$8 \cdot 38$ Find the largest integer $x$ such that $4^{27}+4^{1000}+4^{x}$ is a perfect square.
(6th All-Soviet Union Mathematical Olympiad, 1972)
[Solution]
$$
\begin{aligned}
& 4^{27}+4^{1000}+4^{x} \\
= & 2^{54}+2^{2000}+2^{2 x} \\
= & 2^{54}\left(1+2^{1946}+2^{2 x-54}\right) \\
= & 2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)
\end{aligned}
$$

Obviously, when $2^{2 x-54}=\left(2^{1945}\right)^{2}$, i.e.,
$$
\begin{array}{c}
2 x-54=2 \cdot 1945 \\
x=1972
\end{array}
$$

the original expression $=2^{54}\left(1+2^{1945}\right)^{2}$ is a perfect square.
If $x>1972$, then
$$
2^{2 x-54}<1+2 \cdot 2^{1945}+2^{2 x-54}<\left(2^{x-27}+1\right)^{2} \text {. }
$$

In this case, the known number lies between two consecutive squares $\left(2^{x-27}\right)^{2}$ and $\left(2^{x-27}+1\right)^{2}$, and cannot be a perfect square.
Therefore, the largest integer $x$ that makes $4^{27}+4^{1000}+4^{x}$ a perfect square is $x=1972$.