# Finite solutions of Brocard’s problem

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1. Mar 15, 2015

### secondprime

x^2=n!+1⇒ (x+1)(x-1)=n! where (x+1)/2 and (x-1)/2 are consecutive integers and have consecutive primes as factor ,let ,y and z (respectively) so it can be written y-1=z. Consider prime counting function π(z),π(2z-1) that count primes less than the variable or argument. It can be seen that f(z) is equals to the numbers of z-k when z-k is prime where 0<k<z+1.Same way , the total number of primes in between z+1 and 2z-1 is also the number of primes of the form z+k in between that interval. So, it can be written that, π(z)-w(z)= π(2z-1)- π(z)(ignoring the prime 2) where w(z) is the number of distinct prime divisor of z. so,
2π (z)- π(2z) = w(z).now form prime number theorem and an asymptotic formula for w(n), it can be written 2* ln⁡2 * z =ln⁡ln⁡z *ln⁡2z* ln⁡z +f(z) (where f(z) is error term.)
⇒2* ln⁡2 * z<<ln⁡ln⁡z *ln⁡2z* ln⁡z,which means after a certain z, there will be no intersection point.

It is sort of outline,I would be happy if you find any flaw and post it here,thanks.

I have a stack exchange link regarding the solution-
real analysis - Number of solutions of asymptotic funtion's equation. - Mathematics Stack Exchange

I explained the idea, but no response to that. Initially it was put on hold due to "unclear" tag. Any thoughts?

2. Mar 16, 2015

### mathman

The first sentence is fraught.
Other than n=4, are there any situations where it makes sense?
You state y and z are consecutive and prime. True only for 2 and 3.
Are there any even integers (other than 4 and 6) whose product is a factoral?

3. Mar 17, 2015

### TeethWhitener

I figured secondprime meant that y and z were consecutive and that they had consecutive prime factors...which as mathman points out would have to be 2 and 3. I'm not sure if that's ok. For $n≥3$, $n!+1$ must be odd and coprime with $n!$, so that means that $x^2$ (and therefore $x$) is not divisible by 2 or 3. So $x-1$ and $x+1$ are therefore both even and exactly one of them must be divisible by 3. So exactly one of $$S= \Big\{ \frac{x+1}{2}, \frac{x-1}{2} \Big\}$$ is divisible by 2, and exactly one of S is divisible by 3. The problem I'm having is, how do you prove that the one which is divisible by 2 isn't also the one divisible by 3? For instance, if $x-1=22$ and $x+1=24$, then $\frac{x+1}{2}$ is divisible by both 2 and 3, and $\frac{x-1}{2}$ is divisible by neither.

4. Mar 17, 2015

### TeethWhitener

Trivially, for $n≥4$, $(n-2)!$ and $n(n-1)$ are even integers which multiply to give $n!$.

5. Mar 17, 2015

### secondprime

the product of (x+1)/2 and (x-1)/2 has consecutive primes as factor, sorry for bad Eng.

But that should not stop you to check remaining part.

6. Mar 17, 2015

### TeethWhitener

The remaining part is very hard to understand.
What are the "numbers of z-k?" What does f(z) stand for?
Are you saying that the primes are symmetric about z (p = z ± k)?
I think I see what you're going for: you're looking for an upper bound for a formula to show that Brocard's problem has only finitely many solutions. (At least that's where I think this is going). But I'm having a lot of trouble understanding how exactly you're getting to that formula.

7. Mar 17, 2015

### secondprime

1. that should be π(z) not f(z), f(z) is the error term used later.
2. for any integer, z, and k 0<k<z+1, there exist prime number in the form of z-k, if you count all positive integers that is equal to prime counting function. Same way, you will find primes in the form of z+k, I claim, the total number of primes in between z+1 and 2z-1 is equal to the number of primes of the form z+k in between that interval.Please check and inform if it is not true.
.3. w(z) is the number of distinct prime divisor of z, in equation, it omits the repetition of prime.
if you reach up to the equation , then rest of the part is easy.
4. Not entirely what you are suggesting ,but in a way... it would be clear if you agree up to the equation.

*** i don't see any edit option here, where is it??

Last edited: Mar 17, 2015
8. Mar 18, 2015

### TeethWhitener

For any integer $z$, $\pi(z)$ is the number of primes less than $z$. No information is added by saying that the primes are of the form $z-k$. I have no idea how you got that Brocard's problem implies $\pi(z) - \omega(z) = \pi(2z) - \pi(z)$ (We can set $\pi(2z-1)=\pi(2z)$ because we don't count any more primes by including $2z$ in the set). Let's consider the case where $z=35$ (this is equivalent to $n=7$ in your formulation of Brocard's problem). $\pi(35)=11$ and $\pi(70)=19$, so if your equation holds, then $8=11-\omega(35)$. But $\omega(35)=2$ because there are only 2 distinct prime factors of 35 (namely 5 and 7). So your equation doesn't hold (8≠9).

9. Mar 18, 2015

### secondprime

1. all primes(consecutive) of n! are in z, (z+1) [Brocard's problem implies this]
2. Consider (z+k) and (z-k) where 0<k<z+1, if both are prime or composite at the same time then,
π(z)= π(2z-1)- π(z) is true where π(2z-1)- π(z) is the number of prime in between z and (2z-1)
if one is prime and another is not, then the equation is not true but I claim both have to be prime when k | (z+1).
let, the assumption is false
then (z+k)=cd where (z,k)=1

c,d does not divide z,k since k<z+1 and (z,k)=1
so, at least c | (z+1) implies c| k and leads to a contradiction . Same way,(z-k) is prime respect to the above condition.

3.But the primes of (z+1) are less than z, unbalance the equation (using above argument). So, w(z+1) is subtracted,
π(z)-w(z+1)= π(2z-1)- π(z)
Since I was planning to use asymptotic formula, I wrote w(z) instead of w(z+1).
4. you wrote-
"So your equation doesn't hold (8≠9)."
but in the first post there was a line like-
"So, it can be written that, π(z)-w(z)= π(2z-1)- π(z)(ignoring the prime 2)"
if you ignore prime 2 then......

10. Mar 18, 2015

### TeethWhitener

Are you saying ignore 2 completely, in every aspect of your proof? Because if we go with the example above where z=35, if we let k=2 (which divides z+1) then z+k=37 is prime but z-k=33 is composite. If we let k=3, then both z-k=32 and z+k=38 are composite. If we let k=4, then z-k and z+k are both prime. It seems you can get whatever answer you want this way.

Your mistake is somewhere in here:
But I'm not immediately sure where.

11. Mar 19, 2015

### secondprime

???

12. Mar 19, 2015

### secondprime

why don't you consider z=36, z-1=35 ,the argument I used is true for these pair so,
π(z)-w(z-1)= π(2z-1)- π(z) is true.
this is the only other possible case.

you should check yourself why z=35, z+1=36 didn't work.

"Your mistake is somewhere"... exactly where??

I could be wrong(that is why i shared the solution in this forum ), but you need to tell me why/how.
" It seems you can get whatever answer you want this way."

would you please explain what you meant?

13. Mar 19, 2015

### TeethWhitener

I already gave you a counterexample. What I fear will happen is that you will keep using the escape clause "Except for maybe the number 2" to continue to move the goalposts. As your proof is written right now, it's clearly wrong because I can provide counterexamples. If you have a proof in mind that is correct, that's fine, but you haven't expressed it clearly enough for me not to continue shooting it down with counterexamples.
I gave you a counterexample. I don't want to do you a disservice by telling you exactly why you're wrong because frankly what you've written isn't clear enough for me to determine where exactly your error is. Likewise, if it so happens that you have the kernel of a great idea, I don't want to discourage you from it; I just don't see where it is.
You asserted:
I gave you an example of z=35 where Brocard's equation holds, along with k=3 which divides z+1. I showed that z+k and z-k are both composite, directly contradicting what you asserted. I then showed that if k=2, then z+k is prime while z-k is composite, and if k=4, then z+k and z-k are both prime. Note that 2, 3, and 4 all divide z+1=36. Thus you can get any combination of prime and composite for z±k. Now you're saying that z should have been z-1 all along. I can't keep playing this game with you when you insist on moving the goalposts. Spend some time clarifying your thoughts and testing them with counterexamples, and then present a more rigorous proof.

14. Mar 19, 2015

### secondprime

"I already gave you a counterexample. What I fear will happen is that you will keep using the escape clause "Except for maybe the number 2" to continue to move the goalposts. As your proof is written right now, it's clearly wrong because I can provide counterexamples."
-35 was not a counterexample because it fits to equation, it has to be z-1 not z.

"What I fear will happen is that you will keep using the escape clause "Except for maybe the number 2" to continue to move the goalposts."
-i don't know why you are whining about prime 2, it was in the first post!!! you did not notice, so I informed you that is not "keep using the escape clause"

"I gave you an example of z=35 where Brocard's equation holds, along with k=3 which divides z+1. I showed that z+k and z-k are both composite, directly contradicting what you asserted."
!!! in this case the equation holds!!, problem is when you have one of z+k , z-k is prime and other is not.

" Now you're saying that z should have been z-1 all along."
I knew after your second post but I could not find the edit option, to avoid confusion, i kept things same with respect to first post. The main idea is similar in both (z,z+1) and (z,z-1). ... anyway , I mistyped so... myself to blame.

"I can't keep playing this game with you when you insist on moving the goalposts."
I changed only once, ignoring 2 was in the first post.
lets fix the goal post then , π(z)-w(z-1)= π(2z-1)- π(z) for z, z-1.

if this is not true than there is only one other case and that is for z, z+1. For both the cases π(z)-w(z) is same since asymptotic formula would be used.

Last edited: Mar 19, 2015
15. Mar 19, 2015

### TeethWhitener

Why do you ignore 2 for w(z-1) but not for π(z)? Your equation only holds if you ignore one but not the other.

"-i don't know why you are whining about prime 2,"

This was unnecessary. I'm not getting paid to help you. Your ideas are now significantly clearer and more presentable than the mess in your first post. My point was, from your equation, it's unclear when I'm supposed to include 2 and when I'm not. Apparently this equation only works when I ignore 2 in w(z-1) but when I take it into account for π(z). As it stands, I can't help you any longer. Find someone who will suffer your ingratitude.

Last edited: Mar 19, 2015
16. Mar 19, 2015

### secondprime

Are you pouting???
you can not be serious!!!! You seem to be offended, but that was not my intention!!! I am new to this forum , so might be a little ignorant about the "forum etiquette "
Please, accept my heartiest apology if I have had offended you at any measure.
And yes I messed up in the first post. It has created a lot of confusion due to my English skill and mistyping. But,

You were not a help, you have not pointed out anything new, the whole time you spent on understanding what I wrote. I helped you(though it is my responsibility to help/explain, especially after 1st post), don't have the wrong idea. This a public forum, it is designed to share, to help each other and nobody is getting paid!!!
If you have decided not to participate anymore, I am not going to stop you. Thanks for your time.

One last thing, you said-

"Apparently this equation only works when I ignore 2 in w(z-1) but when I take it into account for π(z).'

wrong, never ignore 2 in w(x), ignore 2 in prime counting function.