<s>[INST] 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.
 [/INST] 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
$$
\begin{align}
 x &= n(n+1)/2\\
y &= n(n-1)/2
\end{align}
$$
</s>