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(abc)^2$

$ Rad(x^p,y^q,z^r)\le xyz<z^3$

$ z^r<z^6$ 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!

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?

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.