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QUESTION 1

Part (i)

We have <math>\left( x_{1},y_{1} \right)=\left( x_{1}^{2}-y_{1}^{2}+1,2x_{1}y_{1}+1 \right)</math>

<math>\Rightarrow \left\{ \begin{align}

 & x_{1}=x_{1}^{2}-y_{1}^{2}+1 \\ 
& y_{1}=2x_{1}y_{1}+1 \\ 

\end{align} \right.</math>

From the second of these we have <math>y_{1}=\frac{1}{1-2x_{1}}</math>

Substituting this into the first of the equations we have :-

<math>x_{1}=x_{1}^{2}-\frac{1}{\left( 1-2x_{1} \right)^{2}}+1</math>

<math>\Rightarrow x_{1}\left( 1-2x_{1} \right)^{2}=x_{1}^{2}\left( 1-2x_{1} \right)^{2}-1+\left( 1-2x_{1} \right)^{2}</math>

<math>\Rightarrow x_{1}\left( 1-4x_{1}+4x_{1}^{2} \right)=x_{1}^{2}\left( 1-4x_{1}+4x_{1}^{2} \right)-1+\left( 1-4x_{1}+4x_{1}^{2} \right)</math>

<math>\Rightarrow x_{1}-4x_{1}^{2}+4x_{1}^{3}=x_{1}^{2}-4x_{1}^{3}+4x_{1}^{4}-1+1-4x_{1}+4x_{1}^{2}</math>

<math>\Rightarrow 0=4x_{1}^{4}-8x_{1}^{3}+9x_{1}^{2}-5x_{1}</math>

<math>\Rightarrow 0=x_{1}\left( 4x_{1}^{3}-8x_{1}^{2}+9x_{1}-5 \right)</math>

<math>\Rightarrow 0=x_{1}\left( x_{1}-1 \right)\left( 4x_{1}^{2}-4x_{1}+5 \right)</math>

Now the equation <math>0=4x_{1}^{2}-4x_{1}+5</math>

has no real roots since 						

<math>\left( -4 \right)^{2}-4\times 4\times 5=-64<0</math>

Therefore the only real solutions to the original equations are:-

<math>x_{1}=0\Rightarrow y_{1}=1,x_{1}=1\Rightarrow y_{1}=-1</math>

We must check that these solutions actually work :-

<math>\left( 0^{2}-1^{2}+1,2\times 0\times 1+1 \right)=\left( 0,1 \right)</math>

(as required)

<math>\left( 1-\left( -1 \right)^{2}+1,2\times 1\times \left( -1 \right)+1 \right)=\left( 1,-1 \right)</math>

(again as required)

Thus the required values are :- <math>\left( 0,1 \right)</math>

and 

<math>\left( 1,-1 \right)</math>

Part (ii)

Now suppose <math>\left\{ \begin{align}

 & \left( x_{2},y_{2} \right)=\left( x_{1}^{2}-y_{1}^{2}+a,2x_{1}y_{1}+b+2 \right) \\ 
& \left( x_{1},y_{1} \right)=\left( x_{2}^{2}-y_{2}^{2}+a,2x_{2}y_{2}+b+2 \right) \\ 

\end{align} \right.</math>

given 

<math>\left( x_{1},y_{1} \right)=\left( -1,1 \right)</math>

Then we have <math>\left\{ \begin{align}

 & \left( x_{2},y_{2} \right)=\left( a,b \right) \\ 
& \left( -1,1 \right)=\left( a^{2}-b^{2}+a,2ab+b+2 \right) \\ 

\end{align} \right.</math>

In other words <math>\left\{ \begin{align}

 & -1=a^{2}-b^{2}+a \\ 
& 1=2ab+b+2 \\ 

\end{align} \right.</math>

The second of these implies that <math>b=-\frac{1}{2a+1}</math>

Substituting this into the first of the equations we obtain :-

<math>-1=a^{2}-\frac{1}{\left( 2a+1 \right)^{2}}+a</math>

<math>\Rightarrow -\left( 2a+1 \right)^{2}=a^{2}\left( 2a+1 \right)^{2}-1+a\left( 2a+1 \right)^{2}</math>

<math>\Rightarrow -\left( 4a^{2}+4a+1 \right)=a^{2}\left( 4a^{2}+4a+1 \right)-1+a\left( 4a^{2}+4a+1 \right)</math>

<math>\Rightarrow -\left( 4a^{2}+4a+1 \right)=4a^{4}+4a^{3}+a^{2}-1+4a^{3}+4a^{2}+a</math>

<math>\Rightarrow 0=4a^{4}+8a^{3}+9a^{2}+5a</math>

<math>\Rightarrow 0=a\left( 4a^{3}+8a^{2}+9a+5 \right)</math>

<math>\Rightarrow 0=a\left( a+1 \right)\left( 4a^{2}+4a+5 \right)</math>

Now the equation <math>0=4a^{2}+4a+5</math>

has no real roots since 			

<math>\left( 4 \right)^{2}-4\times 4\times 5=-64<0</math>

Thus the only real solutions to our equation in ‘ <math>a</math> ’ are :-

<math>a=0\Rightarrow b=-1,a=-1\Rightarrow b=1</math>

However it is clear that the second of these actually gives a constant solution while the first gives us:-

<math>\left( x_{1},y_{1} \right)=\left( -1,1 \right),\left( x_{2},y_{2} \right)=\left( 0,-1 \right),\left( x_{3},y_{3} \right)=\left( 0^{2}-\left( -1 \right)^{2}+0,2\times 0\times \left( -1 \right)+-1+2 \right)=\left( -1,1 \right)</math>

Therefore the required solution is <math>a=0,b=-1</math>