Show that the number $\frac{\sqrt{2021^4+2 \cdot 2020^2-4041^2}}{505 \cdot 1011}$ is natural and a perfect cube.

Question

question

Show that the number $$\frac{\sqrt{2021^4+2 \cdot 2020^2-4041^2}}{505 \cdot 1011}$$ is natural and a perfect cube.

my idea

I tried writing the upper part as a perfect square so that we could get rid of the radical.

I tried using these formulas

$$(a+b)^2=a^2+2ab+b^2$$

or

$$(a-b)^2=a^2-2ab+b^2$$

but could not get anywhere useful.

I also tried writing $4041^2=(2020+2021)^2$

I don't know where to start. I hope one of you can help me! Thank you!

Answer 1

Try to express everything in terms of the number $2020$: $$\begin{align}&2021^4=(2020+1)^4=2020^4+4\cdot2020^3+6\cdot2020^2+4\cdot2020+1\\&4041^2=(2\cdot 2020+1)^2=4\cdot 2020^2+4\cdot 2020+1\end{align}$$ So: $$\begin{align}2021^4+2\cdot 2020^2-4041^2&=2020^4+4\cdot 2020^3+4\cdot 2020^2\\&=2020^2\cdot (2020^2+4\cdot 2020+4)\\&=2020^2\cdot (2020+2)^2\\&=2020^2\cdot 2022^2\end{align}$$ Thus: $$\frac{\sqrt{2021^4+2 \cdot 2020^2-4041^2}}{505 \cdot 1011}=\frac{2020\cdot 2022}{505\cdot 1011}=8$$ which is natural and a perfect cube.

Answer 2

We factor the numerator. Take $a=2020$, so that the part inside the radical becomes $(a+1)^4+2a^2-(2a+1)^2 = (a^2+2a+1)^2-(2a+1)^2+2a^2 = (a^2+4a+2)(a^2)+2a^2 = a^2(a^2+4a+4)=a^2(a+2)^2$

Therefore, $\sqrt{a^2(a+2)^2} = a(a+2)$, and your expression becomes $$\frac{2020\cdot 2022}{505\cdot 1011} = 8$$