Q14 (a)

Assume $x^3+3x+3=0$ has a rational solution, i.e. $\frac pq$ is a solution and the HCF$(p,q)=1$.

$$\left(\frac pq\right)^3+3\left(\frac pq\right)+3=0$$ $$p^3+3pq^2+3q^3=0$$

Method 1:

Case 1: $p$ and $q$ are odd.

$$\begin{aligned} p^3&\text{ is odd}\\ 3q^3&\text{ is odd}\\ 3pq^2&\text{ is odd}\\ \text{odd+odd+odd}&\ne\text{ even} \end{aligned}$$

Therefore, $p^3+3pq^2+3q^3\ne0$ if both $p$ and $q$ are odd.

Case 2: $p$ is odd and $q$ is even.

$$\begin{aligned} p^3&\text{ is odd}\\ 3q^3&\text{ is even}\\ 3pq^2&\text{ is even}\\ \text{odd+even+even}&\ne\text{ even} \end{aligned}$$

Therefore, $p^3+3pq^2+3q^3\ne0$ if $p$ is odd and $q$ is even.

Case 3: $p$ is even and $q$ is odd.

$$\begin{aligned} p^3&\text{ is even}\\ 3pq^2&\text{ is even}\\ 3q^3&\text{ is odd}\\ \text{even+even+odd}&\ne\text{ even} \end{aligned}$$

Therefore, $p^3+3pq^2+3q^3\ne0$ if $p$ is even and $q$ is odd.

Case 4: $p$ and $q$ cannot be both even since HCF$(p,q)=1$.

Therefore,

$$p^3+3pq^2+3q^3\ne0$$

contradicting the assumption.

Method 2:

$$\begin{aligned} p^3&=-3(pq^2+q^3)\\ &=-3q^2(p+q) \end{aligned}$$

$p^3$ is divisible by $3$, therefore $p$ is divisible by $3$. Let $p=3k$.

$$\begin{aligned} (3k)^3&=-3q^2(p+q)\\ -9k^3&=q^2(p+q) \end{aligned}$$

$q^2$ is divisible by $q$, therefore $q$ is divisible by $3$, but HCF$(p,q)=1$.

OR $p+q$ is divisible by $q$.

$$\begin{aligned} \exists n\in\mathbb Z,\quad p+q&=qn\\ q&=qn-p\\ &=qn-3k\\ &=3(3n-k) \end{aligned}$$

and $q$ is divisible by $3$.

Contradiction.