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Main Question or Discussion Point
$$x^{2}+1 \neq n! $$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) .
If x = 3, then x^{2} + 1 = 10, which factors into 10 * 1, 5 * 2, and (3 + i)(3 - i).$$x^{2}+1 \neq n! $$since $$x^{2}+1=(x+i)(x-i) $$so ,$$ x^{2}+1$$
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 said:has only prime of the form of (4k+1) , where n! has prime of the form( 4k-1) and (4k+1) .
It's not clear to me what you're talking about. I was responding to your first post.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.
I found a number x (x = 3) for which x^{2} + 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 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) .
Why is the complex plane necessary to represent x^{2} + 1? As long as x is a positive integer (which you said in post #3), then x^{2} + 1 is also an integer with no imaginary part.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).
That wasn't obvious from your first post or its title.Sir , I was trying to explore complex number, Gaussian integer
Fine, this is clear.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.
This part I don't follow.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).
What does "even number of (4k - 1) prime" mean?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.
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.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).
Or more precise: If an integer can be...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.
That's what I had in mind, but didn't state.Or more precise: If an integer can be...
I'm a mathematician - which means I have an advanced degree in nitpicking.That's what I had in mind, but didn't state.