User joro - MathOverflow

Answer by joro for Help with this Diophantine equation

2013-05-21T14:48:27Z

For $0 < a \le3966887 $ solutions are $(9419, 10418, 8146),(69167, 10776, 87090)$ and (added) $(3966887, 2434179, 4797573)$.

Here is an idea for searching. Loop $a$ from $1$ to certain bound.

You have to solve $x^3 + y^3 = C + 2 a^3 = N$. This is easy to solve if $N$ can be factored since $x^3+y^3$ factors nicely.

Added to the edited question

You have to solve $ a^3+b^3+c^3 + 83449 = 3 d $

Just pick "random" $a,b,c$ such the the lhs is divisible by $3$ like $(300,301,304)$ and $d=27482938$

Here is a pari/gp script which found the solutions.

 {
 jobin1()=
 th=thueinit(x^3+1,1);
 C=36650;
 for(a=1,10^5,
 A=C+2*a^3;
 v=thue(th,A);
 if(v == [],next);
 print([a,v]);
 );
 }
 jobin1()

Answer by joro for Is this combination of generalized polygamma and dilogarithm actually zero? $\Im\;\psi^{(-2)}(1+i)+\frac1{4\pi}\text{Li}_2(e^{-2\pi})-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12$

2013-05-19T09:25:30Z

Edited Maple's $\psi$ disagrees with Wolfram Alpha and your integral, so here are some conjectures with both:

According to Maple -- your equality fails with this definition of psi.

$$ 24 \Im{\psi^{(-2)}}(i)+6 Li_2(e^{-2 \pi}) / \pi + 5 \pi - 12= 0$$

$$ 24 \Im{\psi^{(-2)}}(1+i)+6 Li_2(e^{-2 \pi}) / \pi + 5 \pi + 12= 0$$

Simlarly for $Li_4$,

$$ -1440 \Im{\psi^{(-4)}}(1+i)+ 90 Li_4(e^{-2 \pi}) / \pi^3 - \pi + 220 = 0$$

Checked with precision 1000 decimal digits.

Using your integral and mpmath, these appear to hold for $\psi^{(-2)}(i)$ and $\psi^{(-2)}(2+i)$

$$ -24 \Im{\psi^{(-2)}}(i)-6 Li_2(e^{-2 \pi}) / \pi - 5 \pi + 12 +24 \log{\sqrt{\pi}} + 24 \log{\sqrt{2}} = 0$$

$$ -24 \Im{\psi^{(-2)}}(2+i)-6 Li_2(e^{-2 \pi}) / \pi + \pi - 36 +24 \log{\sqrt{\pi}} + 48 \log{\sqrt{2}} = 0$$

These were found using linear dependencies in real numbers (pari's lindep).

Wolfram Alpha finds another expression for $\psi$.

Closed form for derivatives $\zeta^{(n)}(1/2)$

2013-05-10T10:49:32Z

According to mathworld 41,42. "Derivatives $\zeta^{(n)}(1/2)$ can also be given in closed form" with example for the first derivative.

What is the closed form? References?

The motivation is that this question expresses $\zeta(3)$ in terms of $\zeta(1/2)$ and the first 3 derivatives, so closed form possibly might result in closed form for zeta(3) (unless the closed form is derived by the linked question).

Particaluraly intersted in the second and third derivatives.

On what the derivatives would depend helps too.

zeta(2k+1) is a rational multiple of pi^{2k} zeta'(-2 k) ?

2013-05-07T10:09:19Z

Probably this is well know and elementary and will delete it, but couldn't find it on the web.

Got a sketch of proof and numerical evidence that $\zeta(2k+1)$ is a rational multiple of $\pi^{2k} \zeta'(-2k)$

An identity from Derivatives of the Hurwitz Zeta Function for Rational Arguments p.7

$$ \zeta'(-n,x) + (-1)^n \zeta'(-n,1-x) = \pi i \frac{B_{n+1}(x)}{n+1} + \frac{n!}{(2 \pi) ^ n} e^{-\pi i n / 2} \operatorname{Li}_{n+1}(e^{2\pi i x}). \qquad (21)$$

For even $n$ and $x=\frac12$ (21) is:

$$ 2 \zeta'(-n,\frac12) = \pi i \frac{B_{n+1}(\frac12)}{n+1} \pm \frac{n!}{(2 \pi) ^ n} \operatorname{Li}_{n+1}(-1).$$

The choise of $\pm$ depends on $e^{-\pi i n / 2}$.

According to Wolfram Alpha $Li_{n+1}(-1)$ is an integer multiple of $\zeta(n+1)$ and $B_{n+1}(\frac12)$ vanishes.

$\zeta(s,\frac12) = (2^s-1) \zeta(s) $. Taking derivative and having in mind $\zeta(-2k)=0$ we have $\zeta'(s,\frac12)$ is a rational multiple of $\zeta'(s)$.

For even $n$ and $x=\frac12$ (21) simplifies to:

$$ \mathbb{Q} \zeta'(-n) = \pm \mathbb{Z} \frac{n!}{(2 \pi) ^ n} \zeta(n+1). \qquad (1)$$

In particular,

$$ \zeta(3) = -4 \zeta'(-2) \pi^2$$ $$ \zeta(5) = 4/3 \zeta'(-4) \pi^4$$

The last two hold with precision of 1000 digits.

Is this true?

Is this known?

zeta(3) in terms of derivatives of zeta at 1/2 and pi

2013-05-05T09:53:15Z

Got numerical support that for odd $n$, $\zeta(n)$ might be expressed in terms of the derivatives of $\zeta(\frac12)$.

Based on More Zeta Functions for the Riemann Zeros, Andre Voros , p.12, Table 3:

Conjecture: For odd $n$, $$\zeta(n) = \left(\frac{2}{(n-1)!} (\log(|\zeta|)^{(n)} (\frac12) - 2^n \beta(n))\right)/(2^n-1)$$

$\beta(n)$ is Dirichlet beta function and it is a rational multiple of $\pi^n$ for odd $n$. The derivative can be expressed in terms of $\zeta(\frac12),\zeta^{(k)}(\frac12)$

For $n=3$ get numerical support for:

The last equality holds with precision $10^4$ decimal digits. One can eliminate the first derivative since there is closed form for $\zeta'(\frac12)/\zeta(\frac12)$

Is this result true?

sage/mpmath code in case of typos of the latex.

#run in sage

import mpmath
from mpmath import mpf
mpmath.mp.pretty=True

def zeta3test():

    n=3
    mpmath.mp.dps=10^3
    zeta3=mpmath.zeta(3)

    Pi=mpmath.pi
    gamma=mpmath.euler
    z12=mpmath.zeta(1/mpmath.mpf(2))
    z1=mpmath.zeta(1/mpmath.mpf(2),derivative=1)
    z2=mpmath.zeta(1/mpmath.mpf(2),derivative=2)
    z3=mpmath.zeta(1/mpmath.mpf(2),derivative=3)

    # eliminate the first derivative

    #rh0=1/32*(72*Pi*mpmath.log(2)*mpmath.log(Pi)*z12+144*gamma*mpmath.log(2)*mpmath.log(Pi)*z12+72*mpmath.log(2)*z12*gamma*Pi+24*mpmath.log(Pi)*z12*gamma*Pi-144*z2*mpmath.log(2)-48*z2*mpmath.log(Pi)+216*mpmath.log(2)^3*z12+8*mpmath.log(Pi)^3*z12-48*z2*gamma-24*z2*Pi+8*z12*gamma^3+z12*Pi^3+72*mpmath.log(2)*z12*gamma^2+18*mpmath.log(2)*z12*Pi^2+24*mpmath.log(Pi)*z12*gamma^2+6*mpmath.log(Pi)*z12*Pi^2+216*mpmath.log(2)^2*mpmath.log(Pi)*z12+72*mpmath.log(2)*mpmath.log(Pi)^2*z12+216*gamma*mpmath.log(2)^2*z12+24*gamma*mpmath.log(Pi)^2*z12+108*Pi*mpmath.log(2)^2*z12+12*Pi*mpmath.log(Pi)^2*z12+32*z3+12*z12*gamma^2*Pi+6*z12*gamma*Pi^2)/z12
    z12a=mpmath.fabs(z12)
    rh1= -z3/z12a -3*z2*z1/z12a**2 -2* z1**3/z12a**3

    #print 'rh',mpmath.chop(rh0-rh1)
    #rhs= (mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - mpmath.pi^3 / 4 )/(7)
    rhs= (rh1 - mpmath.pi**3 / 4 )/(7)
    print mpmath.chop(zeta3-rhs)

def conjecture1(n):
    """

    voros, p. 12
    """
    assert n%2==1
    a1= mpmath.zeta(n)
    a2= (2/factorial(n-1) * mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - 2**(n) * dirichletbeta(n))/(2**n-1)
    print mpmath.chop(a1-a2)


def dirichletbeta(s):
    """
    dirichlet beta
    """
    return 4**(-s) * (mpmath.hurwitz(s,1/4)-mpmath.hurwitz(s,3/4))

Answer by joro for zeta(3) in terms of derivatives of zeta at 1/2 and pi

2013-05-06T08:22:32Z

Looks like this is proved by Andre Voros, "Zeta Functions over Zeros of Zeta Functions",p. 69, eq. 7.49:

$$ (\log{|\zeta|})^{(n)} (\frac12) = \frac12 (n-1)![(2^n-1)\zeta(n) + 2^n\beta(n)],\qquad n > 1 , \qquad (7.49) $$

From which the conjecture follows.

Answer by joro for Product over the primes

2013-04-13T09:18:37Z

Modulo errors got unconditionally slightly smaller upper bound than your conjecture. prodeulerrat from Cohen's pari script can compute $\prod_{\text{p prime}}{1-1/O(x^2)}$.

For fixed $p < q < r < L$ define $g(p,q,r)=\prod_{\text{s prime}}{1-24/(pqrs)^2}$

$g$ is efficiently computable by the script and contains the contribution of $s > L$ times extra factors $ < L$. Compute the extra factors the slow way for each triple to get the exact product for $p < q < r < L < s$ for fixed $p,q,r$. Upper bound (since all terms are $<1$) for your product is the product over the exact products times the product $< L$. .

For $L=200$ got $0.9974805703680839724432166987$

Two products over primes

2013-04-12T14:09:14Z

For $k \in \mathbb{N}$ define $$ f(k) = \prod_{\text{p prime}}\left(1+\frac{1}{p^k(p^k+1)}\right)$$ $$ g(k) = \prod_{\text{p prime}}\left(1+\frac{1}{p^k(p^k-1)}\right)$$

By the product for zeta $f(1)=\zeta(2)/\zeta(3)$.

With 100 digits of precision and Cohen's pari script $f(2)=21/(2\pi^2)$ ,$g(1)=315/2\zeta(3)/\pi^4$.

Are the conjectured values correct?

Other closed forms?

$a^5+b^5=c^5+d^5$ and polynomial identities

2013-04-08T13:24:08Z

No nontrivial integer solutions to $$ a^5+b^5=c^5+d^5 \qquad (1)$$ are known.

(1) has infinitely many solutions in an extension of $\mathbb{Z}$ (root of $9-15x+37x^2 $ ) resulting from genus 0 curve and the identity: $$ (-2 y - 1)^5 + (x + \frac{1}{3} y - 1)^5- (\frac{1}{3} y - 1)^5 - (x - 2 y - 1)^5 = $$ $$ \left(-\frac{35}{81}\right) \cdot y \cdot (-3 x + 5 y + 6) \cdot x \cdot (9 x^{2} - 15 x y + 37 y^{2} - 18 x + 30 y + 18)$$

I wonder if some similar genus 0 or 1 curve might have points over $\mathbb{Q}$.

Let $p_1,p_2,p_3,p4 \in \mathbb{Q}[x,y]$, $\deg(p_i)=1, p_i \ne -p_j$ and $p_i$ are distinct.

Let $P=p_1^5+p_2^5-p_3^5-p_4^5$.

Q1. Can $P$ have irredicible factor of degree 3?

Q2. Is there a reasons all degree 2 factors of $P$ to not have infinitely many rational points over $\mathbb{Q}$?

Coulnd't solve Q1 by equating coefficients (couldn't solve the system).

Found a lot of genus 0 factors, but all of them didn't have rational points and linear factors gave trivial solutions.

Answer by joro for irreducible polynomials with arithmetic progression coefficients

2013-03-31T08:53:29Z

Extending Aaron's idea about identities.

Working over $\mathbb{Q}[x]$ this appears degree 3 identity for $c5,c6 \in \mathbb{Z}$: $a=-c5 c6, b= c6 (c5^3+c5^2+c5+1)/(3 c5^2+2 c5+1)$.

$$P=a+(a+b) x+(a+2 b) x^2+(a+3 b)x^3= (-x + c5) \cdot \text{quadratic}$$.

Scale to get integral values.

Found by equating coefficients and suppose higher degree identities exist.

Added Degree 4

a= -c5*c6
b= c6*(c5^4+c5^3+c5^2+c5+1)/(4*c5^3+3*c5^2+2*c5+1)
P=a+(a+b)*x+(a+2*b)*x^2+(a+3*b)*x^3+(a+4*b)*x^4 == (c5-x)*O(x^3)

Examples of interesting false proofs

2012-04-21T14:26:12Z

According to Wikipedia False proof

For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.

The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics.

What are some examples of interesting false proofs?

Answer by joro for Examples of interesting false proofs

2013-03-21T15:18:16Z

Doron Zeilberger proved that P is equal to NP

Abstract: Using 3000 hours of CPU time on a CRAY machine, we settle the notorious P vs. NP problem in the affirmative, by presenting a “polynomial” time algorithm for the NP-complete subset sum problem. Alas the complexity of our algorithm is $O(n^{10^{10000}})$ (with the implied constant being larger than the Skewes number).

Can GRH for complex primitive Dirichlet characters fail with a single non-trivial zero off the critical line?

2013-03-21T09:18:58Z

Can GRH for complex primitive Dirichlet character fail with a single non-trivial zero off the critical line?

For real characters this is impossible because the non-trivial zeros are in quadruples.

On the other hand this paper in Russian constructs a function very similar to zeta with exactly one quadruple of zeros off the critical line.

The motivation: GRH implies a certian efficiently computable sum over zeros of $L(s,\chi)$ must be $0$ and experimentally it is (indistinguishable from) $0$ and I wonder if this result is trivial even if GRH is false. IMO the sum should not be zero if the answer to the question is 'yes' or some cancellation doesn't happen.

A counterexample to a conjecture of Nash-Williams about hamiltonicity of digraphs?

2013-03-11T10:37:10Z

Maybe I am missing something, but found potential counterexample to a conjecture of Nash-Williams.

According to HAMILTONIAN DEGREE SEQUENCES IN DIGRAPHS

The outdegree and indegree sequences of digraph $G$ are $d_1^+ \le \cdots \le d_n^+$ and $d_1^- \le \cdots \le d_n^-$. Note that the terms $d_i^+$ and $d_i^-$ do not necessarily corresponds to the degree of the same vertex of $G$.

Conjecture 1 (Nash-Williams). Suppose that $G$ is a strongly connected digraph on $n \ge 3$ vertices such that for all $i < n/2$

(i) $d_i^+ \ge i + 1$ or $d_{n-i}^- \ge n - i$,

(ii) $d_i^- \ge i + 1$ or $d_{n-i}^+ \ge n - i$,

Then $G$ contains a Hamilton cycle.

The potential counterexample is $G$ on $6$ vertices with edges:

[(0, 3), (0, 5), (1, 4), (1, 5), (2, 3), (2, 4), (3, 0), (3, 2), (3, 4), (3, 5), (4, 0), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)]

$G$ is strongly connected and by inspection the degree sequences satisfy the hypotheses for $i \in [1,2]$ (both degree sequences are $[2, 2, 2, 4, 4, 4]$).

Nonhamitlonicity was shown using exhaustive search, sage 5.6 and Max Alekseyev's hamiltonian cycle counting pari program.

Is this really a counterexample the the conjecture of Nash-Williams?

Drawing of $G$:

Answer by joro for Transformation of a bivariate polynomial into a homogeneous one

2013-02-27T08:23:57Z

Edit corrected major mistake

One approach is to work symbolically and solve a system over the rationals.

Choose bounds for the degrees of $S,T,H$ and write them as $\sum a_m x^i y^j$ where each $a_m$ is a fresh variable and $H$ is homogeneous. $H(S(x,y),T(x,y))$ is a polynomial in $x,y$ with coefficients polynomials in $a_i$. Make a system by equating the coefficients of $P(x,y)=H(S(x,y),T(x,y))$ Solve the system over the rationals.

While this will work in theory, solving the system might be quite hard. Experimenting with your example and degrees $(2,2,3)$, maple found 4 solutions in about 2 minutes.

Partially optimistic might be the fact that the system is overdetermined.

Answer by joro for Ihara zeta function (graph theory) coefficients using a line graph

2013-02-23T10:44:17Z

Not sure if the question is well defined - you can remove 2-cycles in many ways, getting different digraphs.

One possible approach is start with empty set of edges $E$. For all edges $(u,v) \in E(G)$ add $(u,v)$ to $E$ iff $(v,u) \not \in E$.

Here is a sample sage implementation:

 def removedigons(G):
     ed=[]
     for u,v in G.edges(labels=False):
         if (v,u) in ed:  continue
         ed += [(u,v)]

     g=DiGraph(ed)
     return g

Semirings where solving linear systems is in P

2013-02-10T10:27:16Z

Solving linear systems appears hard in semirings.

In $\mathbb{N}_0 (+,\times)$ it is NP-complete via reduction to 1-in-3 SAT.

In the min-plus semiring the complexity is $ NP \cap coNP$ according to this

Are there nontrivial semirings where solving linear systems is in P?

Is they exist I would be interested what constraints they can encode.

Added clarification

By "linear system" mean a system of linear equations of the form: $$ c_1 \otimes x_1 \oplus c_2 \otimes x_2 \oplus \cdots \oplus c_n \otimes x_n = c_{n+1}$$ where $c_i$ are constants and $x_i$ are variables. If the standard definition is other, say $A x = B x$, would accept an answer about the standard definition.

By "solving a linear system" system mean finding at least one solution or claim that no solution exist. Though finding some kind of "basis" for all solutions would be interesting too.

Perfect powers on genus 0 curves (with restrictions)

2013-01-31T09:40:13Z

I suppose a conjecture implies this so there might be an unconditional proof.

Let $F(x,y)=0$ be a curve with infinitely many integral points $(u^m,v^n)$ where $\gcd(u,v)=1,m \ge 3,n \ge 2$. Such curves are easy to construct by starting with a parametrization for example.

For a bivariate polynomial $F$ define $\operatorname{High}(F)$ to be the sum of the highest degree monomials (i.e., $\operatorname{High}(F) = F$ iff $F$ is homogeneous). Let $\gcd(\operatorname{High}(F),xy)=1$.

Under these conditions is $\operatorname{High}(F)$ not square-free?

Maximizing the minimum outdegree of digraph without $m$ cycle

2013-01-27T14:19:17Z

Let $G$ be a simple digraph on $n$ vertices without a directed cycle of length $m$ (it may have directed cycles of length less than $m$. The cycles need not be simple).

How large the minimum outdegree $\delta^+$ can be?

For odd $m$ and even $n$ it is possible $\delta^+ = n/2 $.

$r$ diregular digraphs of girth $g$ can be on $ r (g - 1) + 1$ vertices, but this doesn't use the fact that smaller cycles may exists.

The case $m=3$ is interesting too.

Looks like Caccetta-Häggkvist Conjecture implies $\delta^+ < \frac{m n}{2 m - 1}$ for all $m,n$.

Answer by joro for Polynomial-time algorithm to compare numbers in Conway chained arrow notation

2013-01-21T13:10:49Z

Not sure this will do the job, but try Robert Munafo's hypercalc, the moto is:

Go ahead -- just TRY to make me overflow!

Hypercalc can work with quite big numbers, here is a sample session:

C1 = scale=100
C1 = 27^(86!) - (27^86)!
R1 = 10 ^ (3.467778644301262713584883219782046054843086208195414740688065133320263642461739090290922141022702407 x 10 ^ 130 )

Home page: http://mrob.com/pub/perl/hypercalc.html#versions

Download: http://mrob.com/pub/perl/index.html

Online javascript version: http://www.ylmass.edu.hk/~mathsclub/HyperCalc/HyperCalc.html

Are potential complex zeros not on the critical line of Dedekind zeta function in quadruples?

2013-01-13T08:06:17Z

This question arose from sums over zeros of Dedekind zeta function.

It is known that complex zeros of Dedekind zeta function are in pairs $\rho, 1 - \rho$.

Is it true that potential complex zeros not on the critical line of Dedekind zeta function must be in quadruples $\rho, 1 - \rho, \overline{\rho}, \overline{1 - \rho}$ ?

I am interested in the general case, not for specific number fields (or for number fields for which the answer is "no").

Answer by joro for counting triangle free graphs

2013-01-12T06:40:59Z

Computing the first terms in sage and searching in OEIS gave A006785 Triangle-free graphs on n vertices

The first terms are:

1, 2, 3, 7, 14, 38, 107, 410, 1897, 12172, 105071, 1262180, 20797002, 467871369, 14232552452

There is no formula in OEIS, some of the referenced papers might be useful.

Can the difference of two distinct Fibonacci numbers be a square infinitely often?

2012-01-03T10:54:14Z

Can the difference of two distinct Fibonacci numbers be a square infinitely often?

There are few solutions with indices $<10^{4}$ the largest two being $F_{14}-F_{13}=12^2$ and $F_{13}-F_{11}=12^2$

Probably this means there are no identities between near neighbours.

Since Fibonacci numbers are the only integral points on some genus 0 curves the problem is equivalent to finding integral points on one of few varieties. Fixing $F_j$ leads to finding integral points on a quartic model of an elliptic curve.

Are there other solutions besides the small ones?

[Added later] Here is a link to elliptic curves per François Brunault's comment.

According to DIOPHANTINE EQUATIONS, FIBONACCI HYPERBOLAS,. AND QUADRATIC FORMS. Keith Brandt and John Koelzer.

Fibonacci numbers with consecutive odd indices are the only solutions to $$ x^2-3xy+y^2 = -1 \qquad (1)$$

Fibonacci numbers with consecutive even indices are the only solutions to $$ x^2-3xy+y^2 = 1 \qquad (2)$$

Given $F_n$ and $F_{n+2}$ one can compute $F_{n+k}$ using the linear Fibonacci recurrence and $F_{n+k}$ will be a linear combination $l(x,y)$ of $x,y$. Adding $l(x,y)-x=z^2$ to (1) or (2) gives a genus 1 curve. (Or just solve $l(x,y)-x=z^2$ and substitute in (1) or (2) to get a genus 1 quartic).

The closed form of $l(x,y)$ might be of interest, can't find the identity at the moment.

Probably a genus 0 curve with integral points $F_{2n},F_{2n+1}$ will be better.

Added much later

Does some generalization of the $abc$ conjecture predict something?

For 3 Fibonacci numbers identities are much easier:

$$ F_{4n+1}+F_{4n+3}-F_3 = L_{2n+1}^2$$

Is OEIS A007018 really a subsequence of squarefree numbers?

2013-01-04T12:27:41Z

A comment in A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1 claims

Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004

Is it really so?

As far as I know, it is open problem if a polynomial $f \in \mathbb{Z[x]}$ of degree $\ge 5$ can be squarefree infinitely often (some source require $f$ to be irreducible).

If the OEIS comment is correct, the sequence will give infinite family of (irreducible) polynomials which are squarefree infinitely often.

Let $a_n$ is OEIS A007018. Set $a_n = x$ and $$f(x)=a_{n+4}=x \cdot (x + 1) \cdot (x^{2} + x + 1) \cdot (x^{4} + 2 x^{3} + 2 x^{2} + x + 1) \\ \cdot (x^{8} + 4 x^{7} + 8 x^{6} + 10 x^{5} + 9 x^{4} + 6 x^{3} + 3 x^{2} + x + 1)$$

$f(a_n)=a_{n+4}$ will be squarefree infinitely often (including the irreducible degree 8 factor) and iterating $x \mapsto x^2+x$ will produce infinite family of polynomials with this property.

Added For reference of squarefree values of polynomials the search terms are square free values of polynomials. E.g. here p.1 and here 11. Squarefree values of polynomials.

Curves with infinitely many integral points consecutive Fibonacci numbers

2013-01-02T12:13:52Z

I suspect the Granville-Langevin conjecture implies this.

Let $F(x,y)=0$ be a curve with infinitely many integral points $(F_n,F_{n+1})$.

Is it true that either $x^2+x y - y^2 -1$ or $x^2+ x y -y ^2 +1$ divides $F(x,y)$?

Probably this can be generalized to integral points $(F_{f(n)},F_{g(n)})$.

Answer by joro for Any non-conforming numbers?

2012-12-22T08:16:37Z

Your equation has infinitely many efficiently computable solutions $(m,n,p)=(2,2,p)$ and maybe infinitely many $(m,n,p)=(3,3,2)$ for all integers $a$ s.t. $x^m - y^n \pm z^p =a$.

Every odd integer $b$ is the difference of two squares $(1+b)/2,(b-1)/2$ so take any $p$, any $z$ the opposite parity of $a$ and express $a \mp z^p$ as difference of two squares.

For the $(3,3,2)$ case, pick random $c$ and set $x=u+c,y=u$ and pick the minus sign. Your equation is $x^3-y^3-z^2=3u^2c + 3uc^2 + c^3 -z^2=a$ leading to $ 3 u^{2} c + 3 u c^{2} + c^{3} - z^{2} - a=0$. The last equation is genus $0$ curve which might have infinitely many integral points unless I am missing some obstruction and you can try another $c$. For $c=3 c'^2$ the curve has explicit rational parametrization.

Can FLT fail with a parametrization over some extension of Z?

2012-12-17T10:32:03Z

Not sure if this makes sense, but is it possible Fermat's Last Theorem to fail with a parametrization over some extension of $\mathbb{Z}$, i.e. are there not all constant $x(t),y(t),z(t) \in K[t]$ where $K$ is an extension of $\mathbb{Z}$ s.t. $$x(t)^p + y(t)^p=z(t)^p, x(t)y(t)z(t) \ne 0, p > 2,\gcd(x(t),y(t),z(t))=1 $$

I suppose this would mean in an extension of $\mathbb{Q}$ the curve $x^p + y^p = z^p$ will be of genus $0$.

Tried equating coefficients with $p=3$, got relatively small undetermined system but failed to solve it or compute groebner basis. The system has solutions like $x(t)=0$.

EDIT: The coprimality condition is to avoid scaling a single solution by a polynomial.

2D visualization of sum of divisors using Cantor pairing

2012-11-28T08:52:42Z

Related to Gerhard's question about ascii plots. On the SeqFan mailing list was suggested to plot an integer sequence this way:

Let $F(x,y)= (x+y) (x+y+1)/2+y$ be the Cantor pairing. To plot an integer sequence $a(n)$, for a point $(x,y)$ compute $a(F(x,y))$ and assign color to the integer, e.g. in grayscale smaller is darker, for RGB/HSV there are other choices to map to color.

When $a(n)=\sigma_0(n)$ where $\sigma_0(n)$ is the number of divisors of $n$, the 2D plot shows some structure (hopefully not caused by visual artifacts).

Is there an explanation for the structure in the plot?

Color plot of $\sigma_0(F(x,y))$, smaller is darker (grayscale is quite similar):

When examining the integer values there are some large diagonals indeed.

Polynomial version of the conjecture about Power free-values of polynomials

2012-12-11T15:02:18Z

The conjecture about Power free-values of polynomials is: Let $F(X)$ be a polynomial with integer coefficients and no repeated roots. For any $\epsilon > 0$, there exists a constant $C_{\epsilon,F}$ such that for any integer $n$ $$ |n|^{\deg{(F)}-1-\epsilon} \le C_{\epsilon,F} \operatorname{rad}(F(n))$$

The conjecture implies this polynomial version. For $f(x) , g(x) \in \mathbb{Z}[x]$ and $f(x)$ squarefree,

$$\deg (\operatorname{rad}(f(g(x)))) > \deg(g(x)) (\deg(f(x))-1) \qquad (1)$$

The bound is tight because for Chebyshev polynomials $T_n,U_n$, $T_n(x)^2 - 1 = (x^2-1) U_{n-1}^2(x)$ with $f(x)=x^2-1$.

Is (1) proved for polynomials?

Answer by joro for Maximum size of powers with a given difference

2012-12-11T07:15:44Z

Perfect Powers: Pillai's works and their developments is related to your question.

p.9: Conjecture 3.1. For any $\epsilon > 0$, there exists a constant $\kappa(\epsilon) > 0$ such that, for any positive integers $(a, b, x, y)$, with $x \ge 2, y \ge 2$ and $a^x \ne b^y$ , $$|a^x - b^y | \ge \kappa(\epsilon) \max (a^x , b^y)^{ 1-(1/x)-(1/y)-\epsilon}$$

p.7 In 1986, J. Turk [126] gave an effective estimate from below for $|x^n - y^m |$, which was improved by B. Brindza, J.-H. Evertse and K. Gyory in 1991 and further refined by Y. Bugeaud in 1996:

Let $x$ be a positive integer and $y, n, m$ be integers which are $\ge 2$. Assume $x^n \ne y^m$ . Then $|x^n - y^m | \ge m^{2/(5n)} n^{-5} 2^{-6-42/n}.$

Comment by joro

2013-05-21T15:03:43Z

Not sure if this is correct, check the counterexamples in my answer.

Comment by joro

2013-05-19T06:43:28Z

Wolfram Alpha gives alternative representation for psi, check: wolframalpha.com/input/…

Comment by joro

2013-05-11T05:41:09Z

Thank you Micah. Are you sure you won't have additional terms like $\zeta(2k+1)$? (My approach gets such). Would please give an example for $\zeta'''(1/2)$?

Comment by joro

2013-05-10T14:18:37Z

Hm, does mathematica.stackexchange.com answer questions about mathworld?

Comment by joro

2013-05-07T13:52:43Z

GH, can you express $\zeta''(-2)$ in simpler terms? Might have solution with zeta'(3),zeta(3),log,pi,gamma

Comment by joro

2013-05-07T12:07:08Z

Thank you GH ${}{}$

Comment by joro

2013-05-06T08:38:34Z

Thank you for verifying Per.

Comment by joro

2013-05-05T10:55:57Z

@Kofi: I mean experimentally with precision $10^4$ decimal digits the equality is true. Probably will edit the question. If it is true, it will hold with any precision of course, but experimental results don't prove it. btw, I don't mean the low 10^-4.

Comment by joro

2013-04-23T13:40:42Z

Under RH there is better bound for $|x - \psi(x)|$. ams.org/journals/mcom/1976-30-134/… Sharper bounds for the Chebyshev functions $ \theta (x)$ and $ \psi (x)$. II

Comment by joro

2013-04-20T14:40:06Z

Solving nonlinear systems over F_2 certainly will earn you bitcoins. Probably cryptographers know better tricks.

Comment by joro

2013-04-19T12:00:30Z

Looks like there are more points on the curve. $t$ are: [-4, -13/5, -2, -1, -5/8, -1/2, -8/27, 0, 26/41, 1, 7/4, 2, 36/17, 4]

Comment by joro

2013-04-19T11:06:25Z

ok, I don't claim the answer is correct, just wrote what mpmath returns.

Comment by joro

2013-04-19T08:02:47Z

......for L(-1)

Comment by joro

2013-04-19T07:59:43Z

For m=-1 all sumation methods of mpmath give the rational value 0.0815957190957190957190957190.

Comment by joro

2013-04-14T08:27:51Z

btw, do you know lower bound?