! As I've never seen the method I'll ask the obvious question: Does this generate all possible solutions?

-Dan

It only generates some of the solutions that are multiples of $x_0$ and $y_0$.

And for instance it does not generate any of the multiples of with $x_0$ and $y_0$ reversed (assuming they are distinct).

Oops. You're right.Ok, thank you. I tried so many triples but of course missed this one

But, I like Serena, if you multiply both x0 and yo with the same factor a^k, then x and y aren't relatively prime anymore?

I missed that condition.

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- Mar 5, 2012

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Your problem looks a bit like a pythagorean triplet.But, I like Serena, if you multiply both x0 and yo with the same factor a^k, then x and y aren't relatively prime anymore?

Let's see if we can make it one.

Suppose we pick $z=5^m$, or rather $z=5^{2k+1}$.

Then we get:

$$x^2 + y^2 = 5\left(5^{2k+1}\right)^3$$

$$x^2 + y^2 = \left(5^{3k+2}\right)^2$$

Hmmm... not sure yet if this goes anywhere though...

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- Feb 7, 2012

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Yes, I think that does go somewhere, in combination with the fact that the product of a sum of two squares is a sum of two squares: $$(a^2+b^2)( c^2+d^2) = (ac+bd)^2 + (ad-bc)^2 = (ac-bd)^2 + (ad+bc)^2.$$ Start with $5=2^2+1^2$ and $5^3 = 125 = 11^2 + 2^2$. Using the above formula, you then get $$5(5^3) = 5^4 = 24^2+7^2,$$ $$5(5^2)^3 = 5^7 = 278^2 + 29^2,$$ $$5(5^3)^3 = 5^{10} = 3116^2 + 237^2,$$ $$\ldots\,.$$Your problem looks a bit like a pythagorean triplet.

Let's see if we can make it one.

Suppose we pick $z=5^m$, or rather $z=5^{2k+1}$.

Then we get:

$$x^2 + y^2 = 5\left(5^{2k+1}\right)^3$$

$$x^2 + y^2 = \left(5^{3k+2}\right)^2$$

Hmmm... not sure yet if this goes anywhere though...

In this way, you inductively construct $x_n$, $y_n$ so that $x_n^2 + y_n^2 = 5(5^n)^3$. The only other thing you need to do is to ensure that $x_n$ and $y_n$ are relatively prime. But their only possible common prime factor is $5$, so you want to choose the $\pm$ signs in the formula $(ac\pm bd)^2 + (ad\mp bc)^2$ in such a way as to avoid that. That is always possible, since (in the inductive construction) it cannot happen that both $ac+bd$ and $ac-bd$ are multiples of $5$.

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Thank you both. Amazing how you think of this!!!

I get it until your last sentence: 'That is always possible, since (in the inductive construction) it cannot happen that both ac+bd and ac−bd are multiples of 5 .'

Can you please give me some explanation on this?

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- Feb 7, 2012

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In more detail, the inductive hypothesis is "There exist co-prime integers $x_n,y_n$ such that $x_n^2+y_n^2 = 5^{3n+1}$." The inductive step then takes the form $5^{3n+4} = 125(5^{3n+1}) = (11^2 + 2^2)(x_n^2+y_n^2) = (11x_n\pm 2y_n)(2x_n\mp 11y_n)$. But neither $x_n$ nor $y_n$ can be a multiple of $5$, and therefore $4y_n$ is also not a multiple of $5$. Suppose that $11x_n+2y_n$ is a multiple of $5$. Then $11x_n-2y_n = (11x_n+2y_n) - 4y_n$ is not a multiple of $5$. So by a suitable choice of $+$ and $-$ we can define $x_{n+1} = 11x_n\pm 2y_n$ and $y_{n+1} = 2x_n\mp 11y_n$ so that $x_{n+1}^2 + y_{n+1}^2 = 5^{3n+4}$ and $x_{n+1}$ is not a multiple of $5$. It then follows that $y_{n+1}$ is also not a multiple of $5$. But $x_{n+1}$ and $y_{n+1}$ cannot have any other common prime factor and are therefore coprime.I get it until your last sentence: 'That is always possible, since (in the inductive construction) it cannot happen that both ac+bd and ac−bd are multiples of 5 .'

Can you please give me some explanation on this?

- Mar 22, 2013

- 573

Very nice, Opalg. This particular superfermat form was much harder, although Cohen reckons that there should be infinitely many solutions for any spherical superfermat coordinate.

Now, it is a very interesting thing to discuss with this nice little diophantine form. There are lots of problem related to this one, for example, is there any (at least partial) parameterization for this?* For another, is there any genuine lower bound for the solutions?** If so, then what is the order of magnitude?

(*) : I think there is a way by analysis in \(\displaystyle \mathbb{Z}\)

(**) : Note that a couple of posts back, my analysis partially settles the matter. A better treatment of the sum of squares function and a probabilistic approach should show that there are only finitely many N exceeding \(\displaystyle \delta N^\epsilon \log N\) for some doable \(\displaystyle \delta, \epsilon \geq 1\).

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- Mar 5, 2012

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Perhaps. But I for one did not understand your analysis.Note that a couple of posts back, my analysis partially settles the matter.

Could you clarify?

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- Feb 7, 2012

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Now, it is a very interesting thing to discuss with this nice little diophantine form. There are lots of problem related to this one, for example, is there any (at least partial) parameterization for this?* For another, is there any genuine lower bound for the solutions?** If so, then what is the order of magnitude?

(*) : I think there is a way by analysis in \(\displaystyle \mathbb{Z}\). I hope someone can bring this one down, I am too busy with the incoming examinations to do anything with it.

(**) : Note that a couple of posts back, my analysis partially settles the matter. A better treatment of the sum of squares function and a probabilistic approach should show that there are only finitely many N exceeding \(\displaystyle \delta N^\epsilon \log N\) for some doable \(\displaystyle \delta, \epsilon \geq 1\).

I conjecture that a necessary and sufficient condition for the problem to have a solution (with $x$ and $y$ coprime) is that all the prime factors of $z$ should be congruent to $1$ mod $4$.

The first case where $z$ has a prime factor other than $5$ is when $z=13$. We then have $5(13)^3 = 10985 = 101^2 + 28^2.$

- Mar 22, 2013

- 573

Never mind my previous post, I did a mistake. It should have been \(\displaystyle N^{4/3}/\log N\). For my defense, it was 3:40 am here when I was posting it.Perhaps. But I for one did not understand your analysis.

Could you clarify?

My approach was probabilistic. The usual way to show a weaker result is that the number of integers of the form \(\displaystyle x^2\) inside the interval \(\displaystyle [-N, N]\) is at most \(\displaystyle \mathcal{O}\left (N^{1/2}\right )\). Similarly, integers of the form \(\displaystyle y^2\) and \(\displaystyle -5z^3\) are \(\displaystyle \mathcal{O}\left (N^{1/2}\right )\) and \(\displaystyle \mathcal{O}\left (N^{1/3}\right )\), respectively.

So, the sum of them to be equal to 0, a doable heuristic would be that there are \(\displaystyle \mathcal{O} \left ( N^{1/2 + 1/2 + 1/3} \right )\) or, \(\displaystyle \ll N^{4/3}\) solutions smaller than N.

My approach was to think of \(\displaystyle x^2 + y^2\) and \(\displaystyle -5z^3\) rather than each term individually. So that we get \(\displaystyle \mathcal{O} \left ( N^{4/3} / \log N\right )\)

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- Mar 22, 2013

- 573

Since $x$ and $y$ has opposite parity, it's quite straightforward that your conjecture holds.I conjecture that a necessary and sufficient condition for the problem to have a solution (with $x$ and $y$ coprime) is that all the prime factors of $z$ should be congruent to $1$ mod $4$.

Balarka

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- #19

- Feb 7, 2012

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It will take a lot more than that to prove the conjecture. First, you have to show that if $z$ has any prime factor of the form $4k+3$ thenSince $x$ and $y$ has opposite parity, it's quite straightforward that your conjecture holds.I conjecture that a necessary and sufficient condition for the problem to have a solution (with $x$ and $y$ coprime) is that all the prime factors of $z$ should be congruent to $1$ mod $4$.

- Mar 22, 2013

- 573

Oh, I see what you're asking. Right, then I have merely proved that z is 1 modulo 4.It will take a lot more than that to prove the conjecture. First, you have to show that if $z$ has any prime factor of the form $4k+3$ theneitherno solution existsor, if it does, then in every such solution $x$ and $y$ must have a nontrivial common factor. Then you also have to show that if all the prime factors of $z$ are of the form $4k+1$ then a solution does exist with $x$ and $y$ coprime.

For the other part of the conjecture, that it is sufficient to have z in the 1 mod 4 congruence class, I expect this to be almost surely false.

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- Mar 5, 2012

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(*) : I think there is a way by analysis in \(\displaystyle \mathbb{Z}\). I hope someone can bring this one down, I am too busy with the incoming examinations to do anything with it.

Given the original problem with $x^2+y^2=5z^3$ and gcd(x,y)=1 and x,y,z positive.

We can factorize the left hand side in $\mathbb Z

$$(x+iy)(x-iy)=5z^3$$

Let d be the gcd(x+iy, x-iy).

Then d divides the sum and the difference: d|2x and d|2iy. Therefore d is one of 1, 1+i, 2 up to a unit.

Only d=1 pans out, so (x+iy) and (x-iy) are co-prime in $\mathbb{Z}

We can factorize 5 into (1+2i)(1-2i) which are also co-prime.

So we might construct x+iy as $(1+2i)(a+ib)^3$ and x-iy as$ (1-2i)(a-ib)^3$.

Multiplying this out, and separating the real and imaginary parts gives us x and y, and from their choice also z:

\begin{aligned}

x&=a^3-6a^2b-3ab^2+2b^3 \\

y&=2a^3+3a^2b-6ab^2-b^3 \\

z&=a^2+b^2

\end{aligned}

That leaves how to find a and b such that x and y are co-prime and positive.

I still have some trouble finding those.

- Mar 22, 2013

- 573

Finding a parameterization for the general case is easy, the problem occurs when one wants (x, y) = 1.

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- Mar 5, 2012

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I think it should suffice if $x+iy=(1+2i)(a+ib)^3$ and $x-iy = (1-2i)(a-ib)^3$ are co-prime.Finding a parameterization for the general case is easy, the problem occurs when one wants (x, y) = 1.

And to get that, it should suffice that $\gcd(a+ib, 1-2i)=\gcd(a+ib, a-ib)=1$.

And finally, we don't have to find any $a$ and $b$, only that there are infinitely many.

I guess I'll have to practice a bit with gaussian integers.

- Mar 22, 2013

- 573

Probably true, perhaps. You shouldn't pay much attention to my "baseless" heuristics and criticism now, since I am having a very less time of doing math outside of the syllabus since my examinations are coming (a day to go).

And by the way, I think Darmon & Granville must have something to say about these particular equations. I believe there exists a general theorem for all spherical genrealized Fermat-Catalan equations, a theorem on infinitude of the solutions. A similar theorem concerning the finitude of parabolic ones (appearing in the same paper) has been explicitly worked out there, as I have heard, and a general construction of parametric solution is given also.

Balarka

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- #25

- Mar 5, 2012

- 8,908

Gotcha!

Let $N(u+iv)$ be the gaussian integer norm. That is, $N(u+iv)=u^2+v^2$.

Now suppose $d=\gcd(a+ib, 1-2i)$.

Then d|1-2i and N(d)|N(1-2i)=5.

So we can pick a and b such that $5\not|a^2+b^2$ to make sure a+ib and 1-2i are co-prime.

Suppose $e=\gcd(a+ib, a-ib)$.

Pick a and b co-prime.

Then e|2a and e|2ib, therefore e=1, 1+i, or 2.

Also pick $a \not\equiv b \pmod 2$, then we get that e=1.

There are infinitely many a and b such that $\gcd(a,b) =1, \quad 5 \not| a^2 + b^2, \quad a \not\equiv b \pmod 2$.

With such a and b, the chosen x+iy and x-iy are co-prime.

It follows that x and y are co-prime, because if they have a common divisor > 1, that divisor would also divide both x+iy and x-iy.

Therefore $x^2+y^2=5z^3$ with $\gcd(x,y)=1$ and $x,y,z \in \mathbb N$ has infinitely many solutions. $\qquad \blacksquare$