1:38 AM
@TheSimpliFire How do we get a pole at zero just by making the change of variable from z to w = 1/z?
@Peter You may give this a look too.
I have underlined the line I can't understand^

6 hours later…
8:10 AM
@Archer The only singular point of the function $f(1/w)/w^2$ is $w=0$.
The change of variable $z=1/w$ ensures that all singularities that lie in $C$ lie outside $\tilde C$, and conversely $z=\infty$ is brought into $\tilde C$.

8:36 AM
@TheSimpliFire $\infty$ is brought to 0?

@Archer Yes, you can think of it as $w=1/z=1/\infty=0$

Thanks :D

4 hours later…
12:20 PM
@TheSimpliFire I found one 14-digit and two 15-digit solutions of sigma(n) = sigma(n+1) , the triple equation sigma(n) = sigma(n+1) = sigma(n+2) cannot be solved with them. I have a code in which some parameters can be adjusted which might be able to find even larger solutions. Are you interested ?

@Peter I have taken a look at your question, it looks interesting

My new task is to find the smallest solution beyond the OEIS limit which is 10^13

12:57 PM
0

Find $Res ({f},-1)$ for $f = \dfrac{z^{1/4}}{z+1}$ Attempt: Since the denominator has a simple zero, the residue is $(-1)^{1/4}$. Now, this has four values: $e^{i\pi\left(\frac{1+2n}{4}\right)}$ for $n = 0,1,2,3$. How to decide which value I should choose?

4 hours later…
4:58 PM
@Mathphile Hi, do you at the moment have access to a fast computer ?

5:18 PM
hello @Haran

5:33 PM
Hi

3

A while ago, I asked a similar question For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true? where users mathworker21 and WE Tutorial School proved that for non-square $k$, $n\le k$ is always true when $n!+k$ is sqaure. Recently I got the idea to check if the property is...

@Haran Can you prove that sigma(n) = sigma(n+1) = sigma(n+2) has no solution ?

@Haran can we prove this holds for prime $k$?

@Mathphile Hasn't this been proven ?

@Peter That was for $n!+k$ is a perfect square
I am not sure if the same argument holds for the perfect power version

5:40 PM
@Mathphile This problem got solved?
@Peter Erdos conjectured that an arbitrarily large number of consecutive numbers can have equal sum of divisors.
So I doubt we will be able to solve the problem for three consecutive integers

@Haran No but it was proven that if $n!+k$ is a perfect square for prime $k$, then $n\le k$

Didn't we already show this before?
We only had square values of k left

@Haran yes we did but it was shown for the perfect square version not any perfect power

@Mathphile ?
Oh ok

I am not sure if the prime $k$ argument also works for the perfect power version

5:47 PM
@Mathphile You are probably right, the Jacobi symbol - argument only applies to squares.

Wait, isn't that the thing that's new?

@Haran What ? Erdoes conjectured the existence of arbitary many consecutive integers with the same divisor - sum ? Incredible !

@Haran what thing?

Yeah, surprised me too. Wonder what the heuristic argument to conjecture such a counter-intuitive statement is...

@Mathphile To be honest, I even did not understand the argument with the quadratic residues, but I think, the proof is right.

5:52 PM
13

In Dabrowski's paper, he showed that it would follow from the abc conjecture that the equation $$n!+k=m^2$$ has a finite number of solutions $n, m$ for any given $k$ which was my motivation to find solutions for different values of $k$. Using PARI/GP, I observed that for any $k \gt 1$, if $n!+k$...

@Haran in the above question only the case of square $k$ was left
We were able to prove that the prime $k$ case holds

But the case k = 1 is still open !

@Peter I was checking all $k \gt 1$

which range ?

3

A while ago, I asked a similar question For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true? where users mathworker21 and WE Tutorial School proved that for non-square $k$, $n\le k$ is always true when $n!+k$ is sqaure. Recently I got the idea to check if the property is...

@Haran Bunyakovsky's conjecture might also be counterintuitive, if we consider the generalized version with multiple polynomials.

5:56 PM
does someone has some ideas how to better formulate this question (or it is OK?)

https://math.stackexchange.com/questions/3600525/how-far-apart-can-be-solutions-of-varphim-varphin-while-avoiding-multip

@Haran in this question we can prove that if all composite $k\ne a^b$ if $n!+k=a^b$ holds $n\le k$
But I am not able to prove the prime $k$ case
@Haran Do you understand what I am asking now?
For any prime $k \gt 3$, if $n!+k$ is a perfect power then does there exist any $n\gt k$?

@Ante What is q in this question ?

@Peter natural number, if that´s what you ask
i am not sure that with last condition i also excluded some "legal" solutions

@Mathphile I would formulate it as follows : "Given a posivite integer n and a prime k with 3<k<n. Can n!+k be a perfect power ?"
for(n=1,500,forprime(k=4,n-1,if(ispower(n!+k)>0,print([n,k]))))

IDK how to proceed here

6:04 PM
@Ante What is the problem in plain text ? We search m and n with the same totient value with a given minimum distance w , so far correct ?

Seems like the older square prob, except it's generalized.

@Peter we try to find for every w solutions m and n such that phi(m)=phi(n) and m and n are AT LEAST w apart, but avoiding multiplicativities, so that if phi(m_0)=phi(n_0) then all the solutions with phi(pm_0)=phi(pn_0) are excluded

@Haran yes but like we proved the prime $k$ case in the square prob can we prove it in the generalized prob?

Quadratic reciprocity is out of the options, so I'm a little skeptical.

@Ante I understand that multiplicating with p gives infinite many other solutions. How do we know whether a solution qualifies ?
Suppose, we have m and n with large difference and phi(m) = phi(n) , what is not allowed to happen ?

6:13 PM
hmmm

@Mathphile Already cubic residues are far more complicated, I think.

@Peter you check if $\dfrac{m_w}{n_w} = \dfrac {m_q}{n_q}$ holds, if it holds one of the solutions is excluded

What if we just choose the smallest solution that satisfies a given w ?
or what about demanding that m and n are coprime ?

you can also demand that
you´ll find arbitrarily large distances, i think
@Peter

yes, but I assume that the relative distance is very small.

6:19 PM
@TheSimpliFire i am thinking of asking a question on MO related to your $a^b+b^a$, do you need to be mentioned? you decide

? n=151043;print(eulerphi(n)==eulerphi(n+10^5)," ",gcd(n,n+10^5))
1 1
?

@Ante What in particular do you want to ask about
Are you thinking of a repost

@TheSimpliFire no, about a more general problem of $a^b+b^a=w^r$ and set of solutions

? n=1168061;print(eulerphi(n)==eulerphi(n+10^6)," ",gcd(n,n+10^6))
1 1
?

I don't see how it can be answered given r = 2 has not been resolved @Ante

6:23 PM
@TheSimpliFire that´s MO

You can try
@Ante Here is the code for a general power $r$:
rthfun(a=1000,b=1000,r)={for(i=2,a,for(j=2,b,if(ispower(i^j+j^i,r)==1,print(i," ",j))));}

@TheSimpliFire i will think more about it, since Pillai´s conjecture is not yet solved, i think

? n=12976163;print(eulerphi(n)==eulerphi(n+10^7)," ",gcd(n,n+10^7))1 1
?

So for $\sqrt{a^b+b^a}\in\Bbb Z^+$ it is just rthfun(,,2)

output somehow messed. A distance 10^7 with coprime m and n is apparently still no problem.

6:26 PM
For rthfun(,,3) it seems that $a=b$, in which case we have $2a^a=c^3$
The first few are $a=2,2^2,2^7,2^8,2\cdot7^3$

@Ante Shall we try to solve Pillai or eulerphi now ?

Lehmer´s @Peter

I know a Lehmer conjecture , but not related to what I see here ...

Isn't it that phi(n) = k cannot have exactly one solution ?

6:31 PM
nope, but when phi(n) divides n-1
for non-primes
apparently never

OK, this one
If 3|n , we have an awfully large lower bound
n must be a Carmichael number

@Haran Got a problem for you
Can the integral solution set of $2a^a=c^3$ be characterised?
for(i=2,+oo,if(ispower(2*i^i,3)==1,print(factor(i)[,1]," ",factor(i)[,2])))

@TheSimpliFire a^a must be 4 times a cube right ?

Yeah
$2a^a=c^3\iff a^a=4(c/2)^3$ same thing

OK, so a must be even, and we should first consider the valuation with respect to 2

6:37 PM
$(2^mk)^{2^mk}=4d^3$ ($a:=2k$, $d:=c/2$ and $k>1$ odd)
$2^{2^mkm-2}\cdot k^{2^mk}=d^3$
So $3\mid2^mkm-2$

if v is the valuation of a with respect to 2 , the valuation of a^a is v * a + 1, this must be divisible by 3.

and $3\mid 2^mk$, hang on
Seemingly this gives a contradiction

For example a = 2^7 , 7 * 2^7 + 1 = 897 which is divisible by 3

There's no contradiction there as $k=1$
But I get solutions like $2^4\cdot5^3$

What about the other valuations ? In this case, we have v * a , hence for other primes , 3 must simply divide v * a , hence v or a.

6:43 PM
Apparently $2\cdot2000^{2000}$ is a cube, so $(2^4\cdot5^3)^{2^4\cdot5^3}=4d^3$
Nvm

this correct because of 4 * 2000 + 1 = 8001
Hm, if a is divisble by 3, then v * a + 1 cannot be. This implies that all valuations must be multiples of 3 except that of 2
OK ?

I figured out the fallacy; $k$ is a cube so it is not necessary that $3\mid2^mk$
From the solutions I'm getting, it seems that either $a=2^m$ or $a=2^m\cdot p^n$.

@Ante ? n=110172347;print(eulerphi(n)==eulerphi(n+10^8)," ",gcd(n,n+10^8))
1 1
?

If $a=2^m$ then we have $2^{2^mm-2}=d^3$ so $2^{m-1}m=3r+1$
From which we can deduce that $m\equiv1,2\pmod6$.
So all $2^{6k+1}$ and $2^{6k+2}$ are solutions.

is this compatible with my criterion ? m * 2^m + 1 must be divisible by 3.

6:52 PM
If $a=2^m\cdot p^n$ then $2^{2^mp^nm-2}\cdot p^{2^mp^nn}=d^3$

if m is even, then we get m + 1 , hence m = 2 mod 3 , hence m = 2 mod 6

So $2^{m-1}p^nm=3r+1$ and $n=3s$

if m is odd, then we get -m + 1 , hence m = 1 mod 3 , hence m = 1 mod 6

Yes :)

OK, so all exponents must be divisible by 3 except that of 2
If this exponent is v , then v * a + 1 must be divisible by 3. That's it.
in particular , a cannot be divisible by 3.

6:57 PM
For the $2^m\cdot p^n$ case the equation reduces to $2^{m-1}p^{3s}m=3r+1$

Even simpler : If a is congruent to 1 mod 3, v must be congruent to 2 mod 3 and vice versa.
how do you get this output (10^3) ?

The current solutions are $[2\cdot7^3]$, $[2^2\cdot7^3]$, $[2^4\cdot5^3]$, $[2^5\cdot5^3]$, $[2\cdot13^3]$, $[2^2\cdot13^3]$, $[2\cdot19^3]$, $[2^4\cdot11^3]$, $[2^2\cdot19^3]$, $[2\cdot5^6]$, $[2^5\cdot11^3]$, $[2^7\cdot7^3]$, $[2\cdot31^3]$, $[2^2\cdot5^6]$, $[2^4\cdot17^3]$, $[2^8\cdot7^3]$, $[2\cdot37^3]$, $[2^2\cdot31^3]$ and $[2^{10}\cdot5^3]$.
28 mins ago, by TheSimpliFire
for(i=2,+oo,if(ispower(2*i^i,3)==1,print(factor(i)[,1]," ",factor(i)[,2])))

but why does the program not factorize into the primes ?
The condition for the exponent 2 cannot be further simplified because it depends on the residue of a mod 3.

@Peter ?
@Peter $m\equiv1,2\pmod6$ generates all the solutions for $a=2^m$, it is as simple as it gets.

I though there were (10,3) somewhere

7:06 PM
@Peter They correspond to the powers.
factor(i)[,1] gives the primes and factor(i)[,2] gives the powers corresponding to the primes.

I know, but I wondered about 10^3, maybe I misread it.
you mean the exponents , right ?

Yes, [2, 5]~ [10, 3]~ means 2^10 . 5^3

Ah, now I got it :)

Now it seems in $a=2^m\cdot p^n$, we have $m=1$ only when $p\equiv1\pmod6$.
Oh that's obvious from the equation (as $p^{3s}=3r+1$)

Just use my criterion, we only need a mod 3 and v mod 3
if one is 1 and the other 2, the number passes the "2-test"
so, we can in fact easily classify the positive integers a , such that 2a^a is a cube.

7:12 PM
@Ante I conjecture that there are no solutions for which $a^b+b^a=c^{2d}$ for $d>1$. Trivial constraints $a,b>1$ apply.

How far did you check when a^b + b^a is a perfect power at all ?
I guess that a = 1 or b = 1 is ruled out.
a perfect square seems only possible, if (a,b) = (2,6)
This is an even stronger conjecture implying the given one.

@TheSimpliFire most probably

@Ante By the way, I search a solution of phi(m) = phi(m+10^9) with (m,10^9) = 1

7:29 PM
@Peter you should find something, even with higher differences
@TheSimpliFire you can rule out some cases only by looking at last digits
a^b+b^a has some patterns

@Ante It seems that a^b + b^a can even only be a perfet square for (a,b)=(2,6). I verified it upto a,b = 3000

yes, as expected

If all bases are coprime, Beal's concecture states a = 2 or b = 2.
? n=1037466893;print(eulerphi(n)==eulerphi(n+10^9)," ",gcd(n,n+10^9))
1 1
?
@Ante With Beal I refer not to my stronger conjecture, but to TheSimpliFire's conjecture.
? for(a=2,10^6,if(issquare(2^a+a^2)==1,print(a)))
6
?
Maybe, the case a = 2 can be solved.

it is solved

In fact ?
And what about the case that not all bases are coprime ?

7:39 PM
i think, then it must be gcd(a,b)<=2

which is equivalent to the case that a and b are not coprime.
This does not refer to your comment.
If we also knew that a and b must be coprime (except a=2), then Beal's conejcture would be at least an evidence.
OK, see you tomorrow.

ok