Question
This question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree 3 and 4.
The cubic equation $x^3+p x^2+q x+r=0$, where $p, q, r \in \mathbb{R}$, has roots $\alpha, \beta$ and $\gamma$.
Consider the equation $x^3-7 x^2+q x+1=0$, where $q \in \mathbb{R}$.
Noah believes that if $p^2 \geq 3 q$ then $\alpha, \beta$ and $\gamma$ are all real.
Now consider polynomial equations of degree 4 .
The equation $x^4+p x^3+q x^2+r x+s=0$, where $p, q, r, s \in \mathbb{R}$, has roots $\alpha, \beta, \gamma$ and $\delta$.
In a similar way to the cubic equation, it can be shown that:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma+\delta) \\
& q=\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta \\
& r=-(\alpha \beta \gamma+\alpha \beta \delta+\alpha \gamma \delta+\beta \gamma \delta) \\
& s=\alpha \beta \gamma \delta
\end{aligned}
$$
The equation $x^4-9 x^3+24 x^2+22 x-12=0$, has one integer root.
a. By expanding $(x-\alpha)(x-\beta)(x-\gamma)$ show that:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma) \\
& q=\alpha \beta+\beta \gamma+\gamma \alpha \\
& r=-\alpha \beta \gamma
\end{aligned}
$$
b.i.Show that $p^2-2 q=\alpha^2+\beta^2+\gamma^2$.
b.iiHence show that $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2=2 p^2-6 q$.
c. Given that $p^2<3 q$, deduce that $\alpha, \beta$ and $\gamma$ cannot all be real.
d. Using the result from part (c), show that when $q=17$, this equation has at least one complex root.
e.i. By varying the value of $q$ in the equation $x^3-7 x^2+q x+1=0$, determine the smallest positive integer value of $q$ required to show that Noah is incorrect.
e.ii.Explain why the equation will have at least one real root for all values of $q$.
f.i. Find an expression for $\alpha^2+\beta^2+\gamma^2+\delta^2$ in terms of $p$ and $q$.
f.ii. Hence state a condition in terms of $p$ and $q$ that would imply $x^4+p x^3+q x^2+r x+s=0$ has at least one complex root.
g. Use your result from part (f)(ii) to show that the equation $x^4-2 x^3+3 x^2-4 x+5=0$ has at least one complex root.
h.i. State what the result in part (f)(ii) tells us when considering this equation $x^4-9 x^3+24 x^2+22 x-12=0$.
h.ii.Write down the integer root of this equation.
h.iiiBy writing $x^4-9 x^3+24 x^2+22 x-12$ as a product of one linear and one cubic factor, prove that the equation has at least one complex root.
▶️Answer/Explanation
a. attempt to expand $(x-\alpha)(x-\beta)(x-\gamma) \quad$ M1
$$
\begin{aligned}
& =\left(x^2-(\alpha+\beta) x+\alpha \beta\right)(x-\gamma) \text { OR }=(x-\alpha)\left(x^2-(\beta+\gamma) x+\beta \gamma\right) \\
& \left(x^3+p x^2+q x+r\right)=x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma
\end{aligned}
$$
comparing coefficients:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma) \quad \text { AG } \\
& q=(\alpha \beta+\beta \gamma+\gamma \alpha) \quad \text { AG } \\
& r=-\alpha \beta \gamma \quad \text { AG }
\end{aligned}
$$
Note: For candidates who do not include the $\boldsymbol{A} \boldsymbol{G}$ lines award full marks.
[3 marks]
b.i. $p^2-2 q=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
attempt to expand $(\alpha+\beta+\gamma)^2$
(M1)
$=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)-2(\alpha \beta+\beta \gamma+\gamma \alpha)$ or equivalent
A1
$=\alpha^2+\beta^2+\gamma^2 \quad \boldsymbol{A G}$
Note: Accept equivalent working from RHS to LHS.
[3 marks]
c. $p^2<3 q \Rightarrow 2 p^2-6 q<0$
$\Rightarrow(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2<0$
A1
if all roots were real $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2 \geq 0 \quad \boldsymbol{R 1}$
Note: Condone strict inequality in the $\boldsymbol{R} \mathbf{1}$ line.
Note: Do not award AOR1.
$\Rightarrow$ roots cannot all be real $\quad A G$
[2 marks]
d. $p^2=(-7)^2=49$ and $3 q=51$
A1
so $p^2<3 q \Rightarrow$ the equation has at least one complex root
R1
Note: Allow equivalent comparisons; e.g. checking $p^2<6 q$
[2 marks]
e.i. use of GDC (eg graphs or tables)
(M1)
$$
q=12
$$
A1
[2 marks]
e.ii.complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).
OR
a cubic curve always crosses the $x$-axis at at least one point.
R1
[1 mark]
f.i. attempt to expand $(\alpha+\beta+\gamma+\delta)^2$
(M1)
$$
\begin{aligned}
& (\alpha+\beta+\gamma+\delta)^2=\alpha^2+\beta^2+\gamma^2+\delta^2+2(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta) \\
& \Rightarrow \alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta) \\
& \left(\Rightarrow \alpha^2+\beta^2+\gamma^2+\delta^2=\right) p^2-2 q
\end{aligned}
$$
A1
[3 marks]
f.ii. $p^2<2 q$ OR $p^2-2 q<0$
A1
Note: Allow FT on their result from part (f)(i).
[1 mark]
g. $4<6$ OR $2^2-2 \times 3<0$
R1
hence there is at least one complex root.
AG
Note: Allow $\boldsymbol{F T}$ from part (f)(ii) for the $\boldsymbol{R}$ mark provided numerical reasoning is seen.
[1 mark]
h.i. $\left(p^2>2 q\right)(81>2 \times 24)$ (so) nothing can be deduced
R1
Note: Do not allow $\boldsymbol{F T}$ for the $\boldsymbol{R}$ mark.
[1 mark]
h.ii.-1
A1
[1 mark]
h.iiiattempt to express as a product of a linear and cubic factor M1
$$
(x+1)\left(x^3-10 x^2+34 x-12\right)
$$
A1A1
Note: Award A1 for each factor. Award at most A1A0 if not written as a product.
since for the cubic, $p^2<3 q(100<102) \quad \quad R 1$ there is at least one complex root $\quad \boldsymbol{A G}$
[4 marks]
Question
This question asks you to explore some properties of polygonal numbers and to determine and prove interesting results involving these numbers.
A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers.
For example, a triangular number is a number that can be arranged in the shape of an equilateral triangle. The first five triangular numbers are $1,3,6,10$ and 15 .
The following table illustrates the first five triangular, square and pentagonal numbers respectively. In each case the first polygonal number is one represented by a single dot.
For an $r$-sided regular polygon, where $r \in \mathbb{Z}^{+}, r \geq 3$, the $n$th polygonal number $P_r(n)$ is given by
$$
P_r(n)=\frac{(r-2) n^2-(r-4) n}{2} \text {, where } n \in \mathbb{Z}^{+} .
$$
Hence, for square numbers, $P_4(n)=\frac{(4-2) n^2-(4-4) n}{2}=n^2$.
The $n$th pentagonal number can be represented by the arithmetic series
$$
P_5(n)=1+4+7+\ldots+(3 n-2) .
$$
a.i. For triangular numbers, verify that $P_3(n)=\frac{n(n+1)}{2}$.
a.ii.The number 351 is a triangular number. Determine which one it is.
b.i. Show that $P_3(n)+P_3(n+1) \equiv(n+1)^2$.
b.ii.State, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.
b.iiiFor $n=4$, sketch a diagram clearly showing your answer to part (b)(ii).
c. Show that $8 P_3(n)+1$ is the square of an odd number for all $n \in \mathbb{Z}^{+}$.
d. Hence show that $P_5(n)=\frac{n(3 n-1)}{2}$ for $n \in \mathbb{Z}^{+}$.
e. By using a suitable table of values or otherwise, determine the smallest positive integer, greater than 1 , that is both a triangular number and a pentagonal number.
f. A polygonal number, $P_r(n)$, can be represented by the series
$$
\sum_{m=1}^n(1+(m-1)(r-2)) \text { where } r \in \mathbb{Z}^{+}, r \geq 3 .
$$
Use mathematical induction to prove that $P_r(n)=\frac{(r-2) n^2-(r-4) n}{2}$ where $n \in \mathbb{Z}^{+}$.
▶️Answer/Explanation
a.i. $P_3(n)=\frac{(3-2) n^2-(3-4) n}{2}$ OR $P_3(n)=\frac{n^2-(-n)}{2}$
A1
$P_3(n)=\frac{n^2+n}{2} \quad$ A1
Note: Award AOA1 if $P_3(n)=\frac{n^2+n}{2}$ only is seen.
Do not award any marks for numerical verification.
so for triangular numbers, $P_3(n)=\frac{n(n+1)}{2} \quad \boldsymbol{A G}$
[2 marks]a.i.METHOD 1
uses a table of values to find a positive integer that satisfies $P_3(n)=351$
(M1)
for example, a list showing at least 3 consecutive terms (. . 325, $351,378 \ldots)$
Note: Award (M1) for use of a GDC’s numerical solve or graph feature.
$n=26$ (26th triangular number)
A1
Note: Award $\mathbf{A O}$ for $n=-27,26$. Award $\boldsymbol{A O}$ if additional solutions besides $n=26$ are given.
METHOD 2
attempts to solve $\frac{n(n+1)}{2}=351\left(n^2+n-702=0\right)$ for $n$
(M1)
$n=\frac{-1 \pm \sqrt{1^2-4(1)(-702)}}{2}$ OR $(n-26)(n+27)=0$
$n=26$ (26th triangular number) $\quad$ A1
Note: Award $\mathbf{A O}$ for $n=-27,26$. Award $\boldsymbol{A O}$ if additional solutions besides $n=26$ are given.
[2 marks]
b.i. attempts to form an expression for $P_3(n)+P_3(n+1)$ in terms of $n$
M1
EITHER
$$
\begin{aligned}
& P_3(n)+P_3(n+1) \equiv \frac{n(n+1)}{2}+\frac{(n+1)(n+2)}{2} \\
& \equiv \frac{(n+1)(2 n+2)}{2}\left(\equiv \frac{2(n+1)(n+1)}{2}\right) \quad \text { A1 }
\end{aligned}
$$
OR
$$
\begin{aligned}
& P_3(n)+P_3(n+1) \equiv\left(\frac{n^2}{2}+\frac{n}{2}\right)+\left(\frac{(n+1)^2}{2}+\frac{n+1}{2}\right) \\
& \equiv\left(\frac{n^2+n}{2}\right)+\left(\frac{n^2+2 n+1+n+1}{2}\right)\left(\equiv n^2+2 n+1\right) \quad \text { A1 }
\end{aligned}
$$
A1
THEN
$$
\equiv(n+1)^2 \quad A G
$$
[2 marks]
b.ii.the sum of the $n$th and $(n+1)$ th triangular numbers is the $(n+1)$ th square number
A1
[1 mark]
b.iii.X $\quad X \quad X \quad X \quad X$
O $\mathrm{X} \times \mathrm{X}$
O $O \quad X \quad X \quad X$
A1
O
Note: Accept equivalent single diagrams, such as the one above, where the 4 th and 5 th triangular numbers and the 5 th square number are clearly shown.
Award $\boldsymbol{A 1}$ for a diagram that show $P_3(4)$ (a triangle with 10 dots) and $P_3(5)$ (a triangle with 15 dots) and $P_4(5)$ (a square with 25 dots).
[1 mark]
c. METHOD 1
$$
8 P_3(n)+1=8\left(\frac{n(n+1)}{2}\right)+1(=4 n(n+1)+1)
$$
attempts to expand their expression for $8 P_3(n)+1$
(M1)
$$
\begin{aligned}
& =4 n^2+4 n+1 \\
& =(2 n+1)^2
\end{aligned}
$$
and $2 n+1$ is odd
AG
METHOD 2
$$
8 P_3(n)+1=8\left((n+1)^2-P_3(n+1)\right)+1\left(=8\left((n+1)^2-\frac{(n+1)(n+2)}{2}\right)+1\right) \boldsymbol{A 1}
$$
attempts to expand their expression for $8 P_3(n)+1$
(M1)
$$
8\left(n^2+2 n+1\right)-4\left(n^2+3 n+2\right)+1\left(=4 n^2+4 n+1\right)
$$
$$
=(2 n+1)^2
$$
A1
and $2 n+1$ is odd
AG
Method 3
$8 P_3(n)+1=8\left(\frac{n(n+1)}{2}\right)+1\left(=(A n+B)^2\right)\left(\right.$ where $\left.A, B \in \mathbb{Z}^{+}\right)$
A1
attempts to expand their expression for $8 P_3(n)+1$
(M1)
$$
4 n^2+4 n+1\left(=A^2 n^2+2 A B n+B^2\right)
$$
now equates coefficients and obtains $B=1$ and $A=2$
$$
=(2 n+1)^2
$$
and $2 n+1$ is odd
AG
[3 marks]
d. EITHER
$u_1=1$ and $d=3$
(A1)
substitutes their $u_1$ and their $d$ into $P_5(n)=\frac{n}{2}\left(2 u_1+(n-1) d\right) \quad$ M1
$P_5(n)=\frac{n}{2}(2+3(n-1))\left(=\frac{n}{2}(2+3 n-3)\right) \quad$ A1
OR
$u_1=1$ and $u_n=3 n-2$
(A1)
substitutes their $u_1$ and their $u_n$ into $P_5(n)=\frac{n}{2}\left(u_1+u_n\right) \quad$ M1
$$
P_5(n)=\frac{n}{2}(1+3 n-2)
$$
A1
OR
$$
\begin{aligned}
& P_5(n)=(3(1)-2)+(3(2)-2)+(3(3)-2)+\ldots 3 n-2 \\
& P_5(n)=(3(1)+3(2)+3(3)+\ldots+3 n)-2 n(=3(1+2+3+\ldots+n)-2 n)
\end{aligned}
$$
substitutes $\frac{n(n+1)}{2}$ into their expression for $P_5(n) \quad$ M1
$$
\begin{aligned}
& P_5(n)=3\left(\frac{n(n+1)}{2}\right)-2 n \\
& P_5(n)=\frac{n}{2}(3(n+1)-4)
\end{aligned}
$$
A1
OR
attempts to find the arithmetic mean of $n$ terms
(M1)
$$
=\frac{1+(3 n-2)}{2}
$$
A1
multiplies the above expression by the number of terms $n$
$$
P_5(n)=\frac{n}{2}(1+3 n-2)
$$
THEN
so $P_5(n)=\frac{n(3 n-1)}{2} \quad$ AG
[3 marks]
e. METHOD 1
forms a table of $P_3(n)$ values that includes some values for $n>5$
(M1)
forms a table of $P_5(m)$ values that includes some values for $m>5$
(M1)
Note: Award (M1) if at least one $P_3(n)$ value is correct. Award (M1) if at least one $P_5(m)$ value is correct. Accept as above for $\left(n^2+n\right)$ values and $\left(3 m^2-m\right)$ values.
$n=20$ for triangular numbers
(A1)
$m=12$ for pentagonal numbers
(A1)
Note: Award (A1) if $n=20$ is seen in or out of a table. Award (A1) if $m=12$ is seen in or out of a table. Condone the use of the same parameter for triangular numbers and pentagonal numbers, for example, $n=20$ for triangular numbers and $n=12$ for pentagonal numbers.
210 (is a triangular number and a pentagonal number)
A1
Note: Award all five marks for 210 seen anywhere with or without working shown.
METHOD 2
EITHER
attempts to express $P_3(n)=P_5(m)$ as a quadratic in $n \quad$ (M1) $n^2+n+\left(m-3 m^2\right)(=0)$ (or equivalent) attempts to solve their quadratic in $n$
(M1)
$$
n=\frac{-1 \pm \sqrt{12 m^2-4 m+1}}{2}\left(=\frac{-1 \pm \sqrt{1^2-4\left(m-3 m^2\right)}}{2}\right)
$$
OR
attempts to express $P_3(n)=P_5(m)$ as a quadratic in $m$
(M1) $3 m^2-m-\left(n^2+n\right)(=0)$ (or equivalent) attempts to solve their quadratic in $m$
(M1)
$$
m=\frac{1 \pm \sqrt{12 n^2-12 n+1}}{6}\left(=\frac{1 \pm \sqrt{(-1)^2+12\left(n^2+n\right)}}{6}\right)
$$
THEN
$n=20$ for triangular numbers
$m=12$ for pentagonal numbers
(A1)
210 (is a triangular number and a pentagonal number)
A1
METHOD 3
$$
\frac{n(n+1)}{2}=\frac{m(3 m-1)}{2}
$$
let $n=m+k(n>m)$ and so $3 m^2-m=(m+k)(m+k+1) \quad$ M1 $2 m^2-2(k+1) m-\left(k^2+k\right)=0 \quad$ A1
attempts to find the discriminant of their quadratic
and recognises that this must be a perfect square $\quad M 1$
$$
\begin{aligned}
& \Delta=4(k+1)^2+8\left(k^2+k\right) \\
& N^2=4(k+1)^2+8\left(k^2+k\right)(=4(k+1)(3 k+1))
\end{aligned}
$$
determines that $k=8$ leading to $2 m^2-18 m-72=0 \Rightarrow m=-3,12$ and so $m=12$
A1
210 (is a triangular number and a pentagonal number)
A1
METHOD 4
$$
\frac{n(n+1)}{2}=\frac{m(3 m-1)}{2}
$$
let $m=n-k(m<n)$ and so $n^2+n=(n-k)(3(n-k)-1) \quad$ M1
$$
2 n^2-2(3 k+1) n+\left(3 k^2+k\right)=0 \quad \text { A1 }
$$
attempts to find the discriminant of their quadratic
and recognises that this must be a perfect square $\quad M 1$
$$
\begin{aligned}
& \Delta=4(3 k+1)^2-8\left(3 k^2+k\right) \\
& N^2=4(3 k+1)^2-8\left(3 k^2+k\right)(=4(k+1)(3 k+1))
\end{aligned}
$$
determines that $k=8$ leading to $2 n^2-50 n+200=0 \Rightarrow n=5,20$ and so $n=20$
A1
210 (is a triangular number and a pentagonal number)
A1
[5 marks]
f. Note: Award a maximum of R1MOMOA1M1A1A1RO for a ‘correct’ proof using $n$ and $n+1$.
consider $n=1: P_r(1)=1+(1-1)(r-2)=1$ and $P_r(1)=\frac{(r-2)\left(1^2\right)-(r-4)(1)}{2}=1$
so true for $n=1$
R1
Note: Accept $P_r(1)=1$ and $P_r(1)=\frac{(r-2)\left(1^2\right)-(r-4)(1)}{2}=1$.
Do not accept one-sided considerations such as ‘ $P_r(1)=1$ and so true for $n=1$ ‘.
Subsequent marks after this $\boldsymbol{R} 1$ are independent of this mark can be awarded.
Assume true for $n=k$, ie. $P_r(k)=\frac{(r-2) k^2-(r-4) k}{2} \quad$ M1
Note: Award $\boldsymbol{M O}$ for statements such as “let $n=k$ “. The assumption of truth must be clear.
Subsequent marks after this $\boldsymbol{M 1}$ are independent of this mark and can be awarded.
Consider $n=k+1$ :
$\left(P_r(k+1)\right.$ can be represented by the sum
$\sum_{m=1}^{k+1}(1+(m-1)(r-2))=\sum_{m=1}^k(1+(m-1)(r-2))+(1+k(r-2))$ and so
$P_r(k+1)=\frac{(r-2) k^2-(r-4) k}{2}+(1+k(r-2))\left(P_r(k+1)=P_r(k)+(1+k(r-2))\right) \quad$ M1
$=\frac{(r-2) k^2-(r-4) k+2+2 k(r-2)}{2}$ $=\frac{(r-2)\left(k^2+2 k\right)-(r-4) k+2}{2}$
A1
$=\frac{(r-2)\left(k^2+2 k+1\right)-(r-2)-(r-4) k+2}{2} \quad$ M1
$=\frac{(r-2)(k+1)^2-(r-4) k-(r-4)}{2}$
$=\frac{(r-2)(k+1)^2-(r-4)(k+1)}{2}$
A1
hence true for $n=1$ and $n=k$ true $\Rightarrow n=k+1$ true $\quad \boldsymbol{R 1}$
therefore true for all $n \in \mathbb{Z}^{+}$
Note: Only award the final $\boldsymbol{R} 1$ if the first five marks have been awarded. Award marks as appropriate for solutions that expand both the LHS and (given) RHS of the equation.
Question
A Gaussian integer is a complex number, $z$, such that $z=a+b \mathrm{i}$ where $a, b \in \mathbb{Z}$. In this question, you are asked to investigate certain divisibility properties of Gaussian integers.
Consider two Gaussian integers, $\alpha=3+4 \mathrm{i}$ and $\beta=1-2 \mathrm{i}$, such that $\gamma=\alpha \beta$ for some Gaussian integer $\gamma$.
Now consider two Gaussian integers, $\alpha=3+4 \mathrm{i}$ and $\gamma=11+2 \mathrm{i}$.
The norm of a complex number $z$, denoted by $N(z)$, is defined by $N(z)=|z|^2$. For example, if $z=2+3 \mathrm{i}$ then $N(2+3 \mathrm{i})=2^2+3^2=13$.
A Gaussian prime is a Gaussian integer, $z$, that cannot be expressed in the form $z=\alpha \beta$ where $\alpha, \beta$ are Gaussian integers with $N(\alpha), N(\beta)>1$.
The positive integer 2 is a prime number, however it is not a Gaussian prime.
Let $\alpha, \beta$ be Gaussian integers.
The result from part (h) provides a way of determining whether a Gaussian integer is a Gaussian primea. Find $\gamma$.
b. Determine whether $\frac{\gamma}{\alpha}$ is a Gaussian integer.
c. On an Argand diagram, plot and label all Gaussian integers that have a norm less than 3.
d. Given that $\alpha=a+b$ i where $a, b \in \mathbb{Z}$, show that $N(\alpha)=a^2+b^2$.
e. By expressing the positive integer $n=c^2+d^2$ as a product of two Gaussian integers each of norm $c^2+d^2$, show that $n$ is not a Gaussian prime.
f. Verify that 2 is not a Gaussian prime.
g. Write down another prime number of the form $c^2+d^2$ that is not a Gaussian prime and express it as a product of two Gaussian integers.
h. Show that $N(\alpha \beta)=N(\alpha) N(\beta)$.
i. Hence show that $1+4 \mathrm{i}$ is a Gaussian prime.
j. Use proof by contradiction to prove that a prime number, $p$, that is not of the form $a^2+b^2$ is a Gaussian prime.
▶️Answer/Explanation
a.$$
(3+4 i)(1-2 i)=11-2 i
$$
(M1)A1
[2 marks]
$\frac{\gamma}{\alpha}=\frac{41}{25}-\frac{38}{25} \mathrm{i} \quad$ (M1)A1
(Since $\operatorname{Re} \frac{\gamma}{\alpha}\left(=\frac{41}{25}\right)$ and/or $\operatorname{Im} \frac{\gamma}{\alpha}\left(=-\frac{38}{25}\right)$ are not integers) $\frac{\gamma}{\alpha}$ is not a Gaussian integer $\quad \boldsymbol{R 1}$
Note: Award R1 for correct conclusion from their answer.
[3 marks]
c. $\pm 1, \pm \mathrm{i}, 0$ plotted and labelled
A1
$1 \pm \mathrm{i},-1 \pm \mathrm{i}$ plotted and labelled
A1
Note: Award A1AO if extra points to the above are plotted and labelled.
[2 marks]
d. $|z|=\sqrt{a^2+b^2}$ (and as $N(z)=|z|^2$ )
A1
then $N(\alpha)=a^2+b^2 \quad A G$
[1 mark]
e. $c^2+d^2=(c+d \mathrm{i})(c-d \mathrm{i})$
A1
and $N(c+d \mathrm{i})=N(c-d \mathrm{i})=c^2+d^2$
R1
$N(c+d \mathrm{i}), N(c-d \mathrm{i})>1$ (since $c, d$ are positive)
R1
so $c^2+d^2$ is not a Gaussian prime, by definition
AG
[3 marks]
f. $2\left(=1^2+1^2\right)=(1+\mathrm{i})(1-\mathrm{i})$
(A1)
$$
N(1+\mathrm{i})=N(1-\mathrm{i})=2
$$
so 2 is not a Gaussian prime
AG
[2 marks]
g. For example, $5\left(=1^2+2^2\right)=(1+2 \mathrm{i})(1-2 \mathrm{i})$
(M1)A1
[2 marks]
h. METHOD 1
Let $\alpha=m+n \mathrm{i}$ and $\beta=p+q \mathrm{i}$
LHS:
$$
\begin{aligned}
& \alpha \beta=(m p-n q)+(m q+n p) \mathrm{i} \quad \text { M1 } \\
& N(\alpha \beta)=(m p-n q)^2+(m q+n p)^2 \quad \text { A1 } \\
& (m p)^2-2 m n p q+(n q)^2+(m q)^2+2 m n p q+(n p)^2 \\
& (m p)^2+(n q)^2+(m q)^2+(n p)^2 \quad \text { A1 }
\end{aligned}
$$
RHS:
$$
\begin{array}{ll}
N(\alpha) N(\beta)=\left(m^2+n^2\right)\left(p^2+q^2\right) & \text { M1 } \\
(m p)^2+(m q)^2+(n p)^2+(n q)^2 & \text { A1 }
\end{array}
$$
LHS $=$ RHS and so $N(\alpha \beta)=N(\alpha) N(\beta) \quad$ AG
A1
METHOD 2
Let $\alpha=m+n \mathrm{i}$ and $\beta=p+q \mathrm{i}$
LHS
$$
\begin{aligned}
& N(\alpha \beta)=\left(m^2+n^2\right)\left(p^2+q^2\right) \quad \text { M1 } \\
& =(m+n \mathrm{i})(m-n \mathrm{i})(p+q \mathrm{i})(p-q \mathrm{i}) \quad \text { A1 } \\
& =(m+n \mathrm{i})(p+q \mathrm{i})(m-n \mathrm{i})(p-q \mathrm{i}) \\
& =((m p-n q)+(m q+n p) \mathrm{i})((m p-n q)-(m q+n p) \mathrm{i}) \quad \text { M1A1 } \\
& =(m p-n q)^2+(m q+n p)^2 \quad \text { A1 } \\
& N=((m p-n q)+(m q+n p) \mathrm{i}) \quad \text { A1 } \\
& =N(\alpha) N(\beta) \text { (= RHS) } \quad \text { AG }
\end{aligned}
$$
A1
[6 marks]
i. $N(1+4 \mathrm{i})=17$ which is a prime (in $\mathbb{Z}) \quad \boldsymbol{R} 1$
if $1+4 \mathrm{i}=\alpha \beta$ then $17=N(\alpha \beta)=N(\alpha) N(\beta) \quad \boldsymbol{R 1}$ we cannot have $N(\alpha), N(\beta)>1 \quad \boldsymbol{R 1}$
Note: Award $\boldsymbol{R 1}$ for stating that $1+4 \mathrm{i}$ is not the product of Gaussian integers of smaller norm because no such norms divide 17
so $1+4 \mathrm{i}$ is a Gaussian prime
AG
[3 marks]
j. Assume $p$ is not a Gaussian prime
$\Rightarrow p=\alpha \beta$ where $\alpha, \beta$ are Gaussian integers and $N(\alpha), N(\beta)>1 \quad$ M1
$\Rightarrow N(p)=N(\alpha) N(\beta) \quad$ M1
$p^2=N(\alpha) N(\beta) \quad$ A1
It cannot be $N(\alpha)=1, N(\beta)=p^2$ from definition of Gaussian prime $\quad \boldsymbol{R 1}$ hence $N(\alpha)=p, N(\beta)=p \quad \boldsymbol{R 1}$
If $\alpha=a+b \mathrm{i}$ then $N(\alpha)=a^2+b^2=p$ which is a contradiction
R1
hence a prime number, $p$, that is not of the form $a^2+b^2$ is a Gaussian prime
AG
[6 marks]