Finding smallest possible integer to satisfy a perfect cube

Question

The sum of four consecutive, positive, odd integers is a perfect cube. What is the smallest possible integer that could be the least of the four?

I tried approaching this as follows:

Let $n$ be an odd integer, then we can represent the four odd integers as $n, n+2, n+4, n+6$.

We want the sum of these integers equal to $k^3$, where $k\in \mathbb{Z^+}$. So we can write this as

$n+n+2+n+4+n+6=k^3$ $\Leftrightarrow$ $4n+12=k^3$, but from here I don't see how I could continue. I could probably divide the expression by $4$ and get $n+3= \frac{k^3}{4}$ and then deduce that $k$ has to be something of the form $4n$? I'm not sure. Any tips would be helpful.

Answer 1

You have that $n=2m+1$ because $n$ is odd and that $ 2m+1+2m+3+2m+5+2m+7 = k^3$ so $ 8m+16 = k^3$ and so $ 8m = k^3-16$ so $ m = \frac{k^3}{8}-2$ let $k = 2 u$ and we have that $m = u^3 -2 $ so we have that $ n=2(u^3-2)+1 = 2 u^3-3$ for $u \in \mathbb{N}$ and so the first few $n$'s are $ n= -1,13,51,125,\cdots$

Answer 2

It is an obscure (but provably true) fact that the odd numbers can be separated into consecutive strings whose length increases by $1$, and the sum of each string will be the cube of the number of members in the string: $1=1^3,\ 3+5=2^3,\ 7+9+11=3^3,\ 13+15+17+19=4^3$ etc. A string of consecutive odd numbers with $n$ members will sum to a cube when the first member of that string is $2k+1$ where $k=\sum_{i=1}^{n-1}i$

The example with four members begins with $2(\sum_{i=1}^3i)+1=2\cdot 6+1=13$

Answer 3

Let the four numbers be:

$((2p^3-3),(2p^3 -1),(2p^3+1),(2p^3+3))$

Adding them we get $sum=(2p)^3$

which is a cube.

Since we need the smallest odd number we have,

$(2p^3-3)>0$

which implies $p=2$ is the smallest.

Hence the number's are:

$(13,15,17,19)$ , & they sum up to $(4)^3$