Let's find such positive integers $x$ and $y$ for which the following expressions are perfect squares:
a) $4^{10}+4^{16}+4^{x}$
b) $4^{16}+4^{21}+4^{2}$.
I. solution. a) The unknown term of the expression $K_{1}$ is itself a square number in any case: $4^{x}=\left(2^{x}\right)^{2}$, so our task is to represent the number $4^{10}+4^{16}$ as the difference of two square numbers. We are looking for a solution where $x \geqq 10$. According to this requirement,

\[
\begin{gathered}
K_{1}=4^{10}\left(1+4^{6}+4^{x-10}\right)=N^{2} \\
1+4^{6}+4^{x-10}=4097+\left(2^{x-10}\right)^{2}=\left(\frac{N}{4^{5}}\right)^{2} \\
4097=n^{2}-z^{2}=(n-z)(n+z)
\end{gathered}
\]

where $n$ and $z$ are natural numbers, and $z$ can only be a non-negative integer power of 2. Factoring $4097$ into prime factors, we get $4097=1 \cdot 4097=17 \cdot 241$, so the possible pairs of $n, z$ values are

\[
\begin{aligned}
& \left.\begin{array}{l}
n-z=1 \\
n+z=4097
\end{array}\right\} \text { gives } z=2048=2^{11}, n=2049 \\
& \left.\begin{array}{l}
n-z=17 \\
n+z=241
\end{array}\right\} \text { gives } z=112, \text { which does not fit. }
\end{aligned}
\]

Thus, $x-10=11, x=21$ is a suitable exponent, and indeed,

\[
4^{10}+4^{16}+4^{21}=4^{5} \cdot 2049^{2}
\]

Since the problem does not require finding all suitable $x$ values, we can consider the question of $K_{1}$ solved.

From the obtained result, we can also read a solution for the second question of the problem, with $y=10$, $K_{2}$ is a perfect square.

b) Using the same method, under the constraint $y \geqq 16$, we have

\[
\begin{gathered}
K_{2}=4^{16}\left(1+4^{5}+4^{y-16}\right)=\left(4^{8}\right)^{2}\left\{1025+\left(2^{y-16}\right)^{2}\right\}=M^{2} \\
1025+u^{2}=m^{2}
\end{gathered}
\]

where $u=2^{y-16}$ and $m=M / 4^{8}$, two suitable sets of $u, y, m, M$ numbers are obtained:

|  | $m-u$ | $m+u$ | $u$ | $m$ | $y$ | $M$ |
| ---: | :---: | ---: | ---: | :---: | :---: | :---: |
| I. | 1 | 1025 | $512=2^{9}$ | 513 | 25 | $4^{8} \cdot 513$ |
| II. | 25 | 41 | $8=2^{3}$ | 33 | 19 | $4^{3} \cdot 33$. |

With these, we have found three suitable $y$ values: $y=10,19$ and 25.

II. solution. a) We try to identify the terms of $K_{1}$ individually with the terms of the expansion $(c+d)^{2}=c^{2}+d^{2}+2 c d$, using the fact that the 2 in the $2 c d$ term is a divisor of the bases in $K_{1}$. We assign the roles of $4^{10}, 4^{16}$, and $4^{x}$ to the $2 c d$ term one after the other.
x) $4^{16}=c^{2}$ and $4^{x}=d^{2}$, i.e., $c=2^{16}$ and $d=2^{x}$, then $2 c d=2^{1+16+x}=4^{10}=2^{20}$, which holds for $x=3$, giving us a new solution:

\[
4^{10}+4^{16}+4^{3}=(c+d)^{2}=\left(2^{10}+2^{3}\right)^{2}
\]

ß) The attempt with $c=2^{10}$ and $d=2^{x}$ yields the $x=21$ found in the first solution.

ү) The attempt with $c=2^{10}$ and $d=2^{16}$ does not yield a solution because in this case $2 c d$ is an odd power of 2, which is not an integer power of 4.

b) The same method for $K_{2}$ yields solutions in all three roles, the ones found in the first solution. The difference from the a) case is that in the a) case both known exponents are even, unlike in the b) case.

Remark. 1. It can be shown that there are no other solutions besides the ones found, for example, by the substitution $4=3+1$ and using the binomial theorem, it follows that

\[
3 \mid x, \text { or } 3 \mid(y-1)
\]

György Pálfalvi (Győr, Révai M. Gymnasium, IV. grade)

2. Another way to show this is if in the method of the first solution, we factor the sum $4^{10}+4^{16}$, and the sum $4^{16}+4^{21}$ in every possible way into the product of two factors. The cases $x<10$ and $y<16$ result in a long series of failures, which is only interrupted by one successful case: $x=3$, and $y=10$, which is why we omitted them.

III. solution. a) If for some $x$, $K_{1}$ is a square number, say $K_{1}=N^{2}$, then for this $x$,

\[
4^{10}+4^{16}=N^{2}-4^{x}=\left(N-2^{x}\right)\left(N+2^{x}\right)=M\left(M+2^{x+1}\right)
\]

where $M=N-2^{x}$. Therefore, $\left(4^{10}+4^{16}\right)$ can be represented as the product of a number $M$ and a number that is $2$ raised to some suitable power greater than $M$. We are sure to get such a representation if we choose the square root of $4^{10}$ or $4^{16}$ for $M$:

\[
\begin{aligned}
& M=4^{5} \text { gives } 4^{10}+4^{16}=4^{5}\left(4^{5}+4^{11}\right) \text {, and } x=21 \\
& M=4^{8} \text { gives } 4^{10}+4^{16}=4^{8}\left(4^{8}+4^{2}\right), \quad \text { and } \quad x=3
\end{aligned}
\]

b) Similarly, we get that

\[
\begin{aligned}
& M=4^{8} \quad \text { gives } \quad 4^{16}+4^{21}=4^{8}\left(4^{8}+4^{13}\right), \quad \text { and } \quad y=25 \\
& M=2^{21} \quad \text { gives } \quad 4^{16}+4^{21}=2^{21}\left(2^{21}+2^{11}\right), \quad \text { and } y=10
\end{aligned}
\]