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secondary 3 | A Maths
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Date Posted: 3 years ago
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From your later step :
N = 2ᵖ(2ᵖ - 2)
N = 2ᵖ(2(2ᵖ⁻¹ - 1))
N = 2ᵖ⁺¹(2ᵖ⁻¹ - 1)
Since p is an integer and p > 1 , then p ≥ 2. And therefore p - 1 ≥ 1
This means that 2ᵖ⁻¹ ≥ 2¹ and therefore 2ᵖ⁻¹ is a positive integer ≥ 2
So 2ᵖ-¹ - 1 ≥ 1 and therefore it is a positive integer as well.
And 2ᵖ⁺¹ ≥ 2³ since p + 1 ≥ 3 . So 2ᵖ⁺¹ is a positive integer ≥ 8
So both 2ᵖ⁺¹ and 2ᵖ⁻¹ - 1 are positive integers and as N is a product of them then 2ᵖ⁺¹ is factor of N since it divides N to give you a positive integer 2ᵖ⁻¹ - 1
N = 2ᵖ(2ᵖ - 2)
N = 2ᵖ(2(2ᵖ⁻¹ - 1))
N = 2ᵖ⁺¹(2ᵖ⁻¹ - 1)
Since p is an integer and p > 1 , then p ≥ 2. And therefore p - 1 ≥ 1
This means that 2ᵖ⁻¹ ≥ 2¹ and therefore 2ᵖ⁻¹ is a positive integer ≥ 2
So 2ᵖ-¹ - 1 ≥ 1 and therefore it is a positive integer as well.
And 2ᵖ⁺¹ ≥ 2³ since p + 1 ≥ 3 . So 2ᵖ⁺¹ is a positive integer ≥ 8
So both 2ᵖ⁺¹ and 2ᵖ⁻¹ - 1 are positive integers and as N is a product of them then 2ᵖ⁺¹ is factor of N since it divides N to give you a positive integer 2ᵖ⁻¹ - 1
Thanks so much
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