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1:59 AM
@Peter, i solved this, i think, i checked and it seems there are no errors:

https://math.stackexchange.com/questions/1309427/can-a-number-be-equal-to-the-sum-of-the-squares-of-its-prime-divisors
Actually, it seems that there CANNOT be any errors, since the proof is elementary.
 
 
5 hours later…
7:16 AM
@MrPie thank you :)
 
 
2 hours later…
9:04 AM
But if I write the proof then I should give myself bounty of mine, is that an issue?
 
@Grešnik Then, post an answer. I noticed your comment, but why is there no solution if the given inequality holds ?
 
You can't yourself a bounty
 
@Peter No, this proof is new and is not based on that idea.
So the rules are that someone cannot give himself a bounty?
 
But this is a situation, if my proof is right, where it would only be fair if I give myself the bounty?
 
9:08 AM
The bounty period lasts 7 days. Bounties must have a minimum duration of at least 1 day. After the bounty ends, there is a grace period of 24 hours to manually award the bounty. Simply click the bounty award icon next to each answer to permanently award your bounty to the answerer. (You cannot award a bounty to your own answer.)
 
Okay, so this situation actually isn´t allowed because some would give themselves bounties that are not actually earned, generally.
Or why if not because of that?
@Peter I mean I did not prove it with the help of that inequality, if everything is right now I obtained a contradiction that I do not remember i obtained ever before, it is not usual.
 
@Grešnik Unlucky, that you offered the bounty and then found a solution. Seems that you have to wait until the period ends.
 
I am trying now to find one more way to prove it.
 
9:23 AM
@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
 
@Peter If you managed n>392 then did you solve the case with n=66?
 
I have not factored it for n = 66 yet, neither for n = 62
 
@Peter i can't see a reason why the expression cannot be prime for odd n either
 
@Peter How many cases >392 did you check?
 
Or why would even´s give more primes than odd´s.
 
9:28 AM
I only wonder because the odd cases mostly have a small factor. I passed $n=8\ 000$
I determined the candidates upto $25\ 000$ (no factor below $10^7$) , I did not count the number of odd and even candidates, but looking at the list, I noticed far more even candidates.
 
9:48 AM
Better that I didn´t write a proof, because it´s not a proof at all. I did check it once more.
It holds only on two assumptions on the powers of the primes in the factorization, but one of them should never be true.
 
10:05 AM
Different approach yields this:

$(c-1)(bp_k-ap_1)=(p_1-p_k)(a+b)$ and I do not know how to proceed.
Does someone can conclude something about this equation?
By starting with assumption that $p_1 \neq p_k$
 
10:21 AM
@Grešnik can cancel ap[1] and bp[k]
 
Then we obtain $c(bp_k-ap_1)=bp_1-ap_k$
$c$ must be natural number.
 
Is p[1]=2?
 
No, but any prime.
 
then (a + bc) | (ac + b)
 
I just need to conclude that if $c$ is natural and $p_1 \neq p_k$ then it is not possible that $c=\frac {bp_1-ap_k}{bp_k-ap_1}$
haha :(
 
10:32 AM
Hang on
We have p[k] = (ac + b) p[1] / (a + bc)
This forces ac + b = a + bc
since p[k] =/= p[1]
so a(c - 1) = b(c - 1)
So either c = 1 or a = b
 
hahahah :D
 
If c = 1 then b(p[k] - p[1]) = a(p[1] - p[k]) so a = -b
If a = b then c(p[k] - p[1]) = p[1] - p[k] so c = -1 which is impossible
Done @Grešnik
(c = 1 is the only possible solution)
And if a,b > 0 then there are no solutions at all :)
Actually, it's impossible anyway since p[k] cannot be divisible by p[1]
 
But why "this forces ac+b=a+bc"?
 
10:54 AM
We are CLOSE, but still we do not have a proof.
 
11:24 AM
@Grešnik because if (ac + b) / (a + bc) > 1 then p[k] will have more than two factors so will not be prime.
It can't happen anyway. Even if (ac + b) / (a + bc) = 1, we have p[k] = p[1] which is a contradiction.
 
No, I actually need to obtain the contradiction and conclude that it must be p_1=p_k, i have done that step few minute ago on the paper, i think, now it seems that the problem is completely settled.
 
11:37 AM
Bounty awarded!!!
 
12:20 PM
@TheSimpliFire Check it:

https://math.stackexchange.com/questions/1309427/can-a-number-be-equal-to-the-sum-of-the-squares-of-its-prime-divisors/3289859#3289859
 
12:40 PM
@Grešnik I did not spot an error
 
@TheSimpliFire Do you want that I make an edit and mention your help?
@Peter There is none.
 
@Mathphile $$401^{402}+402^{403}$$ is the smallest odd case without a known factor. Can you try to find a factor ?
 
@Grešnik It's fine, you don't need to
 
okay i'll run it on pari
 
@TheSimpliFire Did you check the proof, do you have any objections?
 
12:45 PM
pari does not display intermediate results as far as I know. yafu would be better.
 
@Peter i have downloaded yafu
 
Great !
Just enter "factor(401^402+402^403)"
 
how do we use it?
@Peter okay
 
@Grešnik Valid proof :)
 
The smallest factor seems to have more than 20 digits
@Grešnik Good job!
 
12:54 PM
The methodology is extremely elementary but clever
 
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
 
@Mathphile I guess, we still are at C1050 ?
 
ecm: 118/214 curves on C1050, B1=50K, B2=gmp-ecm default, ETA: 4.2 min
 
OK, still no factor found as with my yafu (which runs of course much slower)
 
what does B1 mean?
 
1:00 PM
Did you ever hear of "elliptic curves" ?
 
@MrPie @TheSimpliFire people have decided im a boar :P
in The Sphinx's Lair, 9 mins ago, by Rubio
Well. I’ve always found certain people rather boring so maybe they’re the boar.
 
heard but don't know much
 
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
 
@OmegaKrypton and so is everyone who clicks into the link :P
 
@TheSimpliFire then that proves i am boring :P
 
1:03 PM
Here, we need that some "order" is smooth (has no factor more than 50k apart from a single one that can be larger, but with a limit)
What is important : B1=50k is best for around 25 digit prime factors
 
okay
 
Since my program already runs much longer, I relatively soon finish the 50k-phase, around 9 minutes away.
next is 250k which is optimal for about 30 digits
In the file "factor" you can see the found factors.
 
We could make Peter´s life a little easier if we find efficient factoring algorithm.
 
@Grešnik i think ecm is very efficient
 
@Grešnik Or even better a fast "pretest" for huge numbers whether they have a good chance to be prime.
@Mathphile The number is too large for SIQS , hence ECM is the only promising method.
 
1:13 PM
what is siqs?
 
self-initializing quadratic sieve. Can factor numbers upto about 100 digits in a reasonable time
 
is there any reason why $n!+1$ cannot be a perfect square for $n \gt 7$?
 
Is that one of the open problems?
 
oh yes
 
I know. :)
 
1:17 PM
searching it up i found out that this is brocard's problem
 
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for which n ! + 1 = m 2 , {\displaystyle n!+1=m^{2},} where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan. == Brown numbers == Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers: (4,5), (5,11...
Yes.
 
Sure that it is still open ?
I think I recently heard it has been solved, but I am not sure.
 
I do not know surely, I just know that I tried to solve it also.
 
finally , my yafu switched to 250k ...
 
@Peter what is the pari function to check if a number is a square?
 
1:21 PM
issquare()
 
@Mathphile The factors of n! and (m-1)(m+1) after n>7 tend "not to coincide".
 
@Grešnik I think I proved that m cannot be in the sequences A224473 or A224474 from the OEIS.
 
I wonder which is the smallest natural number not being in the factordb-database.
@Mathphile Still C1050, I guess ...
 
@Peter yes
 
Did anyone check for some reasonable range of n´s how close are n!+1 to the squares?
 
1:31 PM
not a bad idea. Identifying "near-misses" could shed some light on this problem.
 
You'd be better off checking how often m^2 - 1 is close to n!
 
can $n^n+1$ be a perfect square?
 
No because of the now proven Catalan-conjecture.
It cannot even be a perfect power
 
i see
 
@Mathphile Does the program for the orders still run ? I mean the "power-summer-problem"
 
1:36 PM
what about $n^n+2$?
@Peter no i stopped it
 
@Mathphile I recall that you mentioned $n^n\pm k$ a few months ago
or was it $p^p\pm k$
 
@TheSimpliFire yes
although that was for primes
 
Upto $n=10^4$ , $n^n+2$ is no perfect power. Now I run the range upto $10^5$
 
no perfect power for $10^5$ either @Peter
 
Related to Pillai's conjecture
 
1:40 PM
Let's just consider squares for now
We want to solve $n^n+1=(m-1)(m+1)$
n and m have the same parity
 
Why not n!+1=(m-1)(m+1), i would rather try with an open problem.
 
$n^n+1$ has been proven by Catalan's conjecture
 
@Grešnik Are you planning to tackle Brocard's problem
 
Yes.
 
let's try $n^n+2$
 
1:41 PM
@OmegaKrypton is this a sighting of you?
 
even n gives residue 2 mod 4, cannot be a perfect square.
 
@Grešnik Seeing as I have failed previously, I might as well share what I've done
 
yes @MrPie
 
n = 4k+1 is also impossible (residue mod 4 woule be 3)
 
:D
 
1:43 PM
I speak of $n^n+2$
 
@OmegaKrypton
 
@Peter yes
 
$n^n+2$ for natural $n$ or prime $n$ (like last time?)
 
@MrPie general $n$
 
1:46 PM
Do we have the formula for the exact number of factors of n! as a function of n (in any form)?
Well, not the exact exactly, at least some good bounds.
The m^2-1 also should have some of its own laws.
 
This was my partial proof
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
 
@Peter It also seems that $n^n+3$ is only a perfect square for $n=1$
 
@TheSimpliFire pretty :P
 
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
 
Whether we can proof it ?
@Mathphile
 
1:53 PM
@Grešnik See above ^^
 
@Mathphile
Professor: "So, show me what you got."
Mathphile: "Kapoor's Conjecture 1, 2, 3, 4, 5 ...... 100, 101, 102, 103 .... 2930, 2931. All done sir!"
Mathphile: "oh, and just came up with one right now! :D"
 
@MrPie hahahahahah
 
@Mathphile I have come up with the term "Konjecture". It is a verb, and to "konjecture" means to makes 10 conjectures in one day
 
@MrPie lol i could do that although the conjectures would all be pretty bad xD
 
@TheSimpliFire I´m tired.
What do we know about sums of four squares of naturals?
I mean sleepy.
 
2:01 PM
heh
read it whenever you want
@Mathphile Hmm, we can write $n^n+3=m^2$ as $n^n-1=(m-2)(m+2)$
LHS is also factorable
If $n$ is even then $k(n-1)(n+1)=(m-2)(m+2)$
 
@TheSimpliFire you mean $(m-\sqrt{2}))(m+\sqrt2)$
 
No
(m - 2)(m + 2) = m^2 - 4
Bring the 4 across and you get 4 - 1 = 3
 
@TheSimpliFire sorry don't know what i was thinking
 
Who wants to listen to a song I chose?
 
@Mathphile Regarding the $n^n+3$ thing, I think I proved that $n$ must be odd
$k^2\equiv 0,1\pmod 4\,\forall k \in\mathbb{Z}$

$\therefore n^n+3\equiv 0,1\pmod 4\Leftrightarrow n^n\equiv -3,-2\equiv 1,2\pmod 4$.

If $n^n\equiv 1\pmod 4$ then $n\equiv 1\pmod 4$. So $n$ is odd.

If $n^n\equiv 2\pmod 4$ then if $n=2a$ for some $a\in\mathbb{Z}$, we have $2^{n-1}a^{n}\equiv 1\pmod 2$. But this must mean that $n=1$, which isn't true because $n$ is even.

Therefore, $n$ must be odd.
 
2:08 PM
Great @MrPie
 
@TheSimpliFire thanks :P
 
@MrPie nice
 
Which then means that $$(m-2)(m+2)=(n-1)(n^{n-1}+n^{n-2}+\cdots+1)$$
 
So really, $n^n\equiv 1\pmod 4$ or $n\equiv 1\pmod 4$ (otherwise it would be even).
 
Forcing $m$ to be even
 
2:10 PM
@TheSimpliFire does $m$ represent $k$ here?
 
@TheSimpliFire smart what you did :P
 
Then RHS is divisible by 4
 
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
 
i gtg guys
be back in an hour
 
2:12 PM
@Mathphile cya
 
bye @Mathphile
Now $n^r$ is odd for all positive integers $r$
 
$n$ has to be odd
 
But the cardinality of $\{1,\cdots,n-1\}$ must be even
Therefore, $4\mid (n-1)$
 
Exactly. Similarly, $n^n\equiv -3\pmod 4 \implies n^n\equiv 1^n \pmod 4$ so you're right
How to prove now that $n:=1$ :/
 
We could consider last digits of n^n
It is easy when $n$ ends in $1,5$
If $n\equiv5\pmod{10}$, then $n^n+3\equiv8\pmod{10}$ which can never be a square
 
2:18 PM
Aha!
@TheSimpliFire genius! :D
 
thanks, I'm no genius though :)
It is harder with other odd numbers
 
$n\equiv 1\pmod {10}$ then $n^n +3\equiv 4\pmod {10}$
 
the powers are rather inconsistent though cyclic
@MrPie I deliberately avoided that since 4 is possible (e.g. 2^2, 8^2)
$n\equiv9\pmod{10}$ is also impossible
As $n$ is odd, $n^n\equiv9\pmod{10}$ so $n^n+3\equiv2\pmod{10}$ (contradiction)
 
@TheSimpliFire well, since $n^n+3\equiv 4\pmod {10} \equiv 0\pmod 4$ then it goes to show that the only solution to suffice is $n^n+3=4$.
 
@MrPie You can't do that
You'd need the Chinese Remainder Theorem
 
2:22 PM
@TheSimpliFire why not?
 
CRT allows you to combine different modulos
 
Chinese Remainder Theorem?
 
Nvm
I'll give an example instead
4 mod 10 = 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, ...
 
yeah
 
0 mod 4 = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, ...
then we see the pairs match when it is 4, 24, 44, 64, 84, ...
i.e. 4 mod 20
We now have modulo 20 since (by CRT), gcd(4, 10) = 20
 
2:24 PM
oh
 
So $n\equiv1\pmod{10}$ is still a tricky case to solve
We have $n\equiv3\pmod{10}$ and $n\equiv7\pmod{10}$ remaining
For $n\equiv3\pmod{10}$, when $n\equiv1\pmod4$, we have $n^n+3\equiv3+3\equiv6\pmod{10}$ which is possible
For $n\equiv3\pmod{10}$, when $n\equiv-1\pmod4$, we have $n^n+3\equiv7+3\equiv0\pmod{10}$ which is also possible
For the latter case, this forces $100\mid n^n+3$ if it is a square
 
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
 
@MrPie You have just discovered the process of CRT :)
That is essentially how it works
 
@TheSimpliFire thanks for walking me through it, though :P
 
np
Question is: can $n^n+3$ end in two zeroes?
 
2:30 PM
No
$n^n\equiv -3\pmod {100}$ otherwise
$\therefore n^n\equiv 1\pmod 4$ (I think) but we said it was $0\pmod 4$
Wait, did we? no
Because $n$ was odd. Wait.
 
4 | (n-1)
n = 4k + 1
Then that gets rid of n = -1 mod 4 immediately!
So we must have $n^n+3\equiv6\pmod{10}$
 
I'm confused
How is it important that $n\not\equiv -1\pmod 4$?
 
19 mins ago, by TheSimpliFire
Therefore, $4\mid (n-1)$
 
Yeah, I know, but I'm asking how $n\not\equiv -1\pmod 4$ implies $n^n+3\equiv 6\pmod {10}$
 
8 mins ago, by TheSimpliFire
For $n\equiv3\pmod{10}$, when $n\equiv1\pmod4$, we have $n^n+3\equiv3+3\equiv6\pmod{10}$ which is possible
 
2:35 PM
Oh SH**
My bad! That last digit thing, I got confused with
 
You were midway through typing your version of CRT when I sent it :P
 
Ah. I get it now, because we're working in base 10.
@TheSimpliFire I'm gonna think of that next time. Stuck on divisibility? Consider last digit :P
 
:) mod 10 sometimes works very well
 
Thanks (again) xD
Just realised it was 12:39am.
 
2:37 PM
Y'know what, Im not actually that bad at math as I think to be
I think it's just my sleeping habits
 
@MrPie Nobody thinks you're bad at maths :P
"good" is an understatement as long as you have an account on MSE ;)
 
@TheSimpliFire hahah, well, I am far too elementary. I want to be creative.
Like Ramanujan was working on partitions
people thought it couldn't be done
 
In conclusion, we have shown that $n^n+3\equiv0,4,6\pmod{10}$ under certain conditions
 
And then he plucks out $e$ and $\pi$ and wait, what?
ok, yeah, back to topic :P
 
Is it as far as we can go?
(all words less than 4 letters long :)
 
2:41 PM
$n^n$ must be odd. Let $n^n=2N+1$ then $N+1\equiv 0,2,3\pmod 5$
 
N+2 :P
 
@TheSimpliFire what do you mean?
 
n^n + 3 = 2N + 4
 
Oh XD
Forgot to add the 3
 
heheh
 
2:44 PM
$N+2\equiv 0,2,3\pmod 5$
$\therefore N\equiv -2,0,1\pmod 5$
Agh, where to go now.... hmmm..
 
yes
@MrPie But we found that $n=4N+1$
which is stronger
 
So....
$2(N+2)\equiv 0,2,3\pmod 5$
Wait, we can do that with modulos?
 
$N\equiv0,1,3\pmod5$
Typo, 2N+2 instead of 2(N+2)
 
Don't you have to divide the rest of the remainders and the modulus?
man, i am... I need a glass of milk xD
OK shakes head
LET'S DO DIS
$N=\frac{n^n-1=(n-1)(1+n+\cdots n^{n-1})}{2}$
excuse my lazy notation :P
 
Eh, I'll just let you figure out again why n = 4N + 1
 
2:51 PM
@TheSimpliFire I know why
I just don't know what to do with that fact
 
@MrPie Take a break.
 
How do I apply it to the rest of the things we know?
@Grešnik XD
I am doing this
 
We get 4N + 4 = 0, 4, 6 mod 10
subtract 4: 4N = 0, 2, 6 mod 10
 
If it is the last thing I do...
 
3:44 PM
@TheSimpliFire hahah, I just took a shower and then said nothing just to scare you guys ;)
 
I was analysing papers during that time
almost done
@MrPie et al. (heh) I notice that $2^5-5^2=2+5$
@Peter @MrPie @Haran @Grešnik I conjecture that the only solution to the integral equation $$a^b-b^a=a+b$$ is $(a,b)=(2,5)$.
 
Wow
Another conjecture
 
PARI/GP code
for(i=1,1000,for(j=1,1000,if(i^j-j^i-i-j==0,print(i," ",j))))
@MrPie This week has been great with conjectures :)
 
You know what, math is infinity
@TheSimpliFire oh, and your other conejcture with the prime number triplet thingand the exponents
I still got nothing
I think there is no solution ;)
 
someone downvoted the algorithm thing today...
maybe they sensed I was one vote away from a good q badge XD
 
3:57 PM
@TheSimpliFire what??!
 
you can't satisfy everyone in life :P
 
xD You were on 24
And now you're on 23
 
from the bounty I was hoping for some breakthrough on the problem
sometimes votes aren't a good indication of how good a problem is - some conjectures have very few votes but are outstanding
 
@TheSimpliFire yeah, it's all about taste
@TheSimpliFire Omega posted a new puzzle. Imma take a crack at it
 
quick post
0
Q: Conjecture: Is the identity $2^5-5^2=2+5$ unique?

TheSimpliFireYet again a conjecture! Motivated by Catalan's conjecture and a recent question of mine, I conjecture that For distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$ It is of anticipation that there will be much fewer solutions for incr...

cu after dinner
 
4:22 PM
@TheSimpliFire For $1\le a,b\le 2\ 000$, there is no further solution
 
4:59 PM
@Mathphile Found a factor ?
 
ecm: 156/904 curves on C1050, B1=1M, B2=gmp-ecm default, ETA: 11.31 hrs
 
Apparently, no
Now, more than 30 digits are very likely
 
do you want me to continue?
 
5:27 PM
hmm
it seems like $\sum_{k=1}^n n^{n-k}$ can only be a perfect square for $n=1$
 
@Mathphile Note $n^n-1=(n-1)\sum\limits_{k=1}^n n^{n-k}$
Don't know if that helps
 
@TheSimpliFire how did you get this?
 
4 hours ago, by TheSimpliFire
Which then means that $$(m-2)(m+2)=(n-1)(n^{n-1}+n^{n-2}+\cdots+1)$$
I got the idea from your n^n + 3 problem
MrPie and I were working on it whilst you were away
 
5:46 PM
@TheSimpliFire nice
@TheSimpliFire seems like you got a nice answer to this too
 
Yes, I didn't realise it was already asked before :)
It's a surprisingly long proof for what seems a very convincing plot on Desmos
 
mathjax makes it hard to search for similar questions
 
6:00 PM
@Mathphile OK, the last time for today : found a factor ?
 
@Peter no
230/904 curves on C1050, B1=1M, B2=gmp-ecm default, ETA: 9.95 hrs
 
tough number
 
yes
@Peter do you want me to continue?
is it possible that this number is a prime?
 
no, it is definitely composite, if you want continue
 
i think that 1050 digits may be impossible
ETA is 9.82 hours, i think i should stop
 
6:10 PM
Of course, we can have bad luck that the smallest factor is too high. It is your decision whether you want continue.
 
6:32 PM
also there seems to be no square of the form $\lfloor \zeta(-n) \rfloor \gt 0$
neither for the ceil and round versions too
 
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