Expand Messages
• Is n!-1 ever a square? There must exist some k such that n!+/-k is never square, other than k=0. Jon Perry perry@globalnet.co.uk
Message 1 of 7 , Apr 20, 2003
• 0 Attachment
Is n!-1 ever a square?

There must exist some k such that n!+/-k is never square, other than k=0.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -- ... 2) p q =n (could be p=q=n) Conclusion: n
Message 2 of 7 , Apr 20, 2003
• 0 Attachment
'If the Goldbach Conjecture holds, then for every n there exist p,q primes
that verify p+q = 2·n -->

---> 1) p, q < 2·n
2) p<=n --> q>=n (could be p=q=n)

Conclusion: n <= q < 2·n --> Proof for Bertrand's postulate: between n and
2n there is always a prime.'

I think everyone who has ever attempted to solve GC has spotted this.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• everyone please send in a cool short proof based on primes. one of my favorites is , the factorial of n with n 1 can never be a square... using bertrands
Message 3 of 7 , Apr 21, 2003
• 0 Attachment
everyone please send in a cool short proof based on primes.

one of my favorites is , the factorial of n with n > 1 can never be a
square... using bertrands postulate there is always a prime between
k and 2k. From this it follows that there is a prime that divides n!
but whose square does not. because in the prime
factorization of a square all exponents are even, thus n! cannot be a
perfect square.
• ... Here is an old favourite: there is an infinite number of primes. Proof: Define F_n = 2^(2^n) + 1 where n is an integer = 1. Now F_n - 2 = 2^(2^n) - 1 =
Message 4 of 7 , Apr 21, 2003
• 0 Attachment
> everyone please send in a cool short proof based on primes.

Here is an old favourite: there is an infinite number of primes.

Proof:

Define F_n = 2^(2^n) + 1 where n is an integer >= 1.

Now F_n - 2 = 2^(2^n) - 1 = (2^(2^n-1) + 1) * (2^(2^n-1) - 1) by the difference of squares rule.

Hence F_n - 2 = F_{2n-1} * (F_{2n-1} - 2))

So F_n leaves a remainder 2 when divided by F_{n-1} and so F_n is not divisible by any F_m when m<n and so the prime factorization of F_n does not include any F_m with m<n

As n may take on an infinity of values, there are an infinite number of primes.

Paul
• Here comes a little proof of mine (oh, my big ego :P): If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -- ...
Message 5 of 7 , Apr 21, 2003
• 0 Attachment
Here comes a little proof of mine (oh, my big ego :P):

If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -->

---> 1) p, q < 2·n
2) p<=n --> q>=n (could be p=q=n)

Conclusion: n <= q < 2·n --> Proof for Bertrand's postulate: between n and 2n there is always a prime.

Can anyone say if this was observed yet for any author? (I don't like to say that is my proof if it isn't).

Jose Brox
http://espanol.groups.yahoo.com/group/Telecomunicacion/

----- Original Message -----
From: xyzenogenicyx
Sent: Monday, April 21, 2003 8:27 PM

everyone please send in a cool short proof based on primes.

one of my favorites is , the factorial of n with n > 1 can never be a
square... using bertrands postulate there is always a prime between
k and 2k. From this it follows that there is a prime that divides n!
but whose square does not. because in the prime
factorization of a square all exponents are even, thus n! cannot be a
perfect square.

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

[Non-text portions of this message have been removed]
• ... I also think so, this is the reason I ask for the references
Message 6 of 7 , Apr 21, 2003
• 0 Attachment
Jon says:
> I think everyone who has ever attempted to solve GC has spotted this.

I also think so, this is the reason I ask for the references
• What happened to this thread? I enjoyed it. Would be any web publications of the proof recolected? Are any more proofs wanted? Jose Brox ... From:
Message 7 of 7 , May 4, 2003
• 0 Attachment
What happened to this thread? I enjoyed it.

Would be any web "publications" of the proof recolected?

Are any more proofs wanted?

Jose Brox

----- Original Message -----
From: xyzenogenicyx
Sent: Monday, April 21, 2003 8:27 PM

everyone please send in a cool short proof based on primes.

one of my favorites is , the factorial of n with n > 1 can never be a
square... using bertrands postulate there is always a prime between
k and 2k. From this it follows that there is a prime that divides n!
but whose square does not. because in the prime
factorization of a square all exponents are even, thus n! cannot be a
perfect square.

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/