Question:
show that: the beautiful ${\tt sqrt}$-identity:
$$ \left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\ +\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4} =5 $$
Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.) Thank you
Because I found this $$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.
On
Let $x = 4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}$. Then, we have:
(1) $x = 4 - 3 \cdot 5^{1/4} + 2 \cdot 5^{2/4} - 5^{3/4}$
(2) $5^{1/4}x = 4\cdot 5^{1/4} - 3 \cdot 5^{2/4} + 2 \cdot 5^{3/4} - 5$ (Multiply (1) by $5^{1/4}$)
(3) $(5^{1/4}+1)x = -1 + 5^{1/4} - 5^{2/4} +5^{3/4}$ (Add (1) and (2))
(4) $5^{1/4}(5^{1/4}+1)x = -5^{1/4} + 5^{2/4} - 5^{3/4} +5$ (Multiply (3) by $5^{1/4}$)
(5) $(5^{1/4}+1)^2x = 4$ (Add (3) and (4))
Therefore, $x = \dfrac{4}{(5^{1/4}+1)^2}$. Hence, $\left(\dfrac{2}{\sqrt{x}}-1\right)^4 = \left(\dfrac{2}{\tfrac{2}{5^{1/4}+1}}-1\right)^4 = (5^{1/4}+1-1)^4 = 5$.
On
[This is a paraphrase on JimmyK4542's elegant answer earlier]
Let $$b=-5^{\frac 14}$$
Then the long expression under the long square root sign becomes an arithmetico-geometric series: $$k=4+3b+2b^2+b^3$$ Multiplying by $b$: $$\begin{align} b\cdot k&=\quad \qquad 4b+3b^2+2b^3+b^4\\ &=\quad \qquad 4b+3b^2+2b^3+5 \end{align}$$ Subtracting: $$\begin{align} (b-1)k&=1+b+b^2+b^3\\ &=\dfrac{b^4-1}{b-1}\\ &=\dfrac 4{b-1}\\ k&=\dfrac 4{(b-1)^2}\\ \sqrt{k}&=\dfrac 2{1-b}\\ \left( \dfrac 2{\sqrt{k}} -1\right)^4&={(-b)}^4=5 \end{align}$$
Hint: Set $x=5^{1/4}$, then $$ \frac{2}{\sqrt{4-3x+2x^2-x^3}}-1=x $$ This equation simplifies to $$ x(x^4-5)=0 $$ The rest is clear.