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Calculate the sum of all perfect numbers less than 1 0 1 5 0 0 such that it can be written as the sum of two positive integral cubes, ie P = n 1 3 + n 2 3 ; n 1 , n 2 ∈ Z +
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The answer is 28.
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2 solutions
How can you say there are no more such perfect numbers?
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There are more perfect numbers like 6,496,8128 and so on but none of them satisfy the condition of sum of two cubes except 28.
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How can u say other numbers generated will not follow the above rule?
Wikipedia states that
Each even perfect number except for 6 is a centered nonagonal number and is equal to the sum of the first 2 2 p − 1 number of odd cubes.
To get sum of two cubes using the following formula, p needs to be 3 and no other value other than 3 in the expression will satisfy the requirement . So now that we know that p is 3 and it is the only solution, we can evaluate the perfect number corresponding to p using this formula -
2 p − 1 ( 2 p − 1 )
So the answer is 28.
There is one property of perfect numbers,that every perfect number may be written as a cube of consecutive odd natural numbers, For e.g 28=sum of cubes of(1,3) 496=sum of cubes of(1,3,5,7) So all the next perfect numbers will definitely be the sum of cubes but the number of cubes required to obtain that perfect number will be greater than 2,so only one perfect number(i.e.28) will be the sum of cubes of TWO positive integers.but thats all from a property,in case if you have any solution pls do post it :)
28 = 2 x 2 x 7.
Divisors of 28 = 1,2,4,7,14 (and 28). = 1 + 2 + 4 + 14 = 28.
So, 28 is a perfect number and it is sum of 2 cubes (1 and 27).
So, answer is 28.