Given that
\(x^2 - y^2 = \left( x - y \right)^3\)
and that \(x-y = d\) (where \(d \neq 0\)),
express each of \(x\) and \(y\) in terms of \(d\).
Hence find a pair of integers \(m\) and \(n\)
satisfying \(m-n = \left( \sqrt {m} - \sqrt{n} \right)^3\)
where \(m > n > 100\).
Given that \(x^3 - y^3 = \left( x - y \right)^4\)
and that \(x-y = d\) (where \(d \neq 0\)),
show that \(3xy = d^3 - d^2\). Hence show that
\[
2x = d \pm d \sqrt {\frac{4d-1 }{3}}
\]
and determine a pair of distinct positive integers \(m\) and \(n\)
such that \(m^3 - n^3 = \left( m - n \right)^4\).
Solution:
\(\,\) \begin{align*}
&& x^2-y^2 &=(x-y)^3 \\
\Rightarrow && x+y &=d^2 \\
&& x-y &= d \\
\Rightarrow && x &= \tfrac12(d^2+d) \\
&& y &= \tfrac12(d^2-d)
\end{align*}
Therefore consider \(x^2 = m, y^2 = n\), so \(m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2\) so we want \(d^2-d > 20\), so \(d = 6, n = 225, m = 441\).
\(\,\)
\begin{align*}
&& x^3-y^3 &= (x-y)^4 \\
\Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\
&& d^3 &= (x-y)^2+3xy \\
&& d^3 &= d^2 + 3xy \\
\Rightarrow && 3xy &= d^3 - d^2 \\
\Rightarrow && 3x(x-d) &= d^3-d^2 \\
\Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\
\Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\
&&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\
&&&= d \pm d \sqrt{\frac{4d-1}{3}}
\end{align*}
Therefore we need \(\frac{4d-1}{3}\) to be an odd square. \(y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}\). Since we want positive values, we should take the positive square roots.
\(d = \frac{3 \cdot 3^2 + 1}{4} = 7\) we have \(2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7\)