Can we express sum of cubes in terms of squares That is possible?, can you show me some theorem and who worked on these.
If we have the sum of n cubes, can we express that like the sum of m squares?
Thanks!
 A: Remark: after reading @fleablood's comment, I realized that I am not answering the OP's question. I haven't decided yet if I'll keep this answer or not.
Something like
$$\sum_{k=1}^nk^3=\left(\sum_{k=1}^nk\right)^2=\left(\frac{n(n+1)}{2}\right)^2\quad ?$$
A: Note that
$$
74 = −284650292555885^3 + 66229832190556^3 + 283450105697727^3
$$
is also a sum of three squares:
$$
74=8^2+3^2+1^2.
$$
In general, it is known which numbers can be written as a sum of $m$ squares. In fact, for $m\ge 4$, all positive integers. The case $m=2$ is due to Fermat, and $m=3$ to Gauss. Also, considering Waring's problem, we know much about the sum of cubes. But it is more difficult. It is still open which numbers are the sum of, say, three cubes - see here.
A: +1 to @valfar's question-you are correct my friend,as I have found through calculation that not only the numbers -23064014558266183195584063166306441237779125 ,290509918128065397278363068942306635399616 and 22773504640138117798305700097364134602379583,sum up to 74,but also 64,1 & 9 are the numbers that sum up to 74
