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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
48
votes
Accepted
How to prove that the solution to $x^{x+1}=(x+1)^{x}$ is transcendental?
The number $x$ is transcendental, and your Gelfond-Schneider argument almost works.
Suppose to the contrary that $x$ is algebraic. Then $x+1$ and $x/(x+1)$ are also algebraic, and so the Gelfond-Schne …
40
votes
Accepted
When $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime na...
Yes, there is another solution. The next one I found is a bit big, namely
$$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$
This solution also satisfies the property that
$$ …
38
votes
Accepted
When does a Catalan number equal a Fibonacci number?
A result of the type you seek follows easily from Carmichael's theorem, that if $m > 12$, then there is a prime $p$ that divides $F_{m}$, but does not divide $F_{k}$ for $k < m$.
Suppose $C_{n} = \b …
34
votes
Accepted
Question on a generalisation of a theorem by Euler
I suspect that $k = 4$ is good, but am not sure how to prove it. However, every positive integer $k \geq 5$ is good. This follows from the fact (see the proof of Theorem 1 from this preprint) that for …
32
votes
Only odd primes?
If $k$ is odd and not a perfect square, then the sets are disjoint. In particular, if $\alpha = \frac{k - \sqrt{k}}{\frac{k-1}{2}}$ and $\beta = \frac{k + \sqrt{k}}{\frac{k-1}{2}}$, then $\alpha$ and …
27
votes
Accepted
Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?
Yes, it is true that $\{ x^{4} + y^{2} + z^{2} : x, y, z \in \mathbb{Z}[i] \} = \{ a + 2bi : a, b \in \mathbb{Z} \}$. Indeed, one can even take $x$ to be either $0$ or $1$ in all cases. Because $y^{2} …
26
votes
Accepted
Does the set of square numbers adhere to Benford's law in every base?
No. Benford's law works well for sequences that grow exponentially, and the squares grow too slowly.
In particular, fix a base $b \geq 3$, consider the case of $d = 1$, and choose $n = 2 \cdot b^{2k}$ …
24
votes
Accepted
Norms in quadratic fields
This is false. The smallest counterexample is $d = 34$. Let $K = \mathbb{Q}(\sqrt{34})$. The fundamental unit in $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{34}]$ is $35 + 6 \sqrt{34}$, which has norm $1$, an …
23
votes
Accepted
Prime factorization "demoted" leads to function whose fixed points are primes
A way to get a non-trivial solution to $f(n) = p$ is that every odd number $\geq 7$ can be written as a sum of three primes (by Helfgott's recent work), so if $p \geq 7$ is prime, we can write $p = q …
23
votes
Accepted
Algorithmic (un-)solvability of diophantine equations of given degree with given number of v...
I'm going to take a stab at this. First (as mentioned in Andres Caicedo's answer to this question), Siegel proved in 1972 that there is an algorithm to determine whether a quadratic equation in any nu …
22
votes
Accepted
On Fibonacci numbers that are also highly composite
The largest highly composite Fibonacci number is $F_{3} = 2$.
If $p$ is a prime number, then either $p \mid F_{p-1}$ (if $p \equiv \pm 1 \pmod{5}$), $p \mid F_{p}$ (if $p = 5$), or $p \mid F_{p+1}$ (i …
21
votes
Is there a real nonintegral number $x >1$ such that $\lfloor x^n \rfloor$ is a square intege...
At the request of the OP, I am turning my comment into an answer. It is possible to have $\lfloor x^{n} \rfloor$ close to a square for all positive integers $n$. For example, if $x = \frac{7 + 3 \sqrt …
19
votes
Accepted
Parity of the Prime Counting Function
The limits you conjecture are natural, and they are currently open. I believe the best known result is by Ping Ngai Chung and Shiyu Li, who proved that
$$
\liminf_{n \to \infty} \frac{|E_{n}|}{n}
$$ …
18
votes
Accepted
Would such polynomial identity exist? (related to sum of four squares)
To see that $f_{1}$, $f_{2}$, $\ldots$, $f_{k}$ must be scalar multiples of $f_{k+1}$, plug in the root of $f_{k+1}$ into both sides of the equation.
17
votes
Accepted
A divisor sum congruence for 8n+6
The congruence you state is true for all $m \equiv 6 \pmod{8}$. The proof I give below relies on the theory of modular forms. First, observe that
$$
\sum_{k=1}^{m-1} d(k) d(m-k) = 2 \sum_{k=1}^{\frac{ …
17
votes
Accepted
Are there integer solutions to $3y^2 = 4x^3-1$ other than $(1,1)$ and $(1,-1)$?
The projective form of your curve is $3y^{2} z = 4x^{3} - z^{3}$. This has three obvious points: $(1 : 1 : 1)$, $(1 : -1 : 1)$, and $(0 : 1 : 0)$.
Your curve is isomorphic over $\mathbb{Q}$ to the Fe …
17
votes
Accepted
Existence of rational points on a generalized Fermat quartic
This question is amenable to the use of the Mordell-Weil sieve. (For a good introduction to this technique, see the paper here.) In this situation, there is a simple version of it (using a single prim …
16
votes
Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$
The equation you specify defines a surface $X$ in $\mathbb{P}^{3}$, and this surface is a K3 surface. It is conjectured that if $X$ is a K3 surface, there is a field extension $K/\mathbb{Q}$ over whic …
16
votes
Accepted
Looking for a "clever" argument for a $q$-series identity
Here's a proof that indicates a systematic method for proving such identities. Let $\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$, with $q = e^{2 \pi i z}$. The identity you state in the equation …
16
votes
Accepted
Prove that $1$ is the sum of three tetrahedral numbers infinitely many different ways
There are infinitely many solutions. I'll show below that there are infinitely many positive integers $k$ for which $93k^{2} - 288k + 276 = z^{2}$ for some positive integer $z$. From such a $z$, we ge …
16
votes
Accepted
$x^3+x^2y^2+y^3=7$, and solvable families of Diophantine equations
(a) No. There are no integer solutions. The curve $C$ you give has genus $3$ and it has an obvious automorphism $\phi(x,y) = (y,x)$. The quotient curve is an elliptic curve. In particular, if you let …
16
votes
Accepted
On $\eta(6z)\eta(18z)$ and the splitting / modularity of $x^3 - 2$
Here's an answer to 2. You can tell me if it's the "best method".
Euler's Pentagonal number theorem gives that
$$ \eta(24z) = \sum_{n \in \mathbb{Z}} (-1)^{n} q^{(6n+1)^{2}}. $$
This yields the formu …
15
votes
Accepted
Splitting of polynomials over rational function fields
Your question 1 is open and is equivalent to the problem of determining the rank of an elliptic curve over a number field. (In most "practical cases" it should be solvable.) In particular, we don't ha …
15
votes
Accepted
A Galois extension over $\mathbb{Q}$ with Galois group $A_4$ and with cyclic decomposition g...
Yes. Note that Daniel Loughran's comment to David Speyer's answer to this question states that for any solvable group $G$, there is a Galois extension $L/\mathbb{Q}$ with all decomposition groups cycl …
15
votes
Accepted
Elliptic curves with the same mod $p$ representation
The largest known is for $p = 17$. This can be found by searching Cremona's tables. In particular, in this paper, Tom Fisher mentions the examples of
curves 3675b1 and 47775b1, which are
$E_{1} : y^{ …
15
votes
Accepted
Is $\text{PSL}_2(\mathbb{F}_{p^m})$ known to be a Galois group over $\mathbb{Q}$ for $m>1$?
It is known that ${\rm PSL}_{2}(\mathbb{F})$ can be realized for some $\mathbb{F}$ but definitely not all at present. According to David Zywina's note here, ${\rm PSL}_{2}(\mathbb{F}_{27})$ is the sma …
15
votes
Existence of rational points on some genus 3 curves
There are no rational solutions to curve (b). This curve has the automorphism $(x,y) \mapsto (x,-y)$ and the quotient is the genus one curve
$$ -yz + w^{2} = 0 \quad x^{2} + y^{2} + xz - z^{2} + yw = …
15
votes
Accepted
More elliptic curves for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$?
Given a rational number $m$, let $C_{m}$ be the intersection of two quadrics given by (3) and (4) in the original question. The way to search for points is to test the curve $C_{m}$ to see if it has l …
15
votes
Accepted
Is $\eta(\tau)^2$ a modular form of weight 1 on $\Gamma(12)$?
Yes, it is. There are standard criteria (stated for example in Ken Ono's book "Web of Modularity" - Theorems 1.64 and 1.65) that indicate when an eta quotient is modular on $\Gamma_{0}(N)$, and these …
14
votes
Accepted
Galois representations for the curve $y^{2} = x^{3} - x$
I don't know that this is written down anywhere, but it's possible. It is known in general that $GL(T_{\ell}(E))$ is contained in the normalizer of $R_{\ell}^{\times}$, where $R_{\ell} = \mathbb{Z}[i] …
14
votes
Number of real roots of irreducible polynomials that are solvable by radicals
Klueners and Malle have a database of number fields of degree $\leq 19$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $n$ wi …
14
votes
Accepted
Is there a known primitive recursive upper bound on the nth "Zhang prime"
One does not have to dig too deep into the arguments to answer this. Maynard's formulation (see his preprint here) shows that if $x$ is sufficiently large, there is a Zhang prime (or even a prime $p$ …
13
votes
Are there any simple, interesting consequences to motivate the local Langlands correspondence?
In the case of $GL_{N}$, the $L$-packets are a non-issue, and the surjective map in the local Langlands correspondence becomes a bijection. At that point, we can think of allowing the information to f …
13
votes
Accepted
A criterion for the equation $ax^n+bx+c=0$ not solvable by radicals via $a,b,c$ and $n$
No, the conjecture is false at least for $n = 5$. The irreducible quintic trinomial $f(x) = 85x^{5} - 4x + 1$ satisfies $\gcd(b,5ac) = \gcd(-4,5 \cdot 85 \cdot 1) = 1$. However, the Galois group of $f …
12
votes
Accepted
Extension of $\mathbb Q$ which splits only at primes in $S$
For many choices of $R$ and $S$ the answer is obviously no. For example, if $R$ is empty, then the answer is no, because there are no unramified extensions of $\mathbb{Q}$.
For a more interesting exa …
12
votes
Accepted
Is every $GL_2(\mathbb{Z}/n\mathbb{Z})$-extension contained in some elliptic curve's torsion...
I have two things to add to this discussion.
$\bullet$ For $n = 2$ and $n = 3$, every Galois extension of $\mathbb{Q}$ with Galois group ${\rm GL}_{2}(\mathbb{Z}/n\mathbb{Z})$ does arise from an elli …
12
votes
Accepted
Etale cohomology approach on $\tau(n)$
One of the goals of the development etale cohomology was to generate a cohomology theory that could successfully count points on varieties over finite fields, with one of the main goals of proving the …
11
votes
Accepted
Which criteria for "uniformly splitting" polynomials?
Yes, there are uniformly splitting polynomials of all degrees which are not of the form $\Phi_{n}(x^{k})$. (For example, $f(x) = x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$.)
The Chebotarev density theorem imp …
11
votes
Accepted
What are the strongest conjectured uniform versions of Serre's Open Image Theorem?
I've seen Conjecture 1 and Conjecture 1' stated in the literature in many places. I don't believe I have seen Conjecture 1'' so stated.
I'd also like to point out that (EDIT: a weaker version of) Con …
11
votes
Is $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau\rightarrow\...
The function $x(\tau) = j(\tau)^{1/3}$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$ and $\tau \mapsto \fr …
11
votes
Fermat's cubic equation in quadratic extension of $\mathbb{Q}$
I don't want to toot my own horn, but I coauthored a paper on this topic with Marvin Jones. One direction of our result is conditional on the Birch and Swinnerton-Dyer conjecture (see the first remark …
11
votes
Accepted
Reference request for recurrence relation of division polynomials
These recurrences are stated explicitly in Weber's Lehrbuch der Algebra (published in 1908). See volume 3, section 58, page 200. Weber doesn't give a citation to them, so it's hard to know if they wer …
11
votes
Accepted
Primes dividing $2^a+2^b-1$
This is a heuristic which suggests that the problem is probably quite hard. We have that $p | 2^{a} + 2^{b} - 1$ if and only if there is some integer $k$, $1 \leq k \leq p-1$ with $k \ne \frac{p+1}{2} …
10
votes
Accepted
Prime numbers in a sparse set
Yes, there is a $c > 1$ for which infinitely many numbers of the form $\lfloor k^{c} \rfloor$ are prime. The first result of this type was proven in Ilya Piatetski-Shapiro's Ph.D. thesis (written in 1 …
10
votes
Accepted
Confusion regarding the definition of semistable reduction of an elliptic curve at a prime $p$
The criteria given by Silverman and Husemoller are equivalent, and so determining the "correct definition" is a bit tricky. In particular, on page 361 of Arithmetic of Elliptic Curves, Silverman (the …
10
votes
Accepted
Extension of a formula for the quadratic Gauss sums
No, the relation is not as simple in this case. For example, if $k = 3$ and $p = 7$, the three different cubic Gauss sums are roots of $y^{3} - 21y - 7$, and the three roots of this polynomial do not …
10
votes
Accepted
Space of functions f such that the number of primes in $ [x, x+f(x)] $ remains bounded
There is no such function $f(x)$. First, the prime number theorem implies that the average gap between two primes $p_{n}$ and $p_{n+1}$ is about $\log p_{n}$ and for this reason, if $f(x)$ is smaller …
10
votes
Accepted
Integral points near elliptic curves
You can take $\theta = 0$, even $\theta = -1/6$ works.
Fix an integer $A \ne 0$ and an integer $B$. If $r$ is an integer, the elliptic curve $E : y^{2} = x^{3} + Ax + r^{2} A^{2}$ has the obvious poin …
10
votes
Accepted
Cusp forms with integer Fourier-coefficients
No. Take $k = 24$ and $N = 1$. Then $\Delta^{2} = q^{2} - 48q^{3} + 1080q^{4} + \cdots \in S_{K}(\Gamma_{1}(N),\mathbb{Z})$. However, if we write $\Delta^{2} = c_{1} f_{1} + c_{2} f_{2}$, where $f_{1} …
10
votes
Number of points of elliptic curve over $\mathbb{F}_p$ with CM by $\sqrt{-2}$ when $p\equiv ...
Your conjecture is true and follows from a theorem in Cox's book ''Primes of the form $x^{2} + ny^{2}$.'' (Theorem 14.16 on page 317, although this is from the first edition.) In particular, if $\math …