\section*{Problem 4 - 121214}

All ordered pairs \((x, y)\) of positive integers \(x\) and \(y\) (\(x \leqslant y\)) are to be determined for which the equation \(\sqrt{x}+\sqrt{y}=\sqrt{1980}\) is satisfied.
Assume \(x\) and \(y\) are such positive integers that

\[
\sqrt{x}+\sqrt{y}=\sqrt{1980}=6 \sqrt{5 \cdot 11}
\]

Then it follows that

\[
x+y+2 \sqrt{x y}=36 \cdot 5 \cdot 11
\]

Let \(t\) be the greatest common divisor of \(x\) and \(y\); then \(x=t u\) and \(y=t v\), where \(u\) and \(v\) are positive integers with the greatest common divisor 1, and \(u \leq v\).

Furthermore, because of (2), \(\sqrt{x y}=t \sqrt{u v}\) is a positive integer, which, due to the coprimality of \(u, v\), implies \(u=u_{1}^{2}, v=v_{1}^{2}\), where \(u_{1}\) and \(v_{1}\) are coprime positive integers with \(u_{1} \leq v_{1}\). From (2) it follows that

\[
t\left(u_{1}^{2}+v_{1}^{2}+2 u_{1} v_{1}\right)=t\left(u_{1}+v_{1}\right)^{2}=36 \cdot 5 \cdot 11
\]

Thus, only one of the following four cases can hold:

a) \(t=36 \cdot 5 \cdot 11, u_{1}+v_{1}=1\);

b) \(t=9 \cdot 5 \cdot 11, u_{1}+v_{1}=2\);

c) \(t=4 \cdot 5 \cdot 11, u_{1}+v_{1}=3\);

d) \(t=5 \cdot 11, u_{1}+v_{1}=5\).

Case a) is not satisfiable by \(u_{1}, v_{1}\) with the given properties,

Case b) is only satisfiable by \(u_{1}=1, v_{1}=1\) and thus \(x=495, y=495\),

Case c) is only satisfiable by \(u_{1}=1, v_{1}=2\) and thus \(x=220, y=880\),

Case d) is only satisfiable by \(u_{1}=1, v_{1}=5\) and thus \(x=55, y=1375\).

Therefore, only these values can have the required properties. A check shows that they indeed have these properties.

Thus, the only ordered pairs of positive integers that are solutions to the given equation are: \((55,1375),(220,880),(495,495)\).

\section*{Adapted from \([5]\)}

\section*{Second Solution:}

Assume \((x, y)\) is a solution to the problem. Then

\[
\sqrt{x}+\sqrt{y}=\sqrt{1980} \quad \text { or } \quad x+y+2 \sqrt{x y}=1980
\]

Thus,

\[
2 \sqrt{x y}=1980-(x+y) \quad \text { or } \quad 4 x y=(1980-(x+y))^{2}
\]
and thus

\[
0=(1980-(x+y))^{2}-4 x y=(1980-(x-y))^{2}-4 \cdot 1980 y
\]

from which we get

\[
4 \cdot 1980 y=(1980-(x-y))^{2}
\]

Since \(4 \cdot 1980 y=12^{2} \cdot 5 \cdot 11\), it follows from (2) that

\[
y=55 n^{2}
\]

where \(n\) is a natural number. From (1) it follows that

\[
\sqrt{x}=6 \sqrt{55}-n \sqrt{55}=m \cdot \sqrt{55}
\]

where \(m\) is a natural number. Therefore, the solution has the form \(\left(55 m^{2}, 55 n^{2}\right)\) with \(m+n=6\) and (because \(0<x \leq y\)) \(0<m \leq n\). This implies \(1 \leq m \leq 3\).

Thus, the pairs (55, 1375), (220, 880), (495, 495) can be the only solutions. As one can verify by substitution, they are indeed all three solutions.

Adapted from [2]

\subsection*{9.14.2 Second Round 1972, Class 12}