ACCCGS8 wrote: The difference between the cubes of two consecutive positive integers is equal to $n^2$ for a positive integer $n$. Show that $n$ is the sum of two squares. Solution: We have: $(n+1)^3-n^3=3n^2+3n+1=k^2\Leftrightarrow (2k)^2-3(2n+1)^2=1$ (*) This is Pell type equation $u^2-3v^2=1$, we have sequence of its roots: \[ \begin{matrix} u_0=1, u_1=2, u_{n+2}=4u_{n+1}-u_n \\ v_0=0, v_1=1, v_{n+2}=4v_{n+1}-v_n \end{matrix} ; n=0,1,2,... \] We consider $u_i,v_i$ satisfy $u_i$ is even and $v_i$ is odd, then $i=2z+1$. We have the roots of equation (*) is: \[ \begin{matrix} u'_0=2, u'_1=26, u'_{n+2}=14u'_{n+1}-u'_n \\ v'_0=1, v'_1=15, v'_{n+2}=14v'_{n+1}-v'_n \end{matrix} ; n=0,1,2,... \] The general form of $(u'_i)$ in this sequence is \[ u'_i=(\frac{2+\sqrt{3}}{2})(7+4\sqrt{3})^n+(\frac{2-\sqrt{3}}{2})(7-4\sqrt{3})^n \] Consider sequence $(t_n)$ satisfy \[ t_0=1, t_1=5, t_{n+2}=4t_{n+1}-t_n; n=0,1,2,... \], $t_i$ is odd. The general form of \[ t_n=(\frac{1+\sqrt{3}}{2})(2+\sqrt{3})^n-(\frac{\sqrt{3}-1}{2})(2-\sqrt{3})^n \] Then \[ u'_n=t_n^2+1\Rightarrow 2k_n=t_n^2+1\Rightarrow k_n=(\frac{t_n-1}{2})^2+(\frac{t_n+1}{2})^2 \]. The proof is end.