Question

The number of integers x for which the number $\sqrt{{{x}^{2}}+x+1}$ is rational is
[a] Infinite
[b] one
[c] two
[d] three

Answer 1

Hint: Use the fact that if for some integer x, $\sqrt{{{x}^{2}}+x+1}$ is rational, we have $\sqrt{{{x}^{2}}+x+1}$ is a non-negative integer. Hence let $m\in {{\mathbb{Z}}^{+}}$ be such that $\sqrt{{{x}^{2}}+x+1}=m$. Square both sides and hence form a quadratic equation in x. Use the fact that if the quadratic equation $a{{x}^{2}}+bx+c=0,a,b,c\in \mathbb{Q}$ has integral roots, then $D={{b}^{2}}-4ac$ is a perfect square. Simplify the resulting expression and use the fact that since 3 is prime, the only factors of 3 are 1 and 3. Hence find the number of integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is rational.

Complete step-by-step answer:
Let for some integer x, $\sqrt{{{x}^{2}}+x+1}$ is rational. Since ${{x}^{2}}+x+1$ is an integer, we have $\sqrt{{{x}^{2}}+x+1}$ is rational if and only if $\sqrt{{{x}^{2}}+x+1}$ is an integer. So let $m\in {{\mathbb{Z}}^{+}}$ be such that $m=\sqrt{{{x}^{2}}+x+1}$.
Squaring both sides of the equation, we get
${{m}^{2}}={{x}^{2}}+x+1$
Subtracting ${{m}^{2}}$ on both sides, we get
${{x}^{2}}+x+\left( 1-{{m}^{2}} \right)=0$, which is a quadratic equation in x.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0,a,b,c\in \mathbb{Q}$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Since x is an integer, we have
${{b}^{2}}-4ac$ must be a perfect square.
Here $a=1,b=1$ and $c=1-{{m}^{2}}$
Hence, we have
$1-4\left( 1 \right)\left( 1-{{m}^{2}} \right)={{k}^{2}},k\in \mathbb{Z}$
Simplifying, we get
$4{{m}^{2}}={{k}^{2}}+3$
Subtracting ${{k}^{2}}$ from both sides, we get
$4{{m}^{2}}-{{k}^{2}}=3$
Using ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right),$ we get
$\left( 2m-k \right)\left( 2m+k \right)=3$
Since 3 is prime, the only factors of 3 are 1 and 3.
Hence, we have
$2m-k=3,2m+k=1$ or $2m-k=1,2m+k=3$or $2m-k=-1,2m+k=-3$ or $2m-k=-3,2m+k=-1$
Solving the first system:
We have
$\begin{align}
  & 2m-k=3\text{ (i)} \\
 & 2m+k=1\text{ (ii)} \\
\end{align}$
Adding equation (i) and equation (ii), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (i), we get
$2-k=3\Rightarrow k=-1$
Hence m = 1 and k = -1 is the solution of the system.
Solving the second system:
$\begin{align}
  & 2m+k=3\text{ (iii)} \\
 & 2m-k=1\text{ (iv)} \\
\end{align}$
Adding equation (iv) and equation (v), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (iv), we get
$2-k=1\Rightarrow k=1$
Similarly solving the third system yields m=-1 and k= -1 and solving the fourth system yields m = -1 and k = 1.
Since m is non-negative, we have
(m=-1,k=1) and (m=-1, k =1) are rejected.
Hence m = 1 is the only possible integral value.
When m = 1, we have
${{x}^{2}}+x+1=1\Rightarrow {{x}^{2}}+x=0\Rightarrow \left( x \right)\left( x+1 \right)=0$
Hence x = 0 or x = -1 are the only integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence the number of integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is rational is two.
Hence option [c] is correct.

Note: Verification: In the above question it is important to check the values for which we suppose $\sqrt{{{x}^{2}}+x+1}$ is rational.
When x = 0, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1}=1$
When x = -1, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1-1+1}=1$
Hence at both x= 0 and x = -1 $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence our solution is correct