# Finding smallest possible integer to satisfy a perfect cube

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.

• Trial and error works very quickly. Failing that, you might notice that all those sums are even so the first possible cube is $8$ (which does not work) and the second is $64$ (which does). – lulu May 2 '20 at 11:37

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 $$n= -1,13,51,125,\cdots$$

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 $$2(\sum_{i=1}^3i)+1=2\cdot 6+1=13$$

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 $$(4)^3$$