## Solution to Diophantine equations

I am trying to solve a problem that boils down to finding the number of solutions of a linear Diophantine equation with restrictions. To be precise for equations :

$$ax + by + cz = k$$ and $$x + y + z = n$$

Refer the solution under “Diophantine Systems with Restrictions” on this page. How do I code this mathematical solution? My constraints include $$k$$ ranging from $$10^{-9}$$ to $$10^9$$ and $$n$$ from $$1$$ to $$10^5$$. Also, what would be the run time. I can code this but in a very haphazard and inefficient manner. Is there a neat or maybe a standard solution to this?

## Quadratic diophantine equations and geometry of numbers

Let (for concreteness) $$a = 2$$, $$b = \sqrt{5}$$ and $$\varphi = (\sqrt{5}+1)/2$$. I am interested in solutions $$(w,x,y,z) \in \mathbb{Z}[\varphi]^4$$ of the system

$$w^2 – ax^2 -by^2 + abz^2 = 1$$ $$\lvert w^2 + ax^2 +by^2 + abz^2 \rvert \ll \infty$$ $$\lvert \bar{w}^2 + a\bar{x}^2 -b\bar{y}^2 – ab\bar{z}^2 \rvert \le C$$ for some constant $$C$$. Here $$\overline{\alpha + \beta\sqrt{5}} = \alpha – \beta\sqrt{5}$$ and “$$\ll \infty$$” means that ideally I would like to enumerate solutions in increasing order of this value. (Restriction of scalars turns this problem into a system of two quadratic equations and two inequalities in eight variables in $$\mathbb{Z}$$; if someone wants to see it, I can write it out including potential mistakes).

1. What is the best (or even any practical) way to produce these?

I am aware that there is a lot of classical mathematics associated to this question but I don’t quite manage to put it together. Perhaps a subquestion is:

1. Can one enumerate the squares $$s$$ in $$\mathbb{Z}[\varphi]$$ with $$\lvert \bar{s} \rvert \le C$$ in increasing order of $$\lvert s \rvert$$?

Context: Let $$k = \mathbb{Q}(\varphi)$$ and let $$A$$ be the quaternion algebra $$(\frac{a,b}{k})$$ with norm $$\nu$$. With the above values the algebra $$A$$ is a skew field but tensoring with $$\mathbb{R}$$ in the two possible ways (taking $$\sqrt{5}$$ to $$\pm\sqrt{5}$$) gives an isomorphism with $$M_2(\mathbb{R})$$ which we equip with the map $$\left\lVert\left(\begin{array}{cc}\alpha&\beta\\gamma&\delta\end{array}\right)\right\rVert = \frac{1}{2}\left(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\right).$$ The above system then asks for solutions $$\lambda$$ in the maximal order of $$A$$ for $$\nu(\lambda) = 1$$, $$\lVert\lambda\rVert_{\sqrt{5} \mapsto \sqrt{5}} \ll \infty$$ and $$\lVert \lambda \rVert_{\sqrt{5} \mapsto -\sqrt{5}} \le C$$.

## Optimal Roth-type result in diophantine approximation

Let $$\alpha$$ be a real algebraic number. It is easy to see that if $$\deg(\alpha) = 2$$, that for there exists a number $$c(D(\alpha))$$, where $$D(\alpha)$$ is the discriminant of the primitive integral quadratic polynomial whose root is $$\alpha$$, such that

$$\displaystyle \left \lvert \alpha – \frac{p}{q} \right \rvert > \frac{c(D(\alpha))}{q^2}$$

for all rational numbers $$p/q$$ with $$q > 0$$ and $$\gcd(p,q) = 1$$. In view of Dirichlet’s theorem, this is the best possible result.

In 1958 Klaus Roth proved the following stunning theorem, as a culmination of the approach invented by Axel Thue in 1909 and developed by many subsequent authors including Siegel and Dyson. He proved that for any algebraic number $$\alpha$$ and $$\varepsilon > 0$$ there exists a number $$c(\alpha, \varepsilon)$$ such that for rational numbers $$p/q$$ with $$q > 0$$ and $$\gcd(p,q) = 1$$ we have

$$\displaystyle \left \lvert \alpha – \frac{p}{q} \right \rvert > \frac{c(\alpha, \varepsilon)}{q^{2 + \epsilon}}.$$

The problem is that the dependence of $$c(\alpha, \varepsilon)$$ on either parameter is ineffective. Thus in many applications one has to use a qualitatively worse but effective version of Roth’s theorem to get results. Lang also conjectured that one can replace the $$q^\varepsilon$$ term in the denominator with $$(\log q)^{1 + \varepsilon}$$. Indeed, if we take $$\varepsilon = 1$$ we can ‘remove’ the dependence on $$\varepsilon$$ completely.

I am asking about what is expected conjecturally, in view of the quality of result in the quadratic case. In particular, does one expect to be able to find a number $$c(\alpha)$$ such that

$$\displaystyle \left \lvert \alpha – \frac{p}{q} \right \rvert > \frac{c(\alpha)}{(q \log q)^2}$$

for all $$p/q \in \mathbb{Q}, \gcd(p,q) = 1, q > 0$$, and such that $$c(\alpha)$$ can be made to depend at most polynomially on the height of the algebraic number $$\alpha$$?

## The diophantine equation: $x^5$+$y^3$=$z^3$

Is there any progress on proving that there are no solutions to $$x^5$$+$$y^3$$=$$z^3$$ for $$gcd(x,y,z)=1$$?

Motivation: Bruin proved the $$(3,3,5)$$ case, and Darmon-Merel proved the $$(3,5,5)$$ and $$(5,5,3)$$ case, and FLT (Wiles) handles the $$(3,3,3)$$ and $$(5,5,5)$$ case, so the $$(5,3,3)$$ case along with the others handles everything above an exponent of $$2$$ and below an exponent of $$6$$ for $$p$$,$$q$$, or $$r$$.

Assuming the $$ABC$$ conjecture:

$$c

$$Rad(x^p,y^q,z^r)\le xyz

$$z^r hence $$r\le5$$.

If we have no solutions for $$(5,3,3)$$ also, then $$r\le2$$.

Then, we have a proof for the theorem:

If the $$ABC$$ conjecture is true then Beal’s conjecture is true.

## Strong Approximation for sol’ns to quadratic diophantine equations

Can anyone either direct me to an elementary proof in the literature–or show me why this (Conjecture 1 stated below) is true–or if I am mistaken and it is not true:

1. For any 4-tuple $$\xi = (x_0,x_1,x_2,x_3) \in \mathbb{Z}^4$$, let us define the quadratic form $$Q(\xi) = x_0^2+x_0x_3+2x^2_1 + x_1x_3+13x^2_2+2x^3_2$$.

2. Now let $$q \in \mathbb{N}$$ be a prime power such that $$(26,q)=1$$, and let $$\cal{B}$$ be the set $$\{B=(b_0,b_1,b_2,b_3) \in \mathbb{Z}^4$$; $$Q(B) \equiv_q 1\}$$.

3. Now let $$\cal{A}$$ be the set $$\{A = (a_0,a_1,a_2,a_3); Q(A) = 2^k$$ for some nonnegative integer $$k\}$$.

Conjecture 1: Then using the notation as above, then for any $$B =(b_0,b_1,b_2,b_3) \in \cal{B}$$ there is an $$A =(a_0,a_1,a_2,a_3) \in \cal{A}$$ that satisfies $$a_i \equiv_q b_i$$ for each $$i \in \{0,1,2,3\}$$.

I do need this result for a paper that I am writing. But I must admit to not understanding the Strong Approximation Theorem well at all, nor the associated math around this–which is what I think this looks like. [I am in graph theory].

Many Thanks!

## Quadratic Diophantine Equation in Three Varables

During some free time I had, I was wondering how to find the integer solutions $$(x,y,z)$$ to this generalized equation: $$z^2=axy+bx+cy+d$$ I am specifically looking for ways that do no involve factoring. And $$a,b,c,d$$ are all non-zero integers.

## Generating function for diophantine equation

Consider $$a_1x_1\dots +a_nx_n=n$$ for positve integers $$a_i,n$$. And let $$A_n$$ be the number of positive valued solutions $$(x_1,\dots x_n)$$. I want to understand why the generating function of $$A_n$$ is given by $$\prod_{i=1}^m \frac{1}{(1-x^{a_i})}\,, \vert x\vert <1$$.

I am reading an introduction to diophantine equations by Andreescu, Andrica, Cucurezeanu. But I do not understand its proof. Using a geometric series we have$$\frac{1}{1-x^{a_k}}=1+x^{a_k}+x^{2a_k}+\dots\,, \quad k=1,\dots , m$$, hence $$\prod_{i=1}^m \frac{1}{(1-x^{a_i})}=(1+x^{a_1}+x^{2a_1}+\dots)\cdots (1+x^{a_m}+x^{2a_m}+\dots)=1+A_1x+\dots +A_nx^n+\dots$$ The last equality is not obvious for me, could someone elaborate?

## Diophantine equation $3^a+1=3^b+5^c$

This is not a research problem, but challenging enough that I’ve decided to post it in here:

Determine all triples $$(a,b,c)$$ of non-negative integers, satisfying $$1+3^a = 3^b+5^c.$$

## The system of Diophantine equations with same solution

There is a system of Diophantine equations: $$\begin{equation*} \begin{cases} 368=x^7 (mod 407)\ 389=x^{11}(mod 407) \end{cases} \end{equation*}$$

Obviously, they have same solution $$x=407n+15$$ finding out by Wolfram:)

However, solving each of them by hand is quite a difficult task. The question is: how knowing both of the equations ease the task? How could they be solved manually in that case?

## On the Diophantine equation $3^a + b^2 = 5^{c}$, with $a, b, c \in \mathbb{N}$

Consider the Diophantine equation $$3^a + b^2 = 5^{c},$$ with $$a, b, c \in \mathbb{N}$$.

My question: $$a, b, c$$ must be necessarily even numbers?

Up to now, I only get that $$b$$ is an even number. To show this, I suppose $$b = 2 b’ + 1$$. Remark that $$5^{c} – 3^{a} =(2 c’ + 1) – (2 a’+ 1) = 2(c’- a’)$$ (even number). Thus, $$b$$ cannot be a odd number. On the other hand I did not see an elementar way to prove that $$a, c$$ has the same property.