Rational (!) number

Prove that 

$$\mathscr{A}= \sqrt{1+ 1999\sqrt{1+2000\sqrt{4+2000\sqrt{1+2003\cdot 2005}}}}$$

is rational.

Solution

We set $x=2000$ hence:

$$\begin{align*}
\sqrt{1+\left(x-1 \right)\sqrt{1+x\sqrt{4+x\sqrt{1+\left(x+3 \right)\left(x+5 \right)}}}} &= \\\sqrt{1+\left(x-1 \right)\sqrt{1+x\sqrt{4+x\sqrt{\left(x+4 \right)^{2}}}}} \\
 &= \sqrt{1+\left(x-1 \right)\sqrt{1+x\sqrt{\left( x+2\right)^{2}}}}\\
 &= \sqrt{1+\left(x-1 \right)\sqrt{\left(x+1 \right)^{2}}}\\
 &= \sqrt{1+\left(x-1 \right)\left(x+1 \right)}\\
 &= \sqrt{x^{2}}=x=2000
\end{align*}$$