Find the largest integer $x$ such that $4^{27}+4^{1010}+4^{x}$ is a perfect square.
First, we show that for $x=1992$, the expression is a perfect square. Indeed, $4^{27}+4^{1010}+4^{1992}=\left(2^{27}\right)^{2}+$ $\left(2^{1992}\right)^{2}+2^{2020}=\left(2^{27}\right)^{2}+\left(2^{1992}\right)^{2}+2 \cdot 2^{27} \cdot 2^{1992}=\left(2^{27}+2^{1992}\right)^{2}$.

Next, we prove that for no integer $x>1992$ can we obtain a perfect square. Factoring out $4^{27}-\mathrm{t}$ from the expression:

$$
4^{27}\left(1+4^{983}+4^{x-27}\right) .
$$

Since $4^{27}$ is a perfect square, and $x>1992$, it suffices to examine the value of $1+4^{983}+4^{x-27}$. Suppose this is indeed a perfect square. Since $1+4^{983}+4^{x-27}>\left(2^{x-27}\right)^{2}$, it cannot be smaller than the square of $2^{x-27}+1$:

$$
1+4^{983}+4^{x-27} \geq\left(2^{x-27}+1\right)^{2}=1+4^{x-27}+2^{x-26}
$$

from which

$$
2^{1966} \geq 2^{x-26}
$$

or

$$
1966 \geq x-26>1992-26=1966
$$

which is a contradiction. Therefore, the sought number is 1992.

Remark. After the given solution, it is a legitimate question to ask how one could have come up with the value $x=1992$. Since the three terms of the given trinomial sum are perfect squares, it suggests the idea of considering the expression as the square of a binomial sum, written in the form $a^{2}+b^{2}+2 a b$. In this case, we need to distinguish three cases depending on which of the three terms we choose as $2 a b$. For example, in the case $2 a b=4^{27}$, $\left\{a^{2}, b^{2}\right\}=\left\{4^{1010}, 4^{x}\right\}$, thus $x=-957$. For the choices $2 a b=4^{1010}$ and $2 a b=4^{x}$, similarly $x=1992$ and $x=519$, respectively.