73

This must have been Heinrich Kornblum (1890-1914).
[note by E. Landau in German, my translation]
$^1$ The author, born in Wohlau on August 23, 1890, had before the war
independently made the discovery that Dirichlet's classic proof of the theorem
of prime numbers in an arithmetic progression (along with the later elementary
reasons for the non-vanishing of ...

answered Nov 4 '20 at 15:02

Carlo Beenakker

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55

I'm summarising the discussion in GH from MO's answer as a separate answer for clarity.
The fact that the primes have (natural) density zero can be deduced from a (seemingly) more general statement:
Theorem Let $1 < n_1 < n_2 < \dots$ be a sequence of natural numbers that are pairwise coprime. Then this sequence has zero (natural) density.
Proof ...

51

That was most certainly Heinrich Kornblum. His paper titled 'Über die Primfunktionen in einer arithmetischen Progression' was published in Math. Z. 5 (1919) pp 100–111 (EuDML), see the zbMath revew. In the paper he establishes the analogue of Dirichlet's theorem on primes in arithmetic progressions, in the polynomial setting (with natural density).
His ...

48

There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]:
Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent:
$p$ is a sum of two squares in $\mathbf Z[x]$;
$p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two ...

ag.algebraic-geometry nt.number-theory ac.commutative-algebra polynomials oc.optimization-and-control

answered May 29 '20 at 4:56

R. van Dobben de Bruyn

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46

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-Schneider theorem guarantees that $x = (x+1)^{\frac{x}{x+1}}$ is transcendental as long as $x/(x+1)$ is irrational.
Claim: $x/(x+1)$ is irrational.
Suppose to the ...

40

About two years ago, I spent the better part of a month trying to get to the bottom of this question: What is a Frobenioid? After all, the IUTer's are fond of insisting that it's incumbent upon other mathematicians to invest time in understanding the foundations of IUT simply because Mochizuki's claimed theorems are, if true, so remarkable. I'm not a number ...

36

In general, to compute the class group of a number field $K$ of degree $n$ and discriminant $\Delta$, we need to find some bound $N$ such that the class group of $K$ is generated by primes of norm at most $N$. Unconditionally, we have the Minkowski bound
$$M_K = \sqrt{\lvert \Delta \rvert} \left( \frac{4}{\pi} \right)^{s} \frac{n!}{n^n},$$
where $s$ is the ...

35

This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula
$$ F_n = (A^n-B^n)/(A-B), $$
where $A$ and $B$ are the roots of $T^2-T-1$. So if $\sqrt5$ is in your finite field, then so are $A$ and $B$, and since $AB=-1$, the period ...

answered Aug 24 '20 at 20:15

Joe Silverman

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35

If one starts with the Weierstrass factorisation
$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty (1 + \frac{z}{k})^{-1} e^{z/k}$$
of the Gamma function, applied to $z = -x, -\omega x, -\omega^2 x$ (where $\omega = e^{2\pi i/3}$ is the cube root of unity), and multiplies the three together, one obtains
$$ \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^...

35

There is no such number. Suppose $\alpha>1$ is a real number such that $\lfloor \alpha^n \rfloor$ is a square for all $n\in {\Bbb N}$. Put $\beta=\sqrt{\alpha}$.
Now for each $n$ we have
$$
m^2 + 1 > \alpha^n \ge m^2
$$
for some integer $m$, so that taking square-roots
$$
m + \frac{1}{2m} > \beta^n \ge m.
$$
In other words $\beta^n$ has ...

33

This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.
For each nonzero rational number $a$ (take $a \in \mathbf Z$ if you wish) and each prime $\ell$, let $S_{a,\ell}$ be the set of primes $p$ not dividing the numerator or denominator of $a$ such that $a \bmod p$ has multiplicative order divisible by $\ell$. When $a = \pm 1$, $S_{a,...

32

Consider $Q(x)=x(2x-1)(3x-1)$. This gives an injective map $\mathbb Z\to \mathbb Z$, because $n<m \implies Q(n)<Q(m)$. However, this $Q$ is not injective over $\mathbb Z/p\mathbb Z$ for any $p$ because $Q(x)=0$ has three solutions when $p\geq 5$ and two solutions when $p\in \{2,3\}$.

answered Aug 9 '20 at 17:49

Gjergji Zaimi

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31

Yes it really occurred.
He was introduced in the normal way, something like "Today we have Shinichi Mochizuki..." that mentioned his name, maybe his talk title and the institution he's from (can't remember exactly.)
He annotated those slides with a magnificent glitter-rainbow-colored pen, adding clarificatory details here and there. But yes, (to ...

29

There is a new manuscript on the arXiv by Giulio Bresciani, A higher dimensional Hilbert irreducibility theorem, arXiv:2101.01090, which shows that assuming the weak Bombieri--Lang conjecture, there cannot be a polynomial bijection from $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$.
The author writes that:
Our strategy is essentially the one followed in a &...

28

$$ \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)} = \sum_{1 \leq i,j \leq X} \frac{\mathrm{gcd}(i,j)}{ij} $$
$$ = \sum_{1 \leq i,j \leq X} \frac{\sum_{d|i,j} \phi(d)}{ij}$$
$$ = \sum_{d \leq X} \phi(d) \sum_{1 \leq i,j \leq X: d|i,j} \frac{1}{ij}$$
$$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} \sum_{1 \leq i',j' \leq X/d} \frac{1}{i'j'}$$
$$ = \sum_{d \leq ...

answered Aug 12 '20 at 20:09

Terry Tao

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27

The short answer is no. If anyone were aware of such a record, it would surely have been Carl Siegel, who undertook a careful study of Riemann’s unpublished notes. However, Siegel wrote:
Approaches to a proof of the so-called “Riemann hypothesis” or even to a proof of the existence of infinitely many zeros of the zeta function on the critical line are not ...

26

This identity is true, though somewhat tricky to prove and the infinite series here might only converge conditionally rather than absolutely.
The key lemma is
Lemma 1 (Fourier representation of averages along homogeneous arithmetic progressions). Let $a_n$ be a bounded sequentially summable sequence. Then there exist complex numbers $c_\xi$ for each $\xi \...

24

Now that our paper Geometrization of the local Langlands correspondence with Fargues is finally out (ooufff!!), it may be worth giving an update to Ben-Zvi's answer above. In brief: we give a formulation of Local Langlands over a $p$-adic field $F$ so that it is finally
an actual conjecture, in the sense that it asks for properties of a given construction, ...

24

The letter is here. Thanks to Will Sawin for alerting me to this request.

24

Here is a very much self-contained version of the argument discussed in the posts by GH from MO and Terry Tao.
The claim immediately follows from $H_k:=1+1/2+\ldots+1/k\to \infty$ and the following
Lemma. For a positive integer $k$, and any positive integer $N$, we have $\pi(N)\leqslant N/H_k+k+\text{lcm}(1,\ldots,k)$.
Proof. Choose the maximal integer $M\...

23

The answer is in the affirmative; indeed,
If $S$ is a finite non-empty subset of any abelian group such that every element of $S$ is a sum of two other (possibly, equal to each other) elements, then $S$ has a non-empty, zero-sum subset.
For a complete proof, see this recent preprint by János Nagy, Péter Pach, and myself. The proof is a little too long to ...

23

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 $1$, and therefore, there is no element in $\mathcal{O}_{K}$ with norm $-1$.
However, $\frac{3}{5} + \frac{1}{5} \sqrt{34}$ has norm $-1$, so there is an element of ...

23

I won't address your question about how small the period of $\{F_n \bmod p\}$ can be as $p$ grows, but instead ask if the upper bounds on the period can be achieved infinitely often. For consistency I'll use the notation from Joe Silverman's answer: set $A = (1 + \sqrt{5})/2$ and $B = (1-\sqrt{5})/2$, so $F_n = (A^n - B^n)/(A-B) = (A^n - B^n)/\sqrt{5}$. Note ...

23

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 $y^{2}+z^{2} = (y+iz)(y-iz)$ is reducible, this is analogous to the statement that every integer can be written in the form $x^{2}-y^{2}$ or $x^{2}-y^{2} + 1$ with $x,...

22

The asymptotic frequency of square-free integers is known to be $6/\pi^2$, see [1].
Denote by $P_n$ the uniform distribution on $[1,n]$ and by $E_n$ the corresponding expectation. Then
$$E_n(r)=\sum_{k \le \sqrt{n}} k P_n(r=k) \sim
\sum_{k \le \sqrt{n}} k \cdot \frac{1}{k^2} \cdot\frac{6}{\pi^2} \sim \frac{3}{\pi^2} \log(n) \,,$$
where $A \sim B$ means that ...

21

See OEIS sequences A067760 and A076336. If $n$ is a dual Sierpiński number, there is no $k$ such that $n+2^k$ is prime. There is no prime with Hamming distance $1$ to the Sierpiński number $2131099$, and this may be the least positive integer with this property.

21

That is not what the link says. To quote (emphasis mine):
... in which this conjecture was reduced to a special instance of the Bloch-Kato conjecture for the symmetric square motive of an elliptic curve.
That means something quite different. You could equally say that Wiles "reduced" the proof to the fact that $X(3)$ and $X(5)$ have genus zero,...

21

There are seven prime values (passing a BPSW test) of $c_n$ with $n \le 2 \cdot 10^7$, at indices 457871, 685031, 1029071, 1101431, 9407831, 11769911, and 18437999.
For a writeup about the computations, source code, and the prime numbers themselves, see:
https://github.com/fredrik-johansson/jfunction
The first prime $c_{457871}$ is the following 3689-digit ...

answered Nov 25 '20 at 19:14

Fredrik Johansson

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21

The count is $2$ and $6$ for $p=2$ and $p=3$ respectively,
and otherwise $p-1$ or $p+1$ according as $p$ is $1$ or $-1 \bmod 3$.
More generally, it is well-known that a smooth plane conic
over ${\bf Z} / p {\bf Z}$ has $p+1$ points in the projective plane,
so we need only subtract the number of points at infinity,
which here is the number of square roots of $...

answered Jan 21 at 5:08

Noam D. Elkies

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21

This lim sup indeed goes to $\infty$. We can prove this using exactly the strategy Lucia suggested.
We will count the number of $x,y,z$ in a box with $x^3+y^3+z^3$ not a cubic residue modulo $p$ for a large finite list of primes $p$, all congruent to $1$ mod $3$. We can similarly count the number of $n$ not a cubic residue modulo $p$ for the same finite list ...

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