prove that 10 is a solitary number

1.	prove that 10 is a solitary number
Answer» Letσ(n)σ(n)be the sum of divisor function.For1010to have a friendly pair we have to find another solution to σ(5x)=9xσ(5x)=9x 6σ(x)=9x(Assuming (5,x)=1)6σ(x)=9x(Assuming (5,x)=1) 2σ(x)=3x2σ(x)=3x Now Letx=2kpx=2kpwherep∈Np∈N,(p,z)=1(p,z)=1 2(2k+1−1)σ(p)=3⋅2kp2(2k+1−1)σ(p)=3⋅2kp σ(p)p=3⋅2k−12k+1−1σ(p)p=3⋅2k−12k+1−1 Now as any number has atleast two prime factors one and itself hence , σ(p)p>1σ(p)p>1 ⟹3⋅2k−1>2 Which can be showed impossible fork>0k>0.Now let us assume thatxxis in the formx=5t2kqx=5t2kq 2(5t+1−1)(2k+1−1)σ(q)4=3⋅5t⋅2k⋅q2(5t+1−1)(2k+1−1)σ(q)4=3⋅5t⋅2k⋅q (5t+1−1)(2k+1−1)σ(q)=3⋅5t⋅2k+1⋅q(5t+1−1)(2k+1−1)σ(q)=3⋅5t⋅2k+1⋅q σ(q)q=3⋅5t⋅2k+1(5t+1−1)(2k+1−1)σ(q)q=3⋅5t⋅2k+1(5t+1−1)(2k+1−1) Which by proceeding as above ie. by the factσ(q)q>1σ(q)q>1can be shown not possible.Thus no friendly pair of1010is possible. Thus1010is solitary.

Answer»

Letσ(n)σ(n)be the sum of divisor function.For1010to have a friendly pair we have to find another solution to

σ(5x)=9xσ(5x)=9x

6σ(x)=9x(Assuming (5,x)=1)6σ(x)=9x(Assuming (5,x)=1)

2σ(x)=3x2σ(x)=3x

Now Letx=2kpx=2kpwherep∈Np∈N,(p,z)=1(p,z)=1

2(2k+1−1)σ(p)=3⋅2kp2(2k+1−1)σ(p)=3⋅2kp

σ(p)p=3⋅2k−12k+1−1σ(p)p=3⋅2k−12k+1−1

Now as any number has atleast two prime factors one and itself hence ,

σ(p)p>1σ(p)p>1

⟹3⋅2k−1>2

Which can be showed impossible fork>0k>0.Now let us assume thatxxis in the formx=5t2kqx=5t2kq

2(5t+1−1)(2k+1−1)σ(q)4=3⋅5t⋅2k⋅q2(5t+1−1)(2k+1−1)σ(q)4=3⋅5t⋅2k⋅q

(5t+1−1)(2k+1−1)σ(q)=3⋅5t⋅2k+1⋅q(5t+1−1)(2k+1−1)σ(q)=3⋅5t⋅2k+1⋅q

σ(q)q=3⋅5t⋅2k+1(5t+1−1)(2k+1−1)σ(q)q=3⋅5t⋅2k+1(5t+1−1)(2k+1−1)

Which by proceeding as above ie. by the factσ(q)q>1σ(q)q>1can be shown not possible.Thus no friendly pair of1010is possible. Thus1010is solitary.

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