Perfect Number Frenzy 2

Level 2

Given that 2 n − 1 ( 2 n − 1 ) 2^{n-1}(2^n-1) is a perfect number when ( 2 n − 1 ) (2^n-1) is prime. Find the total of ρ α \rho_{\alpha} for ρ α n × ρ β \rho^n_{\alpha}\times\rho_{\beta} if ρ α ≥ 3 \rho_{\alpha} \geq 3 , n ≥ 2 n \geq2

Unable to calculate Infinity 0

Tan Peng by Tan Peng , 19, Malaysia

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1 solution

Tan Peng
Aug 18, 2018

Since ρ α n × ρ β \rho^n_{\alpha}\times\rho_{\beta} is a perfect number ,so the sum of all its factors is equal to the number. The sum of all factors ρ α n × ρ β \rho^n_{\alpha}\times\rho_{\beta} is ρ α n + 1 − 1 ρ α − 1 \frac{\rho^{n+1}_{\alpha}-1}{\rho_{\alpha}-1} + ρ α n − 1 ρ α − 1 \frac{\rho^{n}_{\alpha}-1}{\rho_{\alpha}-1} × ρ β \times \rho_{\beta}
Therefore ρ α n + 1 − 1 ρ α − 1 \frac{\rho^{n+1}_{\alpha}-1}{\rho_{\alpha}-1} + ρ α n − 1 ρ α − 1 \frac{\rho^{n}_{\alpha}-1}{\rho_{\alpha}-1} × ρ β \times \rho_{\beta} = ρ α n × ρ β \rho^n_{\alpha}\times\rho_{\beta} which can be simplified to ρ β \rho_{\beta} = 1 + − 2 ρ α n + 1 ρ α n + 1 − 2 ρ α n + 1 1+\frac{ -2\rho^n_{\alpha}+1}{\rho^{n+1}_{\alpha}-2\rho^n_{\alpha}+1} Since − 2 ρ α n + 1 ρ α n + 1 − 2 ρ α n + 1 \frac{ -2\rho^n_{\alpha}+1}{\rho^{n+1}_{\alpha}-2\rho^n_{\alpha}+1} will only be a integer when ρ α = 2 \rho_{\alpha}=2 therefore there is no numbers to add.

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