# Algebraic proofs

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#1

A) In a list of 3 consecutive positive integers at least one of the numbers is even and one of the numbers is a multiple of 3
n is a positive integer greater than 1

Prove that n^3 -n is a multiple of 6 for all positive values of n

B) 2^61 -1 is a prime number

Explain why 2^61 +1 is a multiple of 3
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1 month ago
#2
A) Try factorising n^3 - n, then use the information given in A)

B) Consider the 3 consecutive positive integers 2^61 - 1, 2^61, 2^61 + 1 then use the information given in A) and B)
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#3
Thank you I will try that
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#4
(Original post by 4D Chess)
A) Try factorising n^3 - n, then use the information given in A)

B) Consider the 3 consecutive positive integers 2^61 - 1, 2^61, 2^61 + 1 then use the information given in A) and B)
I have factorised it into n(n^2-1) but I do not understand what to do next
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1 month ago
#5
(Original post by Acallanan)
I have factorised it into n(n^2-1) but I do not understand what to do next
This can be factorised further to n(n-1)(n+1), if you were to think about the difference of two squares. You can rewrite this as (n-1)n(n+1) and now notice how that is the product of 3 consecutive integers that aren't negative, which is where you can now use the information in A)
Last edited by 4D Chess; 1 month ago
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#6
(Original post by 4D Chess)
This can be factorised further to n(n-1)(n+1), if you were to think about the difference of two squares. You can rewrite this as (n-1)n(n+1) and now notice how that is the product of 3 consecutive integers that aren't negative, which is where you can now use the information in A)
Thank you, how would I implement the information
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1 month ago
#7
Just think about it. You have (n-1)n(n+1) and the question tells you that at least one of these numbers is a multiple of 2, and one must be a multiple of 3. However these numbers are being multiplied together so...?
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#8
(Original post by 4D Chess)
Just think about it. You have (n-1)n(n+1) and the question tells you that at least one of these numbers is a multiple of 2, and one must be a multiple of 3. However these numbers are being multiplied together so...?
So because one is a multiple of two and the other a multiple of three, it multiplies to become 6 so any number will be a multiple of 6
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