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Exercise 2 · Q4

Q.Verify that the relation φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p) holds true for p=24p = 24.

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The identity φ(p)+τ(p)=σ(p)\varphi(p)+\tau(p)=\sigma(p) is only true for a prime pp; testing p=24p=24 (composite) gives 8+8=16≠60=σ(24)8+8=16\ne 60=\sigma(24), so it fails.

For n=p1a1⋯pkakn=p_1^{a_1}\cdots p_k^{a_k}:

φ(n)=n∏(1−1pi),τ(n)=∏(ai+1),σ(n)=∏piai+1−1pi−1\varphi(n)=n\prod\left(1-\tfrac1{p_i}\right),\quad \tau(n)=\prod(a_i+1),\quad \sigma(n)=\prod\frac{p_i^{a_i+1}-1}{p_i-1}

The stated identity φ(p)+τ(p)=σ(p)\varphi(p)+\tau(p)=\sigma(p) is a property of primes only: for a prime φ(p)=p−1, τ(p)=2, σ(p)=p+1\varphi(p)=p-1,\ \tau(p)=2,\ \sigma(p)=p+1, so (p−1)+2=p+1(p-1)+2=p+1.

  1. Factorise: 24=23×324=2^3\times 3.
  2. φ(24)\varphi(24): φ(24)=24(1−12)(1−13)=24×12×23=8\varphi(24)=24\left(1-\tfrac12\right)\left(1-\tfrac13\right)=24\times\tfrac12\times\tfrac23=8.
  3. τ(24)\tau(24): τ(24)=(3+1)(1+1)=4×2=8\tau(24)=(3+1)(1+1)=4\times 2=8.
  4. σ(24)\sigma(24): σ(24)=24−12−1×32−13−1=15×82=15×4=60\sigma(24)=\dfrac{2^4-1}{2-1}\times\dfrac{3^2-1}{3-1}=15\times\dfrac{8}{2}=15\times 4=60. (Check: 1+2+3+4+6+8+12+24=601+2+3+4+6+8+12+24=60.) …

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