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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 …
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 …
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 …
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 …
5
votes
Accepted
Limit of quotients of polynomials at fixed value
First, cancelling common factors we get
$$p(t,n) = \frac{ \sum_{i=0\atop i\equiv 1\pmod{2}}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} }{ \sum_{i=0}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} …
3
votes
Are there any solutions to this congruence system
We can even get the first congruence hold modulo $p^3$ as well.
For example, $m=3$, $p=5$, $q_1=67$, $q_2=367$, and $q_3=743$.
1
vote
Is there a solution to the a+b^m=b+c^n=c+a^l for l,m,n >1 and a, b, c distinct odd primes?
Here is somewhat interesting feature of such primes.
Without loss of generality, there are two cases to consider:
1) If $a < b < c$ then $a^{\ell} < c^n < b^m$ and thus
$$0 < c^n - a^{\ell} = c-b < …
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 …
3
votes
Accepted
Sum of small divisors with powers
The second sum, can be rewritted as
$$\sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{\ell=\lceil \ …
3
votes
Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions
It can be reduced to Mordell equation:
$$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. …
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 …
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 $z^2\equiv p+q\pmod{pq}$.
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, …
5
votes
Mordell like equation
There are no integral points on this curve as established by SageMath. Here is a code to run
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 $10^{5000}$.