Find the smallest natural number $n$ for which there exists a natural number $x$ such that
$$(x+1)^3 + (x + 2)^3 + (x + 3)^3 + (x + 4)^3 = (x + n)^3.$$
$(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3=[(x+1)^3+(x+4)^3]+[(x+2)^3+(x+3)^3]=2*(2x+5)(x^2+5x+10)$
$\Longrightarrow x+n$ should be even
First case $x=2k, n=2a$
$2*(2x+5)(x^2+5x+10)=(x+n)^3$ becomes
$2(4k+5)(4k^2+10k+10)=(2k+2a)\Longrightarrow 2^2(4k+7)(2k^2+5k+5)=2^3(k+a)$
$\Longrightarrow k=2t+1\Longrightarrow$ Conclusion: $x$ exhibits the form $x=4t+2$
$t=0\rightarrow x=2\rightarrow 2*(2x+5)(x^2+5x+10)=2*9*24=432~~$ (No cube)
$t=1\rightarrow x=6\rightarrow 2*(2x+5)(x^2+5x+10)=2*17*82=2788~~~ $ (No cube)
$t=2\rightarrow x=10\rightarrow 2*(2x+5)(x^2+5x+10)=2*25*160=8000 =20^3\rightarrow (x,n)=(10,10)$
Considering the second case $x=2k+1, n=2a+1$, one arrives similarly at $x=4t+1$
However no cubes turn up for $x=1,5,9$