Question

The number of integers $n(<20)$ for which $n^2-3n+3$ is a perfect square is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C

$2$

Solution:

Step 1: Equating the given expression to ${\mathit{m}}^{\mathbf{2}}$:

Let, the given expression be equal to a perfect square that is $m^2$.

$$\begin{gathered} \Rightarrow n^2-3n+3=m^2 \\ \Rightarrow n^2-m^2=3n-3 \\ \Rightarrow (n-m)(n+m)=3(n-1)\times 1...\left(1\right) \end{gathered}$$

Step 2 : Finding the value of $n$ by equating $\mathit{n}\mathbf{-}\mathit{m}\mathbf{=}\mathbf{3}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{and}\mathbf{}\mathit{n}\mathbf{+}\mathit{m}\mathbf{=}\mathbf{1}\mathbf{:}$:

$$\begin{gathered} n+m=1....\left(2\right) \\ n-m=3(n-1) \end{gathered}$$

We get $m=1-n$ from $\left(2\right)$ in $n-m=3(n-1)$ we get,,

$$\begin{gathered} n-m=3(n-1) \\ \Rightarrow n-(1-n)=3n-3\left[\because fromeqn\left(2\right)\right] \\ \Rightarrow n-1+n=3n-3 \\ \Rightarrow 2n=3n-3+1 \\ \Rightarrow 2n=3n-2 \\ \therefore n=2 \end{gathered}$$

Step 3: Finding the value of $n$ by equating $\mathit{n}\mathbf{-}\mathit{m}\mathbf{=}\mathbf{3}\mathbf{}$and $\mathit{n}\mathbf{+}\mathit{m}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1}$

$$\begin{gathered} n-m=3...\left(3\right) \\ n+m=n-1 \end{gathered}$$

Substituting $m=n-3$ from (3) in $n+m=n-1$ to get the value of $n$.

$$\begin{gathered} n+m=n-1 \\ \Rightarrow n+(n-3)=n-1\left[\because fromeqn\left(3\right)\right] \\ \Rightarrow 2n-3=n-1 \\ \Rightarrow 2n-n=3-1 \\ \Rightarrow n=2 \end{gathered}$$

Therefore, there is only one value for which $n^2-3n+3$ is a perfect square and that is $\mathit{n}\mathbf{=}\mathbf{2}$.

Final answer: Hence, option (C) is correct.