Square of x , added to one is not equal to n

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Discussion Overview

The discussion revolves around the assertion that \(x^{2}+1\) is not equal to \(n!\) for positive integers \(x\) and \(n\). Participants explore the implications of this statement through various mathematical perspectives, including properties of prime numbers, complex numbers, and factorials.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that \(x^{2}+1\) has only primes of the form \(4k+1\), while \(n!\) includes primes of both forms \(4k-1\) and \(4k+1\).
  • There is a suggestion that if \(x\) is a positive integer, then \(x^{2}+1\) cannot equal \(n!\) for \(n > 3\) due to differing numbers of factors.
  • Some participants question the necessity of using the complex plane to analyze \(x^{2}+1\), suggesting it is sufficient to consider \(x\) as a positive integer.
  • One participant references Fermat's theorem on sums of two squares, proposing that only primes of the form \(4k+1\) can be expressed as sums of two squares.
  • There is a discussion about finding counterexamples that include primes of the form \(4k-1\) and whether such examples exist.
  • Participants explore the implications of representing real numbers as products of complex numbers, suggesting that such representations cannot include primes of the form \(4k-1\).
  • One participant presents a modular arithmetic argument to support the claim that \(n!\) cannot be a perfect square for \(n \geq 5\).

Areas of Agreement / Disagreement

Participants express differing views on the properties of primes related to \(x^{2}+1\) and \(n!\). There is no consensus on the necessity of complex numbers in this context, and the discussion remains unresolved regarding the existence of counterexamples and the implications of Fermat's theorem.

Contextual Notes

Some arguments depend on specific definitions of primes and their forms, and there are unresolved mathematical steps related to the claims made about the properties of \(n!\) and \(x^{2}+1\).

secondprime
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$$x^{2}+1 \neq n! $$since $$x^{2}+1=(x+i)(x-i) $$so ,$$ x^{2}+1$$ has only prime of the form of (4k+1) , where n! has prime of the form( 4k-1) and (4k+1) . :oldbiggrin:
 
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I am assuming that ##x\in\mathbb{Z}^+##?
 
yap! and consider odd prime, not 2!
 
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secondprime said:
$$x^{2}+1 \neq n! $$since $$x^{2}+1=(x+i)(x-i) $$so ,$$ x^{2}+1$$
If x = 3, then x2 + 1 = 10, which factors into 10 * 1, 5 * 2, and (3 + i)(3 - i).
secondprime said:
has only prime of the form of (4k+1) , where n! has prime of the form( 4k-1) and (4k+1) . :oldbiggrin:
Of the pairs of factors I show above, 5 is of the form 4k + 1, but 10 and 3 + i (or 3 - i) aren't.
 
yes, but you can see that I am talking about prime (4k+1), 2 is neither (4k+1) nor (4k-1), it is the only even prime. is 10 a prime of the form of (4k+1) ?no, it has prime of the form (4k+1), but not (4k-1).

a good counter example would be to find a number which has (4k-1) prime as factor.:smile:
 
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secondprime said:
yes, but you can see that I am talking about prime (4k+1), 2 is neither (4k+1) nor (4k-1), it is the only even prime. is 10 of the form of (4k+1) no, it has prime of the form (4k+1), but not (4k-1).

a good counter example would be to find a number which has (4k-1) prime as factor.
It's not clear to me what you're talking about. I was responding to your first post.
secondprime said:
##x^{2}+1 \neq n! ## since ##x^{2}+1=(x+i)(x-i)## so, ## x^{2}+1## has only prime of the form of (4k+1) , where n! has prime of the form( 4k-1) and (4k+1) .
I found a number x (x = 3) for which x2 + 1 had factors other than the (x + i) and (x - i) that you list above. I believe that ##x^2 + 1 \neq n!## is a true statement, but the rest of what you're saying is not clear.
 
Sir,

In additive number theory, Pierre de Fermat's theorem on sums of two squares states that an odd prime p is expressible as sums of two squares iff it is of the form (4k+1). that can be proved, using complex number. only prime (4k+1) are possible to express in complex plane, since x^2 +1 is in complex plane ,i assumed it has primes of form (4k+1).

one might argue that even number of (4k-1) prime makes a (4k+1) number(e.g 7*7 +1=2*5*5), in that case I am looking for an example.

Regards.
 
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secondprime said:
only prime (4k+1) are possible to express in complex plane, since x^2 +1 is in complex plane ,i assumed it has primes of form (4k+1).
Why is the complex plane necessary to represent x2 + 1? As long as x is a positive integer (which you said in post #3), then x2 + 1 is also an integer with no imaginary part.
 
so I can say it has prime of the form (4k+1), all of it , not composite (4k+1) but prime(4k+1). Can I do that Sir??
 
  • #10
The title of this thread is "Square of x , added to one is not equal to n!" In other words, ##x^2 + 1 \neq n!##, where x is a positive integer.

This seems obviously true to me. If n > 3, then n! will have at least 3 factors (e.g., 4! = 4 * 3 * 2 -- I am ignoring the factor of 1), whereas x2 + 1 will have no more than two factors, although the pairs of factors might be different.
 
  • #11
Sir , I was trying to explore complex number, Gaussian integer, this post is a "prelude", if my interpretation is right , i might have something more useful to say , for the time being, is my idea wrong , Sir??

it would be helpful if you comment on my idea( or anyone else).
 
  • #12
secondprime said:
Sir , I was trying to explore complex number, Gaussian integer
That wasn't obvious from your first post or its title.

Is what I've quoted below what you're asking about (from post #7)?

secondprime said:
In additive number theory, Pierre de Fermat's theorem on sums of two squares states that an odd prime p is expressible as sums of two squares iff it is of the form (4k+1). that can be proved, using complex number.
Fine, this is clear.
secondprime said:
only prime (4k+1) are possible to express in complex plane, since x^2 +1 is in complex plane ,i assumed it has primes of form (4k+1).
This part I don't follow.
5 is a prime of the form 4k + 1, and 5 = 22 + 12
13 is a prime of the form 4k + 1, and 13 = 32 + 22
17 is a prime of the form 4k + 1, and 17 = 42 + 12
These are examples that use the Fermat theorem that you cited.
What do any of these have to do with the complex plane?
secondprime said:
one might argue that even number of (4k-1) prime makes a (4k+1) number(e.g 7*7 +1=2*5*5), in that case I am looking for an example.
What does "even number of (4k - 1) prime" mean?
 
  • #13
Sir,
only primes of form (4k+1) are possible to express as Gaussian integer, Fermat's thm can be proved using Gaussian integer(i.e as complex number,so those primes are in complex plane) . x^2 +1 can expressed as Gaussian integer too ,so I deduced , it's all prime factors are of the form (4k+1).

7 is the form of (4k-1), 7*7(even number of 7)=49=12*4+1 does not have a prime of the form (4k+1) but itself of the form (4k+1)! ...but if it is confusing please discard what I said about even (4k+1) in previous post.
below links might be helpful-
http://www.had2know.com/academics/gaussian-prime-factorization-calculator.html
http://en.wikipedia.org/wiki/Table_of_Gaussian_integer_factorizations

what I am trying to say that , if a real number can be represented as a complex number product , it can not have a prime of the form (4k-1).
 
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  • #14
secondprime said:
what I am trying to say that , if a real number can be represented as a complex number product , it can not have a prime of the form (4k-1).
I would say it like this: If a real number can be represented as the product of two complex numbers, it can not have a prime factor of the form (4k-1). I don't know how you would prove that, though.
 
  • #15
Mark44 said:
If a real number can be represented as the product of two complex numbers, it can not have a prime factor of the form (4k-1). I don't know how you would prove that, though.
Or more precise: If an integer can be...
 
  • #16
Svein said:
Or more precise: If an integer can be...
That's what I had in mind, but didn't state.
 
  • #17
Mark44 said:
That's what I had in mind, but didn't state.
I'm a mathematician - which means I have an advanced degree in nitpicking.
 
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  • #18
I think this could work too , let n>2.:
For ##n \geq 5 ##, n!-1 ends in 9 (check other cases), so that, if n! were a perfect square, it would have factors of one of the two types:

1)10K+3
2)10K+7

But neither ##(10K+3)^2+1## , nor ##(10K+7)^2 +1## is divisible by 3, so it cannot equal n!:

1')## (10K+3)^2+1 =100K^2+60K+10 ==1+0+K^2== 1+0+0=1 or 1+0+1=2 mod3 ##

2') ## (10K+7)^2+1 =100K^2+140K+50 == 2+2K+K^2 \neq 0 mod3 ## , since the sum will always be even.
 
  • #19
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