Sum of five cubes

True or False :

All integers can be written as the sum of cubes of five integers, not necessarily distinct.

True False

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1 solution

Sathvik Acharya
May 16, 2019

It is true that all integers can be expressed as the sum of cubes of five integers, not necessarily distinct.

Note the identity 6 k = ( k + 1 ) 3 + ( k 1 ) 3 k 3 k 3 6k=(k+1)^3+(k-1)^3-k^3-k^3 We use the above identity for k = n 3 n 6 = n ( n 1 ) ( n + 1 ) 6 k=\large{\frac{n^3-n}{6}=\frac{n(n-1)(n+1)}{6}} which is an integer for all n n as the product of 3 consecutive numbers is divisible by 3 ! = 6 3!=6 .

This gives us n 3 n = ( n 3 n 6 + 1 ) 3 + ( n 3 n 6 1 ) 3 ( n 3 n 6 ) 3 ( n 3 n 6 ) 3 n^3-n=\left(\frac{n^3-n}{6}+1\right)^3+\left(\frac{n^3-n}{6}-1\right)^3-\left(\frac{n^3-n}{6}\right)^3-\left(\frac{n^3-n}{6}\right)^3

Therefore, n = ( n ) 3 + ( n n 3 6 1 ) 3 + ( n n 3 6 + 1 ) 3 + ( n 3 n 6 ) 3 + ( n 3 n 6 ) 3 \boxed{n=(-n)^3+\left(\frac{n-n^3}{6}-1\right)^3+\left(\frac{n-n^3}{6}+1\right)^3+\left(\frac{n^3-n}{6}\right)^3+\left(\frac{n^3-n}{6}\right)^3} .

How about using positive cubes only?

X X - 2 years ago

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Well, then there exist several integers which cannot be expressed as the sum of five positive cubes.

Sathvik Acharya - 2 years ago

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I mean how many at least should we use to make sure every positive integer can be expressed as the sum of those positive cubes?

X X - 2 years ago

It doesn't work for n = 2 ) n=2) .

Aditya Sky - 1 year, 10 months ago

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2 = 1 3 + 1 3 + 1 3 + 0 3 + ( 1 ) 3 2=1^3+1^3+1^3+0^3+(-1)^3

Mr. India - 1 year, 10 months ago

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