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Results tagged with nt.number-theory
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user 7076
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
2
votes
Asymptotic of $Lcm(\binom{2k}k)_{1\le k\le n}$
Kummer's theorem implies the formula:
$$\mathrm{LCM}(\binom{2}{1},\binom{4}{2},\dots,\binom{2n}{n}) = 2^{\lfloor \log_2(n+1)\rfloor} \cdot \prod_{3\leq p\leq 2n} p^{\lfloor \log_p(2n-1)\rfloor},$$
whe …
- 29.5k
2
votes
Accepted
Characters of integers of $4p^2q^2-(p\pm q)^2$ form
It can be seen that both forms can be factored as differences of squares.
Correspondingly, $n$ can be expressed in these forms iff there exists a divisor $d\mid n$ such that $d$ and $\frac{n}{d}$ have …
- 29.5k
2
votes
Squares of the form $2^j\cdot 3^k+1$
It is also possible to reduce to 9 Mordell equations. Denoting $x := 2^{[i/3]}3^{[k/3]}$, we get 9 equations:
$$d x^3 + 1 = y^2$$
indexed by $d\mid 2^23^2$.
Equivalently,
$$(dy)^2 = (dx)^3 + d^2,$$
wh …
- 29.5k
8
votes
Partitioning $2p$, subject to a divisibility condition.
It is possible. Take $p=61$ and $t_1=2$, $t_{120}=1$. Then $l=113$ and $1^{t_1}\cdot t_1!\cdot 120^{t_{120}}\cdot t_{120}!=240$ divides $(2p-l)!=9!=362880$.
Another example: $p=673$, $t_1=t_{672}=2$. …
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4
votes
s(n) = kn or s(n) = n/k?
This is related to some open questions in number theory. See http://oeis.org/A134639 and references/links there.
Conjecture 2 is incorrect, the smallest counterexample is given by $$n=154345556085770 …
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3
votes
Need to require a assertion for two subsets of natural numbers
Consider the set
$$T = \{ (x,y)\ :\ 1\leq x\leq Q,\ 1\leq y\leq P,\ Px+Qy\leq PQ\}.$$
It is easy to see that
$$S = \sum_{(x,y)\in T} (Px+Qy).$$
Switching to double summation in $S$, we get
$$S = \sum …
- 29.5k
4
votes
Is there an asymptotic formula that describes the correlation of multiplicative inverses in ...
Finding asymptotic for $S(X)$ amounts to estimation of $\sum_{(a,q)=1} aa^*$ for any given $q$.
From the rearrangement inequality, it follows that
$$\frac{1}{2}q^2\varphi(q) - T = \sum_{(a,q)=1} a(q …
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1
vote
Euler's theorem for Tetration, Pentation, etc. (Superexponentiation)
Suppose that for a positive integer $m$, there exists a positive integer $a>1$ such that $(a,m)=1$ and $\mathrm{rad}(\mathrm{ord}_m(a))\not|a$. Then $f_2(m)>0$ does not exists.
(Here $\mathrm{ord}_m( …
- 29.5k
4
votes
Accepted
Restriction of Jacobi's four-squares theorem
By inclusion-exclusion principle, the number of representations of $n$ as the sum of squares of four nonzero integers equals:
$$\sum_{k=0}^4 \binom4k (-1)^k r_{4-k}(n).$$
Formulae for $r_k(n)$ are giv …
- 29.5k
4
votes
how can I minimise (n * y) (mod x) for known x and y, and for a given range of n?
What I'm going to say is somewhat similar to Roland's answer but more precise in the case when the range for $n$ is given in the form of upper bound, i.e., $0 < n < N$.
Notice that $ny\bmod x = ny - …
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3
votes
Solutions of the equation $2^{q-1} \equiv q \pmod {4q^2+1 }$ where $q$ is an odd prime
I've checked all primes $q$ below $10^{11}$ with no new solutions found (even if the primality of $p=4q^2+1$ is relaxed).
Notice that $2^{q-1}\equiv q\pmod{4q^2+1}$ implies $2^{2q}\equiv 4q^2\equiv - …
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2
votes
A sum involving mod(n) arithmetic
I believe the formula $|A|=\frac{\varphi(q)}{\mathrm{lcm}(\varphi(d_1),\varphi(d_2))}$ is not quite correct. In particular, for $q=15$, $d_1 = 3$, $d_2=5$, $a=1$, this formula gives $|A|=2$, while, in …
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8
votes
Elementary divisibility problem
Zsigmondy theorem states that for any $p>1$, $a^p+b^p$ has a primitive prime factor, except when $\{a,b\}=\{1,2\}$. Such prime does not divide $a+b$. Hence, $a^p+b^p$ cannot divide $(a+b)^p$ unless it …
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1
vote
The existence of solutions of a system of indeterminate equations
I expect that in many cases the following construction will do the job. By Fermat polygonal number theorem, we have representation of $\frac{m(m-3)}4$ as the sum of 3 triangular numbers:
$$ \frac{m(m- …
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2
votes
Accepted
The existence of solutions of a system of indeterminate equations
In the comments OP proposed a greedy algorithm to represent a given positive integer $A$ as the sum of triangular numbers whose indices sum to $m$, and applied it to $A = \frac{m(m+1)}4$. I will prove …
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29
votes
Harmonic sums and elementary number theory
Graham (1963) proved that any integer $n>77$ can be represented as
$$ n = a_1 + \dots + a_m,$$
where $a_i$ are distinct positive integers with
$$\frac{1}{a_1} + \dots + \frac{1}{a_m} = 1.$$
Hence, a …
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12
votes
Is $441$ the only square of the form $\frac{397\cdot 10^n-1}{9}$?
If $\frac{397\cdot 10^n - 1}9$ is a square then so is $397\cdot 10^n - 1$. Let $y^2=397\cdot 10^n - 1$. Denoting $x:=10^{\lfloor n/3\rfloor}$, we get that
$$y^2 = 397\cdot 10^r\cdot x^3 - 1$$
or
$$(39 …
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9
votes
Does $a_0=6$, $a_{n+1} = (a_n-1)\cdot a_n\cdot (a_n+1)$ define a square-free sequence?
The question is equivalent to asking whether $b(n):=\frac{a(n)}{a(n-1)}$ are squarefree. The sequence $b(n)$ is listed in OEIS A231831 and satisfies the recurrence $b(n+1) = b(n)^3 + b(n)^2 - 1$ with …
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0
votes
Accepted
Is there a general way to solve this modular equation?
There seems to be no simple formula for the solutions to the given congruence. Still, they can be computed iteratively as follows.
First, we notice that $2^m = (3-1)^m \equiv (-1)^m \sum_{i=0}^{N-1} \ …
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2
votes
Are there infinitely many composite $a$ such that $\sum_{k=1}^{a}(k,a)\equiv1\pmod{a-1}$?
I do not know answer to your question, but here is an approach how one can extend a given integer $m$ to a solution $mp$ (if one exists) with a prime $p\nmid m$.
Using the multiplicativity, we want
$$ …
- 29.5k
2
votes
Accepted
A problem on generators and Hensel lifting
Write $g'=g(1+ap)$ so that
$$g'^y\equiv g^y(1 + pya)\pmod{p^2}.$$
It follows that we can take
$$a = \frac{(\ell^2/g^{y}\bmod p^2)-1}{py},$$
which can be computed in polynomial time.
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9
votes
$23005\cdot (2^n-1)\cdot 2^n +1=p^2$
Multiplying the equation by $2^n$ and denoting $X:=2^n$ and $Y:=p2^{\lfloor n/2\rfloor}$, we obtain two elliptic curves (depending on the parity of $n$):
$$(23005(X-1)X+1)X=Y^2,$$
$$(23005(X-1)X+1)X=2 …
- 29.5k
13
votes
Accepted
Integers $b$ such that $n \nmid (b^n-1)$ for $n>1$
$b=2$ is the only almost 2-like number. Indeed, if $n\mid (b^n-1)$ and $p$ is a prime divisor of $(b^n-1)/n$, then $np\mid (b^{np}-1)$. That is, existence of one $n>1$ dividing $b^n-1$ implies existen …
- 29.5k
6
votes
Mid-Square with all bits set
I'd like to point out why the case of middle bits being all ones is somewhat special and different from other fixed values of them. Also, there is an approach for finding a suitable numbers that may n …
- 29.5k
3
votes
Accepted
Runs of consecutive numbers all of which are Murthy numbers
Let $m+1, \dots, m+n$ be a sequence of $n$ consecutive Murthy numbers such that each $m+i$ shares with its reversal $\overline{m+i}$ a prime factor $p_i\equiv 3\pmod4$ such that 10 is a quadratic nonr …
- 29.5k
3
votes
What is the highest power of 2 that divides $3^y(2z-1)-1$?
I doubt there is a simple expression for $x$ as a function of $y,z$. However, it is possible to characterize all cases when $2^k\mid 3^y (2z-1) - 1$ for a given positive integer $k$.
Say, for $k\geq 3 …
- 29.5k
1
vote
A variant of Landau's function
Let $k = \nu_2(g(n))$. Then $g(n)$ is attained at $n = 2^k + \texttt{1s and powers of odd primes}$, implying that $g(n) = h(n-2^k)2^k$.
Hence, $h(n) = \frac{g(n+2^k)}{2^k}$ for some $k$ satisfying $k= …
- 29.5k
2
votes
Accepted
Solutions of a exponential diophantine equation involving the $\sigma$ function
Note that
$$z = \frac{p^{4k+1}}{\sigma(p^{4k+1})} = \frac{1-p^{-1}}{1-p^{-(4k+2)}} > 1-p^{-1}.$$
Clearly, for a fixed p, the larger is $k$ the closer is $z$ to this lower bound.
Since the lower bound …
- 29.5k
7
votes
Accepted
A modification of the Ljunggren-Nagell equation
The equation
$$
\frac{a^m-1}{a-1}=2b^2
$$
does not have solutions in positive integers for $m>2$ as shown below.
First, notice that $a$ must be odd.
Second, one can see that $m$ must be even. Indee …
- 29.5k
2
votes
Simultaneous lcms
For each prime $p|d$, let $q_p$ be the number of $n_i$ with $p|n_i$.
Then the number of ordered but not necessarily distinct solutions $(m_1,\dots,m_r)$ is given by
$$f(r)=\prod_{p|d} S(r,q_p)\cdot q_ …
- 29.5k
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 …
- 29.5k
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 …
- 29.5k
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 …
- 29.5k
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 …
- 29.5k
5
votes
Accepted
Limit of quotients of polynomials at fixed value
First, cancelling common factors we get
$$p(t,n) = \frac{ \sum_{i=0\atop i\equiv 1\pmod{2}}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} }{ \sum_{i=0}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} …
- 29.5k
3
votes
Are there any solutions to this congruence system
We can even get the first congruence hold modulo $p^3$ as well.
For example, $m=3$, $p=5$, $q_1=67$, $q_2=367$, and $q_3=743$.
- 29.5k
1
vote
Is there a solution to the a+b^m=b+c^n=c+a^l for l,m,n >1 and a, b, c distinct odd primes?
Here is somewhat interesting feature of such primes.
Without loss of generality, there are two cases to consider:
1) If $a < b < c$ then $a^{\ell} < c^n < b^m$ and thus
$$0 < c^n - a^{\ell} = c-b < …
- 29.5k
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 …
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3
votes
Accepted
Sum of small divisors with powers
The second sum, can be rewritted as
$$\sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{\ell=\lceil \ …
- 29.5k
3
votes
Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions
It can be reduced to Mordell equation:
$$Y^2 = X^3 + (4ABC)^2$$
with $Y:=4AB(2By^3-C)$ and $X:=-4ABxy$, which was shown by Mordell to have finitely many integer solutions.
ADDED. M. A. Bennett and A. …
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5
votes
Inverting the totient function
See http://oeis.org/A002202 and further references there.
UPDATE: See also my recent paper "Computing the (number or sum of) inverses of Euler's totient and other multiplicative functions", which pre …
- 29.5k
4
votes
Accepted
On the quadratic reciprocity law?
It is not clear what relation you look for.
E.g., $x$ and $y$ may be viewed as reductions of the same residue $z$ modulo $q$ and $p$, respectively, where $z^2\equiv p+q\pmod{pq}$.
- 29.5k
3
votes
Accepted
Can a Lucas Carmichael number also be a Smith number?
Here are all numbers below $10^9$ that are both Smith and Lucas-Carmichael:
8164079, 8421335, 21408695, 30071327, 47324639, 77350559, 103727519, 121538879, 134151479, 202767551, 239875559, 287432495, …
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5
votes
Mordell like equation
There are no integral points on this curve as established by SageMath. Here is a code to run
- 29.5k
4
votes
A Diophantine equation with prime powers
Another counterexample for $a=2$: $p=2288805793$, $q=1321442641$.
There are only 2 counterexamples in primes below $10^{5000}$.
- 29.5k
6
votes
How many primes have the form $(2^p+1)/3$?
Such primes are called Wagstaff primes. Their (in)finiteness is an open question (similarly to Mersenne primes).
See this OEIS entry for further references: http://oeis.org/A000978
- 29.5k
3
votes
Accepted
Congruence properties of $x_1^6+x_2^6+x_3^6+x_4^6+x_5^6 = z^6$?
To answer this question, one needs to understand why Theorem holds. In fact, it is a corollary of a stronger statement: if
$$x_1^k+x_2^k+x_3^k+\dots+x_k^k = z^k\qquad (1)$$
and $k+1$ is prime, then al …
- 29.5k
1
vote
How to solve the following system of diophantine equations?
This system can be routinely reduced to a single univariate polynomial using resultants -- see https://en.wikipedia.org/wiki/Resultant
- 29.5k
1
vote
Bounds for an Egyptian Fraction Inequality
UPDATE. As pointed out in the comments, an upper bound exists for $x_i$ only if $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$. If we have $B\leq \sum_{j=1}^{i-1} \frac{1}{x_j}<A$, then any sufficiently large $ …
- 29.5k
1
vote
Representing the integers with powers of 2 and 3
For $k\leq m-1$, we have $v_k = k-1$. The expression for $x$ is then reduced to
$$x=\frac{1}{3^m}\left( 2^h - \sum_{k=1}^{m-1} 3^{m-k}2^{k-1} - 2^{v_m}\right)=\frac{1}{3^m}\left( 2^h - 3^m + 2^m + 2^{ …
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