Find the number of values for m for which (m+i)^4 is an integer? i

$i = \sqrt{- 1}$

Nov 16, 2017

$\forall m \in \left\{0 , \pm 1\right\} , {\left(m + i\right)}^{4} \in \mathbb{Z} .$

Explanation:

Using the Binomial Theorem, we have, ${\left(m + i\right)}^{4} ,$

$= {\text{_4C_0m^4i^0+""_4C_1m^3i+""_4C_2m^2i^2+""_4C_3m^1i^3+}}_{4} {C}_{4} {m}^{0} {i}^{4} ,$

$= {m}^{4} + 4 {m}^{3} i - 6 {m}^{2} - 4 m i + 1 ,$

$\Rightarrow {\left(m + i\right)}^{4} = \left({m}^{4} - 6 {m}^{2} + 1\right) + \left(4 {m}^{3} - 4 m\right) i .$

In order that ${\left(m + i\right)}^{4}$ be integer (real), equating the Real and

Imaginary parts of both sides, we must have,

$\left({m}^{4} - 6 {m}^{2} + 1\right) \in \mathbb{R} , \mathmr{and} 4 {m}^{3} - 4 m = 4 m \left({m}^{2} - 1\right) = 0.$

Now, $4 {m}^{3} - 4 m = 4 m \left({m}^{2} - 1\right) = 0 \Rightarrow m = 0 , m = \pm 1.$

$\text{Also, "AA m in {0,+-1}, (m^4-6m^2+1) in RR" holds good.}$

$\therefore m \in \left\{0 , - 1 , + 1\right\} .$

Hence, there are $3$ values of $m$ such that, ${\left(m + i\right)}^{4} \in \mathbb{Z} .$