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2:56 AM
Who wants a number theory puzzle: why can $4k^3+1$ never be a perfect square for integral k
And k!=0
 
 
2 hours later…
4:47 AM
@Quintec, I require a small help : math.stackexchange.com/questions/3887105/…
 
 
3 hours later…
7:47 AM
@MartinHopf Is it a Carmichael number ? So far, I found the prime factor $$99907216305808161584859754095439370059601743197802412$$ $$165969811948884041251024167658698583417161881$$ , the cofactor is composite.
In fact , the number is Carmichael ! The prime factors are
gp > print(a)
99907216305808161584859754095439370059601743197802412165969811948884041251024167658698583417161881
gp > print(c)
64040525652023031575895102375176636208204717389791346198386649459234670441906491469225791970400765081
gp > print(b)
73231989552157382441702199751957058253688077763989168117655872158532002237000714893826061644779658041
Form of the number :
gp > (r+1)*(641*r+1)*(733*r+1)==n
%105 = 1
gp > print(r)
99907216305808161584859754095439370059601743197802412165969811948884041251024167658698583417161880
gp >
And thank you for solving the $2685$-case !
@Mathphile Hi !
 
 
1 hour later…
9:31 AM
@TheSimpliFire Hi
 
10:04 AM
@Quintec $4k^3+1=(2n+1)^2\impliedby k^3=n(n+1)\impliedby\sqrt[3]n,\sqrt[3]{n+1}\in\Bbb Z$
 
 
3 hours later…
12:35 PM
@MartinHopf $$643488800783303671523332732373505$$ is a smaller Carmichael number divisible by $2685$ with $8$ factors. After several attemps and approaches, I finally found it. I wonder whether it is the smallest.
 
 
1 hour later…
1:57 PM
@TheSimpliFire Actually isn't that related to the conjecture that $Ax^n-By^m=C$ has only finitely many solutions
Here, $n^2-4k^3=1$
I'm not sure if any particular solved cases are there
 
 
3 hours later…
4:53 PM
@Mourad We don't need to turn to the general result here as the deduction is straightforward
 
@TheSimpliFire But what about $3^3=9 \cdot 3$, but $\sqrt[^3]{9},\sqrt[^3]{3}\notin \mathbb{Z}$?
 
@Mourad That doesn't work as the pair need to be coprime in order for both to be perfect cubes, here (n, n + 1) = 1
 
@Peter Great job! I thought the base form [1,641,733] was a difficult one to find. Not for Peter :) . I wonder if one can construct a Carmichael number difficult to factorise.
 
5:11 PM
@TheSimpliFire Can we use the same logic to show that $k^3=n(2n-1)$ has no other non-trivial solutions?
 
@Peter Smaller than mine anyway! The Erdoes algo cant't find the least Carmichael numbers. With much effort some that are divisible by a given number. I think yor conjecture holds.
 
@Mourad Well we definitely have $\sqrt[3]n,\sqrt[3]{2n-1}\in\Bbb Z$ so we need to check the first few cases as after that the difference between consec will be $\ge n$
 
ah yeah
 
This is an interesting problem: Prove that $(-1)^nL_n(2)\le1$
 
5:40 PM
@TheSimpliFire ding
 
6:13 PM
@MrPie You can simplify this by using $\left(\frac{x-y}{x+y}\right)^{2}=\frac{1}{12}\frac{\left(a-b\right)^{2}}{ab}$
 
@MartinHopf I found the factors by using a nontrivial congruence $x^2\equiv 1\mod n$, one with base $2$ and one with base $3$. I expected that it is probably a Carmichael number , which it actually is. I guess that a Carmichael number never is really hard to factor, if one uses enough bases. But I am not sure.
@MartinHopf Yes, Erdoes is not designed for small Carmichael numbers. I have found out some useful strategies, but still I am not able to prove my conjecture.
 
This gets rid of the $12,12^2,12^3...$ and then bring the $1/6$ to the RHS. You can then define the function $f(x)$ as below to neaten this up
 
@Peter Yes, its more time consuming to construct them then to factorise.
 
Then you can write it more compactly as: $$f(x+1)=\left(x+1\right)^{2}f\left(1\right)+f\left(1\right)^{x+1}+\left(x+1\right)\sum_{n=2}^{x}\frac{1}{n}\binom{x+n}{2n-1}f\left(1\right)^{n}$$
 
At least I have written routines to find the Carmichael numbers with one or two additional prime factors, if some suitable positive integer is given. If you want, you can ask "Is ... the smallest Carmichael number divisible by $2685$ ? Or shall I ask it ?
 
6:22 PM
@Peter If $C$ is too large for factorization, I suggested a probable Carmichael test years ago: math.stackexchange.com/questions/1726016/…
 
Most composite numbers anyway are revealed composite already by a weak $2$-PRP-test. Carmichael numbers are very rare although infinite many exist.
 
Yes, a simple 2-PRP test is the best method to exclude Carmichael numbers.
 
If my routine actually works correctly, and I have not overlooked something, I should have been able to show that there is no Carmichael number with exactly $3$ prime factors, one of which is $1223$
@MartinHopf As said, I invite you to ask about my $2685$. Or do you think, I should do that ?
You can mention that $C$ must have the form $2685\cdot m$ with $m\equiv 273\mod 356$ (This condition is apparently still not enough for a brute force search however)
 
6:39 PM
but $2685$ is solved anyway!?
 
Maybe , someone can find a smaller one.
Or has even an efficient algorithm to solve my conjecture - who knows.
 
so the task would be: find the least lcm(2685,2685*m) ?
 
If your computer is free, you can search for a factor of $2^{2^{2^2}}+3^{3^{3^3}}$ , an old and very tough problem.
The task is to find the smallest m with this form doing the job.
 
Yes, but my computer has to process the priorities. :(
 
Can we apply Erdoes with a smaller start number perhaps ?
 
6:46 PM
Erdoes is greedy concerning the factorbase
 
What is the largest prime factor of a Carmichael number which can be found with this method ?
 
for the prime vector running all primes from 2..10^11 that should last at least an hour.
 
hello
 
Hello
 
Another project : Is $$1909001$$ the largest Carmichael number with at most $3$ distinct digitis in the decimal representation ?
 
6:55 PM
@Quintec A perfect square is either $\equiv 0,1\pmod 4$. So, if $4k^3+1\equiv 1\pmod 4$ is a perfect square then this means $k$ has to be even $2n$. But since $m+1\mid m^3+1$ for any $m$ then this means $(2n)^3\equiv -1\pmod{2n+1}$ And so there exists an integer $p$ such that $(2n)^3=(2n+1)p-1$. Therefore $4k^3=4p(2n+1)-3$ which is not congruent to $0,1\pmod 4$. So it cannot be a perfect square. wait nevermind i am dumb
 
Only one hour for every product with prime factors below $10^{11}$ ? That would be fantastic !
 
I meant $4k^3\color{red}{+1}=4p(2n+1)-3$ of course. But hm, I am stuck.
 
when the prime vector is set with p-1 divides 'a' then the search begins. 'a' is the product of the input factorset.
 
@Quintec actually we can prove that every square number is $\equiv 0,\pm 1\pmod 5$ and every cube number is $0, \pm 1\pmod 7$
 
So we can find, lets say, all Carmichael numbers with no prime factor larger than $10^6$ in a short time ?
 
7:04 PM
with a given factorset, you can only get a small subset of all possible C-numbers.
 
@Mourad ah thanks
didn't know i was only considering a special case lmao
incidentally, i have a geometry problem.
 
@MartinHopf Is there a tight estimate for the number of Carmichael numbers in a range $[a,b]$ ?
 
If $x=\frac 1{3-\sqrt[3]2}$ and $y=\frac 1{\sqrt[3]2-1}$, find $z$.
@@Peter if an estimate did exist, i don't think it would be tight
 
 
3 hours later…
10:20 PM
oop that's tiny for some reason. hope it's legible
 
10:49 PM
@Mourad oh btw in ur post of $f(x)$, I think the denominator should be $(ab)^{2x}$ instead of $(ab)^x$
 

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