TheSimpliFire's Chatroom - 2019-07-11 (page 1 of 2)

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.

Mathphile

7:16 AM

@MrPie thank you :)

Grešnik

9:04 AM

But if I write the proof then I should give myself bounty of mine, is that an issue?

Peter

@Grešnik Then, post an answer. I noticed your comment, but why is there no solution if the given inequality holds ?

TheSimpliFire

You can't yourself a bounty

Grešnik

@Peter No, this proof is new and is not based on that idea.

So the rules are that someone cannot give himself a bounty?

TheSimpliFire

Yes

Grešnik

But this is a situation, if my proof is right, where it would only be fair if I give myself the bounty?

TheSimpliFire

9:08 AM

math.stackexchange.com/help/bounty

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.)

Grešnik

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.

Peter

@Grešnik Unlucky, that you offered the bounty and then found a solution. Seems that you have to wait until the period ends.

Grešnik

I am trying now to find one more way to prove it.

Peter

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.

TheSimpliFire

@Peter If you managed n>392 then did you solve the case with n=66?

Peter

I have not factored it for n = 66 yet, neither for n = 62

Mathphile

@Peter i can't see a reason why the expression cannot be prime for odd n either

TheSimpliFire

@Peter How many cases >392 did you check?

Grešnik

Or why would even´s give more primes than odd´s.

Peter

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.

Grešnik

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.

Grešnik

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 $p_1 \neq p_k$

TheSimpliFire

10:21 AM

@Grešnik can cancel ap[1] and bp[k]

Grešnik

Then we obtain $c(bp_k-ap_1)=bp_1-ap_k$

$c$ must be natural number.

TheSimpliFire

Is p[1]=2?

Grešnik

No, but any prime.

TheSimpliFire

then (a + bc) | (ac + b)

Grešnik

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 :(

TheSimpliFire

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

Grešnik

hahahah :D

TheSimpliFire

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]

Grešnik

But why "this forces ac+b=a+bc"?

Grešnik

10:54 AM

We are CLOSE, but still we do not have a proof.

TheSimpliFire

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.

Grešnik

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.

Grešnik

11:37 AM

Bounty awarded!!!

Grešnik

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

Peter

12:40 PM

@Grešnik I did not spot an error

Grešnik

@TheSimpliFire Do you want that I make an edit and mention your help?

@Peter There is none.

Peter

@Mathphile $$401^{402}+402^{403}$$ is the smallest odd case without a known factor. Can you try to find a factor ?

TheSimpliFire

@Grešnik It's fine, you don't need to

Mathphile

okay i'll run it on pari

Grešnik

@TheSimpliFire Did you check the proof, do you have any objections?

Peter

12:45 PM

pari does not display intermediate results as far as I know. yafu would be better.

Mathphile

@Peter i have downloaded yafu

Peter

Great !

Just enter "factor(401^402+402^403)"

Mathphile

how do we use it?

@Peter okay

TheSimpliFire

@Grešnik Valid proof :)

Peter

The smallest factor seems to have more than 20 digits

@Grešnik Good job!

TheSimpliFire

12:54 PM

The methodology is extremely elementary but clever

Grešnik

@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.

Peter

@Mathphile I guess, we still are at C1050 ?

Mathphile

ecm: 118/214 curves on C1050, B1=50K, B2=gmp-ecm default, ETA: 4.2 min

Peter

OK, still no factor found as with my yafu (which runs of course much slower)

Mathphile

what does B1 mean?

Peter

1:00 PM

Did you ever hear of "elliptic curves" ?

Omega Krypton

@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.

Mathphile

heard but don't know much

Peter

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)

TheSimpliFire

@OmegaKrypton and so is everyone who clicks into the link :P

Omega Krypton

@TheSimpliFire then that proves i am boring :P

Peter

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

Mathphile

okay

Peter

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.

Grešnik

We could make Peter´s life a little easier if we find efficient factoring algorithm.

Mathphile

@Grešnik i think ecm is very efficient

Peter

@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.

Mathphile

1:13 PM

what is siqs?

Peter

self-initializing quadratic sieve. Can factor numbers upto about 100 digits in a reasonable time

Mathphile

is there any reason why $n!+1$ cannot be a perfect square for $n \gt 7$?

Grešnik

Is that one of the open problems?

Mathphile

oh yes

Grešnik

I know. :)

Mathphile

1:17 PM

searching it up i found out that this is brocard's problem

Grešnik

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.

Peter

Sure that it is still open ?

I think I recently heard it has been solved, but I am not sure.

Grešnik

I do not know surely, I just know that I tried to solve it also.

Peter

finally , my yafu switched to 250k ...

Mathphile

@Peter what is the pari function to check if a number is a square?

Peter

1:21 PM

issquare()

Grešnik

@Mathphile The factors of n! and (m-1)(m+1) after n>7 tend "not to coincide".

TheSimpliFire

@Grešnik I think I proved that m cannot be in the sequences A224473 or A224474 from the OEIS.

Peter

I wonder which is the smallest natural number not being in the factordb-database.

@Mathphile Still C1050, I guess ...

Mathphile

@Peter yes

Grešnik

Did anyone check for some reasonable range of n´s how close are n!+1 to the squares?

Peter

1:31 PM

not a bad idea. Identifying "near-misses" could shed some light on this problem.

TheSimpliFire

You'd be better off checking how often m^2 - 1 is close to n!

Mathphile

can $n^n+1$ be a perfect square?

Peter

No because of the now proven Catalan-conjecture.

It cannot even be a perfect power

Mathphile

i see

Peter

@Mathphile Does the program for the orders still run ? I mean the "power-summer-problem"

Mathphile

1:36 PM

what about $n^n+2$?

@Peter no i stopped it

TheSimpliFire

@Mathphile I recall that you mentioned $n^n\pm k$ a few months ago

or was it $p^p\pm k$

Mathphile

@TheSimpliFire yes

although that was for primes

Peter

Upto $n=10^4$ , $n^n+2$ is no perfect power. Now I run the range upto $10^5$

Mathphile

no perfect power for $10^5$ either @Peter

Peter

Related to Pillai's conjecture

TheSimpliFire

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

Grešnik

Why not n!+1=(m-1)(m+1), i would rather try with an open problem.

Mathphile

$n^n+1$ has been proven by Catalan's conjecture

TheSimpliFire

@Grešnik Are you planning to tackle Brocard's problem

Grešnik

Yes.

Mathphile

let's try $n^n+2$

Mr Pie

1:41 PM

@OmegaKrypton is this a sighting of you?

Peter

even n gives residue 2 mod 4, cannot be a perfect square.

TheSimpliFire

@Grešnik Seeing as I have failed previously, I might as well share what I've done

Omega Krypton

yes @MrPie

Peter

n = 4k+1 is also impossible (residue mod 4 woule be 3)

Mr Pie

:D

Peter

1:43 PM

I speak of $n^n+2$

@OmegaKrypton

lel

@Peter yes

$n^n+2$ for natural $n$ or prime $n$ (like last time?)

Mathphile

@MrPie general $n$

Grešnik

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.

TheSimpliFire

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}

Mathphile

@Peter It also seems that $n^n+3$ is only a perfect square for $n=1$

Mr Pie

@TheSimpliFire pretty :P

TheSimpliFire

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. $\square$

Peter

Whether we can proof it ?

@Mathphile

TheSimpliFire

1:53 PM

@Grešnik See above ^^

Mr Pie

@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"

Mathphile

@MrPie hahahahahah

Mr Pie

@Mathphile I have come up with the term "Konjecture". It is a verb, and to "konjecture" means to makes 10 conjectures in one day

Mathphile

@MrPie lol i could do that although the conjectures would all be pretty bad xD

Grešnik

@TheSimpliFire I´m tired.

What do we know about sums of four squares of naturals?

I mean sleepy.

TheSimpliFire

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 $k(n-1)(n+1)=(m-2)(m+2)$

Mathphile

@TheSimpliFire you mean $(m-\sqrt{2}))(m+\sqrt2)$

TheSimpliFire

No

(m - 2)(m + 2) = m^2 - 4

Bring the 4 across and you get 4 - 1 = 3

Mathphile

@TheSimpliFire sorry don't know what i was thinking

Grešnik

Who wants to listen to a song I chose?

Mr Pie

@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.

TheSimpliFire

2:08 PM

Great @MrPie

Mr Pie

@TheSimpliFire thanks :P

Mathphile

@MrPie nice

TheSimpliFire

Which then means that $$(m-2)(m+2)=(n-1)(n^{n-1}+n^{n-2}+\cdots+1)$$

Mr Pie

So really, $n^n\equiv 1\pmod 4$ or $n\equiv 1\pmod 4$ (otherwise it would be even).

TheSimpliFire

Forcing $m$ to be even

Mr Pie

2:10 PM

@TheSimpliFire does $m$ represent $k$ here?

TheSimpliFire

Yes

Mr Pie

@TheSimpliFire smart what you did :P

TheSimpliFire

Then RHS is divisible by 4

Mr Pie

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)

Mathphile

i gtg guys

be back in an hour

Mr Pie

2:12 PM

@Mathphile cya

TheSimpliFire

bye @Mathphile

Now $n^r$ is odd for all positive integers $r$

Mr Pie

$n$ has to be odd

TheSimpliFire

But the cardinality of $\{1,\cdots,n-1\}$ must be even

Therefore, $4\mid (n-1)$

Mr Pie

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 $n:=1$ :/

TheSimpliFire

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

Mr Pie

2:18 PM

Aha!

@TheSimpliFire genius! :D

TheSimpliFire

thanks, I'm no genius though :)

It is harder with other odd numbers

Mr Pie

$n\equiv 1\pmod {10}$ then $n^n +3\equiv 4\pmod {10}$

TheSimpliFire

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)

Mr Pie

@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 $n^n+3=4$.

TheSimpliFire

@MrPie You can't do that

You'd need the Chinese Remainder Theorem

Mr Pie

2:22 PM

@TheSimpliFire why not?

TheSimpliFire

CRT allows you to combine different modulos

Mr Pie

Chinese Remainder Theorem?

TheSimpliFire

Nvm

I'll give an example instead

4 mod 10 = 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, ...

Mr Pie

yeah

TheSimpliFire

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

Mr Pie

2:24 PM

oh

TheSimpliFire

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

Mr Pie

@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

TheSimpliFire

@MrPie You have just discovered the process of CRT :)

That is essentially how it works

Mr Pie

@TheSimpliFire thanks for walking me through it, though :P

TheSimpliFire

np

Question is: can $n^n+3$ end in two zeroes?

Mr Pie

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.

TheSimpliFire

4 | (n-1)

n = 4k + 1

Then that gets rid of n = -1 mod 4 immediately!

So we must have $n^n+3\equiv6\pmod{10}$

Mr Pie

I'm confused

How is it important that $n\not\equiv -1\pmod 4$?

TheSimpliFire

19 mins ago, by TheSimpliFire

Therefore, $4\mid (n-1)$

Mr Pie

Yeah, I know, but I'm asking how $n\not\equiv -1\pmod 4$ implies $n^n+3\equiv 6\pmod {10}$

TheSimpliFire

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

Mr Pie

2:35 PM

Oh SH**

My bad! That last digit thing, I got confused with

TheSimpliFire

You were midway through typing your version of CRT when I sent it :P

Mr Pie

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

TheSimpliFire

:) mod 10 sometimes works very well

Mr Pie

Thanks (again) xD

Just realised it was 12:39am.

TheSimpliFire

:o

Mr Pie

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

TheSimpliFire

@MrPie Nobody thinks you're bad at maths :P

"good" is an understatement as long as you have an account on MSE ;)

Mr Pie

@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

TheSimpliFire

In conclusion, we have shown that $n^n+3\equiv0,4,6\pmod{10}$ under certain conditions

Mr Pie

And then he plucks out $e$ and $\pi$ and wait, what?

ok, yeah, back to topic :P

TheSimpliFire

Is it as far as we can go?

(all words less than 4 letters long :)

Mr Pie

2:41 PM

$n^n$ must be odd. Let $n^n=2N+1$ then $N+1\equiv 0,2,3\pmod 5$

TheSimpliFire

N+2 :P

Mr Pie

@TheSimpliFire what do you mean?

TheSimpliFire

n^n + 3 = 2N + 4

Mr Pie

Oh XD

Forgot to add the 3

TheSimpliFire

heheh

Mr Pie

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..

TheSimpliFire

yes

@MrPie But we found that $n=4N+1$

which is stronger

Mr Pie

So....

$2(N+2)\equiv 0,2,3\pmod 5$

Wait, we can do that with modulos?

TheSimpliFire

$N\equiv0,1,3\pmod5$

Typo, 2N+2 instead of 2(N+2)

Mr Pie

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

TheSimpliFire

Eh, I'll just let you figure out again why n = 4N + 1

Mr Pie

2:51 PM

@TheSimpliFire I know why

I just don't know what to do with that fact

Grešnik

@MrPie Take a break.

Mr Pie

How do I apply it to the rest of the things we know?

@Grešnik XD

I am doing this

TheSimpliFire

We get 4N + 4 = 0, 4, 6 mod 10

subtract 4: 4N = 0, 2, 6 mod 10

Mr Pie

If it is the last thing I do...

TheSimpliFire

:O

Mr Pie

3:44 PM

@TheSimpliFire hahah, I just took a shower and then said nothing just to scare you guys ;)

TheSimpliFire

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 $(a,b)=(2,5)$.

Mr Pie

Wow

Another conjecture

TheSimpliFire

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 :)

Mr Pie

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 ;)

TheSimpliFire

someone downvoted the algorithm thing today...

maybe they sensed I was one vote away from a good q badge XD

Mr Pie

3:57 PM

@TheSimpliFire what??!

TheSimpliFire

you can't satisfy everyone in life :P

Mr Pie

xD You were on 24

And now you're on 23

TheSimpliFire

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

Mr Pie

@TheSimpliFire yeah, it's all about taste

@TheSimpliFire Omega posted a new puzzle. Imma take a crack at it

puzzling.stackexchange.com/questions/86086/…

TheSimpliFire

quick post

0

Q: Conjecture: Is the identity $2^5-5^2=2+5$ unique?

Yet 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...