Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$ Problem: Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$
My attempt:
If $n^3 = x^2-y^2$, then $x = y \mod(n)$, and at the same time $x = -y \mod(n)$, 
So that $2x = 0 \mod(n)$. I could then split the problem into two cases:
If $2^{-1}$ exists, then $x=kn$ for some integer $k$, and so the equation reduces to: $n^2(n-k^2)=-y^2$, so that $n<k^2$. This is about how far I have gotten, not sure how to proceed. Hints appreciated.
If $2^{-1}$ does not exist...
Edit: I made a mistake above, but now that the answer is given I will just leave it up for now.
 A: Factor $x^2-y^2=(x-y)(x+y)$.  Particularly in contest math that should be your first thought on seeing a difference of squares.
If $n$ is odd, so is $n^3$ and you can write $n^3=2k+1$.  Then $x=k+1, y=k$ works based on $2k+1=1\cdot (2k+1)$.  
If $n$ is even you can write $n^3=8k$, note $8k=2\cdot (4k)$ and solve $x-y=2,x+y=4k$ to get $x=2k+1,y=2k-1$ 
This approach does not depend on the fact we are trying to express is a cube.  The first works for all odd numbers, the second for all multiples of $4$.  There is no solution for a number of the form $4k+2$
A: Insipred by Ross Millikan: Factoring, we get $n^3 = (x-y)(x+y)$
Claim: We can always find integers $x, y$ such that $x-y = n$ and $x+y = n^2$. 
Proof: If we eliminate $y$ and solve for $x$, we get $x = \dfrac {n^2+n}{2}$. By considering even/odd cases you can see that $x$ is always an integer.
Next from $x-y = n$ we get $y = \dfrac {n^2+n}{2}-n$
A: Let $n$ be an integer.

Solving the system
$$
\begin{cases}
x+y=n^2\\[4pt]
x-y=n\\
\end{cases}
$$
yields the solution
$$(x,y)=\left(\frac{n^2+n}{2},\frac{n^2-n}{2}\right)$$
where the integrality of $x,y$ is guaranteed since $n$ and $n^2$ are either both even or both odd.
A: One of my favorite observations concerns the sequence of odd numbers, which for any natural number $n$ contains some series of $n$ consecutive odd numbers which sum to any $k^{th}$ power of $n$. Therefore it can be shown the any $k^{th}$ power of $n$ (not just cubes) is the difference of two squares, viz: $$\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}} (2i-1)=n^k=\sum_{i=1}^{\frac{n^{k-1}+n}{2}} (2i-1) -\sum_{i=1}^{\frac{n^{k-1}-n}{2}} (2i-1)=\bigl(\frac{n^{k-1}+n}{2}\bigr)^2-\bigl(\frac{n^{k-1}-n}{2}\bigr)^2$$For $k=3$, the resulting squares are just those in the answer given by quasi.
A: One way to systematically arrive at an answer is to assume that
$$
\begin{align}
x &= a_2 n^2 + a_1 n + a_0\\
y &= b_2 n^2 + b_1 n + b_0
\end{align}
$$
We may assume $a_2,b_2>0$, changing signs if necessary. Then comparing coefficient of $n^4$ in 
$$
n^3 = x^2 - y^2
$$
forces
$$
a_2 = b_2
$$
Next, comparing coefficient of constant forces $a_0=\pm b_0$. Having $a_0 = b_0$ would force $a_1 = b_1$ when comparing coefficient of $n^1$. This in turn forces coefficient of $n^3$ in $x^2 - y^2$ to be zero which is not possible, so $a_0 = -b_0$.  
Now comparing coefficient of $n^1$ gives $b_1 = - a_1$. Comparing coefficients $n^3$ and $n^2$, we arrive at
$$
4a_2a_1 = 1, \quad 4a_2a_0 = 0
$$
Hence $a_0 = 0$. This leads to
$$
\begin{align}
x &= a_2n^2 + n/(4a_2)\\
y &= a_2n^2 - n/(4a_2)
\end{align}
$$
Only $a_2 = 1/2$ gives integral values of $x,y$ so we get
$$
\begin{align}
 x &= n(n+1)/2\\
y &= n(n-1)/2
\end{align}
$$
