STEP2-1998 Section A, Q1

This is the first part of question 1 on the exam paper. I am ignoring the second part as its solution is virtually identical to this.

Show that if $n$ is an integer, such that

$$(n-3)^3+n^3=(n+3)^3,\ \ \ \ \ \ \ (*)$$

then $n$ is even and that $n^2$ is a factor of $54$. Deduce that there is no integer $n$ which satisfies equation $(*)$

We begin by expanding as simplifying:

$$(n-3)=n^3-9n^2+27n-27$$

$$(n+3)=n^3+9n^2+27n+27$$ 

So $(*)$ becomes:

$$n^3-18n^2-54=0\ \ \ \ \ \ \ (1)$$

 We can rewrite $(1)$ as: $n^2(n-18)=54$, which as we are dealing with integers implies that $n^2$ divides $54$ ( and that $n$ is positive and $>18$. Also as the RHS of this rewrite is even then the LHS must also be even which implies that $n$ is even.

There is no $n>18$ such that $n^2 \mid 54$ so there is no integer $n$ which satisfies $(*)$

The second part of this question was:

Show that, if $n$is an integer such that

$$(n-6)^3+n^3=(n+6)^3,\ \ \ \ \ \ \ (**)$$

then $n$ is even. Deduce that there is no integer $n$ which satisfies $(**)$.

This when expanded and simplified becomes:  $n^3-36n^2-432=0$. Which implies $n$ is even and that $n>36$ and that $n^2 \mid 432$, but all $n>36$ have squares greater thatn $432$, hence there is no integer which satisfies $(**)$