For $n=4$, sketch a diagram clearly showing your answer to part (b)(ii).

Prove that the sum of these three integers is always divisible by $3$.

Prove that the sum of the squares of these three integers is never divisible by $3$.

Sketch the graph of $x^2-y^2=1$, stating the coordinates of any axis intercepts and the equation of each asymptote.

Hence prove that the square of any integer can be written in the form $4t$ or $4t+1$, where $t\in {\mathbb{Z}}^{+}$.

Show that ${\left(2n-1\right)}^2+{\left(2n+1\right)}^2=8n^2+2$, where $n\in \mathbb{Z}$.

Show that $P_3\left(n\right)+P_3\left(n+1\right)\equiv {\left(n+1\right)}^2$.

State, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.

Show that $8P_3\left(n\right)+1$ is the square of an odd number for all $n\in {\mathrm{\mathbb{Z}}}^{+}$.

Hence show that ${\left(\alpha -\beta \right)}^2+{\left(\beta -\gamma \right)}^2+{\left(\gamma -\alpha \right)}^2=2p^2-6q$.

By using a suitable substitution for $a$, show that $1-3 \cos 2x+3 {\cos }^2 2x- {\cos }^3 2x=8 {\sin }^6 x$.

Hence, or otherwise, prove that the sum of the squares of any two consecutive odd integers is even.

Express $g\left(x\right)$ in the form $A+\frac{B}{x-2}$ where A, B are constants.

Prove that the sum of the first $n$ terms of this arithmetic sequence is a square number.

Show that $u_1=4$.

Show that ${\left(f\left(t\right)\right)}^2+{\left(g\left(t\right)\right)}^2=f\left(2t\right)$.

$f\left(\text{i}u\right)$.

The hyperbola with equation $x^2-y^2=1$ can be rotated to coincide with the curve defined by $xy=k, k\in \mathrm{ℝ}$.

Find the possible values of $k$.

For triangular numbers, verify that $P_3\left(n\right)=\frac{n\left(n+1\right)}{2}$.

Show that $p^2-2q={\alpha }^2+{\beta }^2+{\gamma }^2$.

By expanding $\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right)$ show that:

$p=-\left(\alpha +\beta +\gamma \right)$

$q=\alpha \beta +\beta \gamma +\gamma \alpha$

$r=-\alpha \beta \gamma$.

Show that $2x-3-\frac{6}{x-1}=\frac{2x^2-5x-3}{x-1}, x\in \mathrm{ℝ}, x\ne 1$.

Hence or otherwise, solve the equation  $2 \sin 2\theta -3-\frac{6}{\sin 2\theta -1}=0$  for  $0\le \theta \le \mathrm{\pi }, \theta \ne \frac{\mathrm{\pi }}{4}$.

The equation $2x^2-5x+1=0$ has two real roots, $\alpha$ and $\beta$.

Consider the equation $x^2+mx+n=0$, where $m, n\in \mathrm{\mathbb{Z}}$ and which has roots $\frac{1}{{\alpha }^3}$ and $\frac{1}{{\beta }^3}$.
Without solving $2x^2-5x+1=0$, determine the values of $m$ and $n$.

Verify that $u=f\left(t\right)$ satisfies the differential equation $\frac{d^{2}u}{d{t}^2}=u$.

Show that ${\left(f\left(t\right)\right)}^2-{\left(g\left(t\right)\right)}^2={\left(f\left(\text{i}u\right)\right)}^2-{\left(g\left(\text{i}u\right)\right)}^2$.

By substituting $Y=\frac{dy}{dt}$, show that $Y=B{\text{e}}^{2t}$ where $B$ is a constant.

Let the two values found in part (c)(ii) be ${\lambda }_1$ and ${\lambda }_2$.

Verify that $y=F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

Prove the identity ${\left(p+q\right)}^3-3pq\left(p+q\right)\equiv p^3+q^3$.

Show that $N\left(\alpha \beta \right)=N\left(\alpha \right)N\left(\beta \right)$.

Consider two consecutive positive integers, $n$ and $n+1$.

Show that the difference of their squares is equal to the sum of the two integers.

$g\left(\text{i}u\right)$.

Hence find, and simplify, an expression for ${\left(f\left(\text{i}u\right)\right)}^2+{\left(g\left(\text{i}u\right)\right)}^2$.

By solving the differential equation $\frac{dy}{dt}=y$, show that $y=A{\text{e}}^t$ where $A$ is a constant.

Show that $\frac{dx}{dt}-x=-A{\text{e}}^t$.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$.

By differentiating $\frac{dy}{dt}=-x+y$ with respect to $t$, show that $\frac{d^{2}y}{d{t}^2}=2\frac{{\displaystyle dy}}{{\displaystyle dt}}$.

Hence find $y$ as a function of $t$.

Hence show that $x=-\frac{B}{2}{\text{e}}^{2t}+C$, where $C$ is a constant.

Show that $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.

Find the two values for $\lambda$ that satisfy $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.