Review question

# Can we find a solution to $(n-3)^3+n^3=(n+3)^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9919

## Solution

Show that, if $n$ is an integer such that $\begin{equation*} (n-3)^3+n^3=(n+3)^3,\label{eq:1sol}\tag{1} \end{equation*}$

then $n$ is even and $n^2$ is a factor of $54$. Deduce that there is no integer $n$ which satisfies the equation $\eqref{eq:1sol}$.

Expand the brackets to obtain $(n+3)^3=n^3+9n^2+27n+27,$ and $(n-3)^3=n^3-9n^2+27n-27.$

Then the equation simplifies to $n^3-18n^2-54=0.$

Now $-18n^2-54$ must be even, so $n^3$ must be even, and so $n$ must be even.

Alternatively we could argue as follows:

if $n$ is odd then $n-3$ and $n+3$ are both even, so $(n-3)^3$ and $(n+3)^3$ are both even.

But then $n^3$ must be even, from our equation, and $n$ must be even, which is a contradiction.

Therefore $n$ cannot be odd, and must be even.

From the rearrangement $n^2(n-18)=54$ we can see immediately, because $n-18$ is an integer, that $n^2$ is a factor of $54$.

If there is an integer $n$ which satisfies the equation then we know it must be even and $n^2$ must be a factor of $54$.

The prime factorisation of $54$ is $54=2\times3^3$ and so the only square factors of $54$ are $1^2$ and $3^2$.

But $1$ and $3$ are not even, so there is no $n$ that satisfies $n^2(n-18)=54$.

Or we could say, the factors of $54$ are $1$, $2$, $3$, $6$, $9$, $18$, $29$ and $54$, and none of these are even squares.

Show that, if $n$ is an integer such that $\begin{equation*} (n-6)^3+n^3=(n+6)^3,\label{eq:2sol}\tag{2} \end{equation*}$

then $n$ is even. Deduce that there is no integer $n$ which satisfies the equation $\eqref{eq:2sol}$.

It is easy to see why $n$ is even here, arguing as for the first part.

So let $n = 2k$, where $k$ is an integer.

So the equation becomes $(2k-6)^3+(2k)^3=(2k+6)^3$, and we can divide by $2^3$ to give $(k-3)^3+k^3=(k+3)^3$.

We know from the first part that this has no integer solutions.