1^{3}=1

1^{3}+2^{3}=9 = 3^{2 }

1^{3}+2^{3}+3^{3}=36 = 6^{2}

1^{3}+2^{3}+3^{3}+4^{3}=100 = 10^{2}

1^{3}+2^{3}+3^{3}+4^{3}+5^{3}=225 = 15^{2}

…. **1**,2,**3**,4,5,**6**,7,8,9,**10**,11,12,13,14,**15**,16….

These are triangular numbers are in bold:

- 1+2 =3
- 1+2+3=6
- 1+2+3+4=10

( n(n+1)/2 )^{2}= n^{2}(n+1)^{2}/4

A question I have in my mind is that Fermat’s last theorem states: If an integer *n* is greater than 2, then the equation *a*^{n} + *b*^{n} = *c*^{n} has no solutions in non-zero integers *a*, *b*, and *c*.

But how about a^{3} + b^{3 }+ c^{3 }= d^{3 } … Are there any integer solutions to this? I ask this because geometrically speaking volume is 1 degree of freedom more than area.

3^{3}+4^{3}+5^{3}=6^{3 }…. = 15^{2 }– 3^{2}

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