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

  1. Mar 31, 2015 #1
    $$x^{2}+1 \neq n! $$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:
     
  2. jcsd
  3. Mar 31, 2015 #2

    RUber

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    I am assuming that ##x\in\mathbb{Z}^+##?
     
  4. Mar 31, 2015 #3
    yap!! and consider odd prime, not 2!!
     
    Last edited: Mar 31, 2015
  5. Mar 31, 2015 #4

    Mark44

    Staff: Mentor

    If x = 3, then x2 + 1 = 10, which factors into 10 * 1, 5 * 2, and (3 + i)(3 - i).
    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.
     
  6. Mar 31, 2015 #5
    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:
     
    Last edited: Mar 31, 2015
  7. Mar 31, 2015 #6

    Mark44

    Staff: Mentor

    It's not clear to me what you're talking about. I was responding to your first post.
    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.
     
  8. Mar 31, 2015 #7
    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.
     
    Last edited: Mar 31, 2015
  9. Mar 31, 2015 #8

    Mark44

    Staff: Mentor

    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.
     
  10. Mar 31, 2015 #9
    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??
     
  11. Mar 31, 2015 #10

    Mark44

    Staff: Mentor

    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.
     
  12. Mar 31, 2015 #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).
     
  13. Mar 31, 2015 #12

    Mark44

    Staff: Mentor

    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)?

    Fine, this is clear.
    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?
    What does "even number of (4k - 1) prime" mean?
     
  14. Mar 31, 2015 #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).
     
    Last edited: Mar 31, 2015
  15. Mar 31, 2015 #14

    Mark44

    Staff: Mentor

    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.
     
  16. Mar 31, 2015 #15

    Svein

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    Or more precise: If an integer can be...
     
  17. Mar 31, 2015 #16

    Mark44

    Staff: Mentor

    That's what I had in mind, but didn't state.
     
  18. Mar 31, 2015 #17

    Svein

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    I'm a mathematician - which means I have an advanced degree in nitpicking.
     
  19. Mar 31, 2015 #18

    WWGD

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    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.
     
  20. Apr 1, 2015 #19
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