Inverse Euler Phi Function

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  • Опубліковано 16 січ 2025

КОМЕНТАРІ • 11

  • @binayakbanik1694
    @binayakbanik1694 Рік тому

    Great explanation ❤

  • @beyond_e_to_the_i_pi
    @beyond_e_to_the_i_pi 2 роки тому +1

    Really helpful video.
    Thank you sir.

  • @heweicao1041
    @heweicao1041 4 роки тому +1

    Thank you

  • @amanpreet269
    @amanpreet269 2 роки тому

    thank you sir

  • @hAsH_learn2code
    @hAsH_learn2code 4 роки тому +1

    Thanks, Subscribed

  • @syifaamustafa954
    @syifaamustafa954 2 роки тому

    hi! thanks for the video! I want to ask what if (p_i) | phi(n). what will be the consequences?

    • @vks1231
      @vks1231  2 роки тому

      if p divides phi(n), and p>2, then p^k divides n for some k>1.

  • @kylieshinez
    @kylieshinez 2 роки тому

    Can you explain again why there are no solutions to phi(theta)= 3 ? you said because phi(x) is even so only odd number allowed is 1. Is that a theorem or ?

    • @vks1231
      @vks1231  2 роки тому +1

      Theta is a natural number, so it can be written as product of powers of prime. If p is prime divisor of theta, then p-1 divides phi(theta). But p-1 is even for any prime except 2, therefore, theta cannot have any prime factors other than 2.
      Now if theta is power of 2, say 2^k, then phi(2^k)=2^{k-1}. So, k cannot be greater than 1 to have an phi(theta)= odd number. However, if k=1, then you have phi(theta)=1. Therefore, the result.

  • @aashita6438
    @aashita6438 2 роки тому

    How find phi(24) ?

    • @vks1231
      @vks1231  2 роки тому

      One of the consequence of definition of phi function is
      phi(mn)=phi(m)phi(n) and phi(p^k)=p^k-p^(k-1).
      Now, phi(24)=phi(3x8)=phi(3)×phi(8)=2×4=8.
      Alternatively, since 24 is a "small" number, find all the numbers less than 24 and co-prime to 24. There are exactly 8 of those: 1, 5, 7, 11, 13, 17, 19, 23.