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
\(6^4 = 8^3 + 28^2\)
Let
\[
T _n =
\left( \sqrt{a+1} + \sqrt a\right)^n\,,
\]
where \(n\) is a positive integer and \(a\) is any given positive integer.
In the case when \(n\) is even, show
by induction that
\(T_n\) can be written in the form
\[
A_n +B_n \sqrt{a(a+1)}\,,
\]
where
\(A_n\) and \(B_n\) are integers (depending on \(a\) and \(n\))
and \(A_n^2 =a(a+1)B_n^2 +1\).
In the case when \(n\) is odd, show by considering
\((\sqrt{a+1} +\sqrt a)T_m\) where \(m\) is even, or otherwise,
that \(T_n\)
can be written in the form
\[
C_n \sqrt {a+1} + D_n \sqrt a \,,
\]
where \(C_n\) and \(D_n\) are integers (depending on \(a\) and \(n\)) and
\( (a+1)C_n^2 = a D_n^2 +1\,\).
Deduce that, for each \(n\), \(T_n\) can be written
as the sum of the square roots of two consecutive integers.
Solution:
Claim: For all \(n \geq 1\) \(T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}\) where \(A_{2n}, B_{2n}\) are integers and \(A_{2n}^2 = a(a+1)B_{2n}^2+1\)
Proof: (By induction)
Base case: \(n =1\).
\begin{align*}
&& T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\
&&&= a+1+2\sqrt{a(a+1)}+a \\
&&&=2a+1+2\sqrt{a(a+1)} \\
\Rightarrow && A_2 &= 2a+1 \\
&& B_2 &= 2 \\
&& A_2^2 &= 4a^2+4a+1 \\
&& a(a+1)B_1^2 + 1 &= 4a^2+4a+1
\end{align*}
Therefore our base case is true. Suppose it is true for some \(n = k\) then consider \(n = k+1\) we must have \(T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}\)
\begin{align*}
&& T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\
&&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\
&&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\
\Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\
&& B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\
&& A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\
&&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\
&&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\
&&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\
&& a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\
&&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\
&&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\
&&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1
\end{align*}
So our relation holds ad therefore by induction we're done.
For even \(n\) \(T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}\)
For odd \(n\), \(T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}\) therefore it is always the sum of the square root of two consecutive integers.
Show that, if \(\l a \, , b\r\) is any point on the curve \(x^2 - 2y^2 = 1\), then \(\l 3a + 4b \, , 2a + 3b \r\,\) also lies on the curve.
Determine the smallest positive integers \(M\) and \(N\) such that, if \(\l a \,, b\r\) is any point on the curve \(Mx^2 - Ny^2 = 1\), then \((5a+6b\,, 4a+5b)\) also lies on the curve.
Given that the point \(\l a \, , b\r\) lies on
the curve \(x^2 - 3y^2 = 1\,\), find positive integers \(P\), \(Q\), \(R\) and \(S\) such that the point \((P a +Q b\,, R a + Sb)\) also lies on the curve.
Solution:
Suppose \(a^2-2b^2=1\) then
\begin{align*}
(3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\
&=a^2-2b^2 \\
&= 1
\end{align*}
Therefore \((3a+4b,2a+3b)\) also lies on the curve.
Suppose \(Ma^2-Nb^2 = 1\) then
\begin{align*}
M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\
&= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2
\end{align*}
Therefore we need \(3M = 2N\) so the smallest possible value would have to be \(M = 2, N = 3\), which does work
Consider \(x + \sqrt{3}y\), then consider \((x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}\). Notice that \((x+\sqrt{3}y)(x-\sqrt{3}y) = 1\) and \((2+\sqrt{3})(2-\sqrt{3}) = 1\) so \(((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1\), so we can take \(P=2,Q=3,R=1,S=2\)