primitive

  1. alexmahone

    Primitive Pythagorean triple

    Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3. My attempt: Let a and b be relatively prime positive integers. If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$, $c^2=a^2+b^2\equiv 1+1\equiv 2...
  2. D

    Primitive Roots

    I have to come up with a number greater than 40 that does not have any primitive roots and explain my reasoning fully. This is for my own research and I am just familiarising myself with the topic of primitive roots as it is new to me. Is there any advice on how I should go about this that...
  3. alexmahone

    Primitive roots

    Is it true that $g$ is a primitive root modulo $p$ if and only if $g^{(p-1)/2} \equiv -1 \pmod p$?
  4. M

    Question about computing primitive roots modulo p for very big prime p

    Greetings, If p is big prime, how do we find a primitive root modulo p? Since p-1 is not prime, it's quite probable that (Z/pZ)* will contain subgroups and that some elements will generate only a subgroup of (Z/pZ)*. The Wikipedia says that there is no exact algorithm and it's all basically...
  5. Matt Westwood

    Primitive of 1 / (p sin a x + q cos a x + r)

    Number 420. Show that: \int \frac {\mathrm d x} {p \sin a x + q \cos a x + r} = \begin{cases} \dfrac 2 {a \sqrt {r^2 - p^2 - q^2} } \arctan \left({\dfrac {p + \left({r - q}\right) \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } }\right) + C & : p^2 + q^2 < r^2 \\ \dfrac 1 {a \sqrt {p^2 + q^2 -...
  6. Matt Westwood

    Primitive of 1 / (cos ax (1 + sin a x))

    Number 410 in my Fiendish Integral List. Show that: \int \frac {\mathrm d x} {\cos a x \left({1 + \sin a x}\right)} = \frac {-1} {2 a \left({1 + \sin a x}\right)} + \frac 1 {2 a} \ln \left\vert{\tan \left({\frac {a x} 2 + \frac \pi 4}\right)}\right\vert + C I've tried a few approaches (e.g...
  7. Matt Westwood

    Primitive of 1 / (p + q cos a x)

    No. 390. This is what the book says: \displaystyle \int \frac {\mathrm d x} {p + q \cos a x} = \begin{cases} \displaystyle \frac 2 {a \sqrt {p^2 - q^2} } \arctan \left({\sqrt {\frac {p - q} {p + q} } \tan \dfrac {a x} 2}\right) + C & : p^2 > q^2 \\ \displaystyle \frac 1 {a \sqrt {q^2 - p^2} }...
  8. Matt Westwood

    Primitive of 1 / (p + q sin a x)^2

    Number 361 in my List of Fiendish Integrals: Prove that: \int \frac {\mathrm d x} {\left({p + q \sin a x}\right)^2} = \frac {q \cos a x} {a \left({p^2 - q^2}\right) \left({p + q \sin a x}\right)} + \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x} I tried the Weierstrass...
  9. Matt Westwood

    Primitive of 1 / (1 - \sin a x)^2

    Here we are again -- let's get cracking on some of these hard integrals again! Number 358 in my source work: \int \frac {\mathrm d x} {\left({1 - \sin a x}\right)^2} which has to be demonstrated to be equal to: \frac 1 {2a} \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 1 {6 a}...
  10. Matt Westwood

    Primitive of Reciprocal of sin^3 ax

    While I do some investigation of the complicated induction work on no. 335 ... No. 352 in my list (there were a load of easy sine formulas): \int \frac {\mathrm d x} {\sin^3 a x} which is supposed to work out to: -\frac {\cos a x} {2 a \sin^2 a x} + \frac 1 {2 a} \ln \left\vert{\tan \frac {a...
  11. Matt Westwood

    Primitive of x^(p-1) / (x^(2m) + a^(2m))

    No. 335 in the list of difficult integrations. Prove that: \int \frac {x^{p - 1} \ \mathrm d x} {x^{2 m} + a^{2 m} } = \frac 1 {m a^{2 m - p} } \sum_{k \mathop = 1}^m \sin \frac {\left({2 k - 1}\right) p \pi} {2 m} \arctan \left({\frac {x + a \cos \left({\dfrac {\left({2 k - 1}\right) \pi} {2...
  12. Matt Westwood

    Primitive of x over (x^3 + a^3)^2

    Okay, this is not such a straightforward assignment ... We have to prove: \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2} = \frac {x^2}{3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {3 a^3} \int \frac {x \ \mathrm d x} {x^3 + a^3} I have a reason for wanting to get it into this specific...
  13. Matt Westwood

    Primitive of Reciprocal of x^4 + a^4

    Number 311 in the ever-more-tricky integrals ... \int \frac {\mathrm d x} {x^4 + a^4} whose posted solution is: \frac 1 {4 a^3 \sqrt 2} \ln \left({\frac {x^2 + a x \sqrt 2 + a^2} {x^2 - a x \sqrt 2 + a^2} }\right) - \frac 1 {2 a^3 \sqrt 2} \left({\arctan \left({1 - \frac {x \sqrt 2} a}\right)...
  14. Matt Westwood

    Primitive of Reciprocal of x^3 + a^3

    Number 299 in my never-ending list of integrals ... Seriously, mental block here, folks: \int \frac {\mathrm d x} {x^3 + a^3} which is supposed to work out to: \frac 1 {6 a^2} \ln \left({\frac {\left({x + a}\right)^2} {x^2 - a x + a^2} }\right) + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a}...
  15. Matt Westwood

    Primitive of (ax^2 + bx + c)^(n + 1/2)

    No. 295 in my list of difficult integrals ... Integral of the general power of a x^2 + b x + c: the object here is to establish the following reduction formula: \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x = \frac {\left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n...
  16. S

    Representation of p-adic integers using primitive roots

    Let p be an odd prime and let r be any positive integer that is a primitive root module p^2. Let X_p = \varprojlim \Bbb{Z} / (p-1)p^n \Bbb{Z} be the inverse limit of the rings \Bbb{Z} / (p-1)p^n \Bbb{Z}. This is useful because for any unit u \in \Bbb{Z}/p^n\Bbb{Z}, there exists K \in \Bbb{Z} /...
  17. A

    Abstract Algebra - Primitive Roots for Polynomials

    The field GF(8) of 8 elements can be thought of as consisting of elements of the form a*x^2 + bx + c, where a, b, c ∈ Z2. (Thus, all coefficients are simplified mod 2.) Moreover, x^3 = x + 1. (a) Compute (x^2 + x + 1) + (x^2 + 1), simplifying your answer to the form a*x^2 + bx + c for a, b, c...
  18. S

    Primitive root

    p prime, If p=1 ( mod 3) then Zp contains primitive cube roots of unity. Now I am considering which p does Zp contains primitive fourth roots of unity. I can prove that if Zp contains primitive fourth roots of unity, then 4|(p-1) . but how about the opposite way? I mean if p=1(mod4) then Zp...
  19. A

    Primitive of (a^2 + x^2)^-1

    the Primitive of (a^2 + x^2)^-1 = \frac{1}{a} tan^-1(x/a) so why does the Primitive of (a^2 + 1)^-1 = tan^-1(a) and not tan^-1(\frac{1}{a})?
  20. U

    prove there is no primitive root mod 3p

    when p odd , p>3. --------- I don't know how to cintinue: a^(p-1)=1(modp) a^2=1(mod3) ....