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Exercise 2 · Q5
Q.

If pp is a prime then φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p). Using the information complete the following table:

PPφ(p)\varphi(p)τ(p)\tau(p)σ(p)\sigma(p)φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p)
2
3
5
7
11
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For a prime pp we always have φ(p)=p−1\varphi(p)=p-1, τ(p)=2\tau(p)=2, σ(p)=p+1\sigma(p)=p+1, and φ(p)+τ(p)=(p−1)+2=p+1=σ(p)\varphi(p)+\tau(p)=(p-1)+2=p+1=\sigma(p); the completed table confirms this for 2,3,5,7,112,3,5,7,11.

For a prime pp (only divisors are 11 and pp):

φ(p)=p−1,τ(p)=2,σ(p)=1+p=p+1\varphi(p)=p-1,\qquad \tau(p)=2,\qquad \sigma(p)=1+p=p+1

so φ(p)+τ(p)=(p−1)+2=p+1=σ(p)\varphi(p)+\tau(p)=(p-1)+2=p+1=\sigma(p).

  1. p=2p=2: φ=1, τ=2, σ=3\varphi=1,\ \tau=2,\ \sigma=3; check 1+2=31+2=3 ✓
  2. p=3p=3: φ=2, τ=2, σ=4\varphi=2,\ \tau=2,\ \sigma=4; check 2+2=42+2=4 ✓
  3. p=5p=5: φ=4, τ=2, σ=6\varphi=4,\ \tau=2,\ \sigma=6; check 4+2=64+2=6 ✓
  4. p=7p=7: φ=6, τ=2, σ=8\varphi=6,\ \tau=2,\ \sigma=8; check 6+2=86+2=8 ✓
  5. p=11p=11: φ=10, τ=2, σ=12\varphi=10,\ \tau=2,\ \sigma=12; check 10+2=1210+2=12 ✓

Completed table: …

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