An algebra problem by Munem Shahriar

$\begin{aligned} m + \frac 1m & = 2 \\ \left(m+\frac 1m \right)^2 & = 2^2 \\ m^2 +2+\frac 1{m^2} &= 4 \\ m^2 +\frac 1{m^2} &= 2 \\ \left(m^2+\frac 1{m^2} \right)^2 & = 2^2 \\ m^4 +2+\frac 1{m^4} &= 4 \\ \implies m^4 +\frac 1{m^4} &= \boxed 2 \end{aligned}$

$m+\frac{1}{m}=2$

so $m=1$

$\large m^{4}+\frac{1}{m^{4}}=\boxed 2$

Because $m+\frac{1}{m}=2$ , this makes the problem easy. It would be better if you used, say, $m+\frac{1}{m}=3$ .

$m^4$ + $\dfrac{1}{m^4}$

$=(m^2)^2 +(\dfrac{1}{m^2})^2$

$= (m^2+\dfrac{1}{m^2})^2 -2 \cdot m^2 \cdot \dfrac{1}{m^2}$

$= \{ (m$ + $\dfrac{1}{m})^2 -2 \cdot m \cdot \dfrac{1}{m}\}^2 - 2$

$= \{ (2)^2 - 2 \}^2 - 2$

$= (4-2)^2- 2$

$= (2)^2 - 2$

$= 4 - 2$

$= 2$

It is given that $\large m+\dfrac{1}{m}= \text{2 So m will be 1}$ So $\large m^4+\dfrac{1}{m^4}=1+1$ $\Rightarrow 1+1=2$

$\begin{aligned}m+\frac{1}{m}&=2\\ \frac{m^2+1}{m}&=2\\m^2+1&=2m\\m^2-2m+1&=0\\(m)^2-2\times m \times 1 +(1)^2&=0\\(m-1)^2&=0\\m-1&=0\\m&=1\end{aligned}$

Therefore,

$\begin{aligned}m^4+\frac{1}{m^4}&=1^4+\frac{1}{1^4}\\&=1+1\\&=\boxed{2} \end{aligned}$

BONUS : If $m+\frac{1}{m}=2$ Then, $m^p+\frac{1}{m^q}=2$ and, $m^p-\frac{1}{m^q}=0$

Another way of solving is by using the AM-GM inequality.

We have: ${m}+\frac{1}{{m}} \geq 2$

Equality holds when m is equal to 1.

Thus, ${m}^4+\frac{1}{{m}^4} = 2$