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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

2 votes

When does the following congruence identity hold?

Let $K=l[nl^{-1}]_m - m[-nm^{-1}]_l$. Notice that $$ K \equiv n \pmod{lm} $$ and $$ -m(l-1) \leq K \leq l(m-1). $$ Let's first assume $n\ge 0$. To guarantee that $K=n$, one needs to have $n\leq l(m-1 …
Max Alekseyev's user avatar
0 votes
Accepted

Integer solution

I doubt that the lower bound $\frac{p-1}{2} - 2cp^{1/3}$ holds for all $p$. Here is a proof for the weaker bound $\frac{p-1}{2} - cp^{1/2}$. First of all, the inequality $\frac{p-1}{2}-cp^{1/2} \leq …
Max Alekseyev's user avatar
6 votes
Accepted

Squarefree Fibonacci Numbers

I assume the traditional definition with $F_0=0$ and $F_1=1$. Most likely there are infinitely many squarefree Fibonacci numbers. A simple way to construct them is to consider a subsequence $F_p$ fo …
Max Alekseyev's user avatar
0 votes

Diophantine equation of a factorial type

The smallest interesting case of $k=2$ reduces to a family of Pell equations paramaterized by $b$: $$(2c-1)^2 - b^3(2a)^2 = 1.$$ This gives infinitely many solutions. For example, for $b=2$, we have a …
Max Alekseyev's user avatar
5 votes
Accepted

Limit of quotients of polynomials at fixed value

First, cancelling common factors we get $$p(t,n) = \frac{ \sum_{i=0\atop i\equiv 1\pmod{2}}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} }{ \sum_{i=0}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} …
Max Alekseyev's user avatar
3 votes

Are there any solutions to this congruence system

We can even get the first congruence hold modulo $p^3$ as well. For example, $m=3$, $p=5$, $q_1=67$, $q_2=367$, and $q_3=743$.
Max Alekseyev's user avatar
1 vote

Is there a solution to the a+b^m=b+c^n=c+a^l for l,m,n >1 and a, b, c distinct odd primes?

Here is somewhat interesting feature of such primes. Without loss of generality, there are two cases to consider: 1) If $a < b < c$ then $a^{\ell} < c^n < b^m$ and thus $$0 < c^n - a^{\ell} = c-b < …
Max Alekseyev's user avatar
2 votes

Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?

See also http://oeis.org/classic/A076445 and this thread on the search for consecutive odd powerful numbers: http://www.mersenneforum.org/showthread.php?t=3474 Similar technique can be used for search …
Max Alekseyev's user avatar
3 votes
Accepted

Sum of small divisors with powers

The second sum, can be rewritted as $$\sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{\ell=\lceil \ …
Max Alekseyev's user avatar
3 votes

Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions

It can be reduced to Mordell equation: $$Y^2 = X^3 + (4ABC)^2$$ with $Y:=4AB(2By^3-C)$ and $X:=-4ABxy$, which was shown by Mordell to have finitely many integer solutions. ADDED. M. A. Bennett and A. …
Max Alekseyev's user avatar
5 votes

Inverting the totient function

See http://oeis.org/A002202 and further references there. UPDATE: See also my recent paper "Computing the (number or sum of) inverses of Euler's totient and other multiplicative functions", which pre …
Max Alekseyev's user avatar
4 votes
Accepted

On the quadratic reciprocity law?

It is not clear what relation you look for. E.g., $x$ and $y$ may be viewed as reductions of the same residue $z$ modulo $q$ and $p$, respectively, where $z^2\equiv p+q\pmod{pq}$.
Max Alekseyev's user avatar
3 votes
Accepted

Can a Lucas Carmichael number also be a Smith number?

Here are all numbers below $10^9$ that are both Smith and Lucas-Carmichael: 8164079, 8421335, 21408695, 30071327, 47324639, 77350559, 103727519, 121538879, 134151479, 202767551, 239875559, 287432495, …
Max Alekseyev's user avatar
5 votes

Mordell like equation

There are no integral points on this curve as established by SageMath. Here is a code to run
Max Alekseyev's user avatar
4 votes

A Diophantine equation with prime powers

Another counterexample for $a=2$: $p=2288805793$, $q=1321442641$. There are only 2 counterexamples in primes below $10^{5000}$.
Max Alekseyev's user avatar

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