Write down a solution of the equation
\[
x^2-2y^2 =1\,,
\tag{\(*\)}
\]
for which \(x\) and \(y\) are non-negative integers.
Show that, if \(x=p\), \(y=q\) is a solution of (\(*\)), then so also is \(x=3p+4q\), \(y=2p+3q\). Hence find two solutions of \((*)\) for which \(x\) is a positive odd integer and \(y\) is a positive even integer.
Show that, if \(x\) is an odd integer and \(y\) is an even integer, \((*)\) can be written in the form
\[
n^2 = \tfrac12 m(m+1)\,,
\]
where \(m\) and \(n\) are integers.
The positive integers \(a\), \(b\) and \(c\) satisfy
\[
b^3=c^4-a^2\,,
\]
where \(b\) is a prime number. Express \(a\) and \(c^2\) in terms of \(b\) in the two cases that arise.
Find a solution of \(a^2+b^3=c^4\), where \(a\), \(b\) and \(c\) are positive integers but \(b\) is not prime.
Solution:
\((x,y) = (1,0)\) we have
Suppose \(p^2-2q^2 = 1\), then
\begin{align*}
&& (3p+4q)^2-2\cdot(2p+3q)^2 &= 9p^2+24pq + 16q^2 - 2\cdot(4p^2+12pq+9q^2) \\
&&&= p^2(9-8) + pq(24-24) + q^2(16-18) \\
&&&= p^2 - 2q^2 = 1
\end{align*}
So we have:
\begin{array}{c|c}
x & y \\ \hline
1 & 0 \\
3 & 2 \\
17 & 12 \\
\end{array}
Suppose \(b^3 = c^4 - a^2 =(c^2-a)(c^2+a)\), since \(b\) is prime and \(c^2 + a > c^2-a\) we must have:
\begin{align*}
&& p = c^2-a && p^2 =c^2 +a \\
\Rightarrow && c^2 = \frac{p+p^2}{2} && a = \frac{p^2-p}2\\
&& 1 = c^2-a && p^3 = c^2+a \\
\Rightarrow && c^2 = \frac{p^3+1}{2} && a = \frac{p^3-1}{2}
\end{align*}
Note that \(c^2 = \frac{p(p+1)}{2}\) is reminicent of our first equation, so suppose \(n = c = 6\) and \(p = m = 8\) then
\(6^4 = 8^3 + 28^2\)