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

[secondprime]

$$x^{2}+1 \neq n! $$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) .

 

[RUber]

I am assuming that $x\in\mathbb{Z}^+$?

 

[secondprime]

yap! and consider odd prime, not 2!

 

[Mark44]

$$x^{2}+1 \neq n! $$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).

has only prime of the form of (4k+1) , where n! has prime of the form( 4k-1) and (4k+1) .

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.

 

[secondprime]

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.

 

[Mark44]

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.

$x^{2}+1 \neq n!$ 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.

 

[secondprime]

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.

 

[Mark44]

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.

 

[secondprime]

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??

 

[Mark44]

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.

 

[secondprime]

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).

 

[Mark44]

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)?

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.

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?

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?

 

[secondprime]

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).

 

[Mark44]

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.

 

[Svein]

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...

 

[Mark44]

Or more precise: If an integer can be...

That's what I had in mind, but didn't state.

 

[Svein]

That's what I had in mind, but didn't state.

I'm a mathematician - which means I have an advanced degree in nitpicking.

 

[WWGD]

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.

 

[secondprime]

.. hmm.. read below links...

http://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares

http://en.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity

http://en.wikipedia.org/wiki/Gaussian_integer


@ WWGD , it can be done in many ways, that is not why I posted what I posted, is my way right??, that is my concern, don't waste your time on this kind of mundane, unimportant problem.
I need to know whether I am right or not, I would appreciate if anyone set me right!