\section*{Problem 1 - 141041}

Let

\[
z=\left(1-\frac{4}{1^{2}}\right) \cdot\left(1-\frac{4}{3^{2}}\right) \cdot\left(1-\frac{4}{5^{2}}\right) \cdot \ldots \cdot\left(1-\frac{4}{199^{2}}\right)
\]

Express the rational number \(z\) in the form \(z=\frac{p}{q}\), where \(p, q\) are integers, \(p\) and \(q\) are coprime, and \(q>0\)!
}

It is

\[
1-\frac{4}{(2 k-1)^{2}}=\frac{4 k^{2}-4 k-3}{(2 k-1)^{2}}=\frac{2 k-3}{2 k-1} \cdot \frac{2 k+1}{2 k-1}
\]

From this it follows that

\[
\prod_{k=1}^{n}\left(1-\frac{4}{(2 k-1)^{2}}\right)=\prod_{k=1}^{n} \frac{2 k-3}{2 k-1} \cdot \frac{2 k+1}{2 k-1}=\frac{-1}{1} \cdot \frac{3}{1} \cdot \frac{1}{3} \cdot \frac{5}{3} \cdot \frac{3}{5} \cdot \frac{7}{5} \cdot \ldots \cdot \frac{2 n-3}{2 n-1} \cdot \frac{2 n+1}{2 n-1}=-\frac{2 n+1}{2 n-1}
\]

Now \(\operatorname{gcd}(2 n+1,2 n-1)=1\), which follows, for example, easily using the Euclidean algorithm:

\(2 n+1=1 \cdot(2 n-1)+2\)

\(2 n-1=(n-1) \cdot 2+1\)

\(2=2 \cdot 1+0\)

The product value \(z\) follows with \(n=100: z=-\frac{2 \cdot 100+1}{2 \cdot 100-1}=-\frac{201}{199}\)

Second solution:

According to the problem,

\[
\begin{aligned}
z & =\frac{1-4}{1^{2}} \cdot \frac{3^{2}-2^{2}}{3^{2}} \cdot \frac{5^{2}-2^{2}}{5^{2}} \ldots \frac{197^{2}-2^{2}}{197^{2}} \cdot \frac{199^{2}-2^{2}}{199^{2}} \\
& =-3 \cdot \frac{(3-2)(3+2)}{3^{2}} \cdot \frac{(5-2)(5+2)}{5^{2}} \ldots \frac{(197-2)(197+2)}{197^{2}} \cdot \frac{(199-2)(199+2)}{199^{2}} \\
& =-3 \cdot \frac{1 \cdot 5}{3^{2}} \cdot \frac{3 \cdot 7}{5^{2}} \ldots \cdot \frac{195 \cdot 199}{197^{2}} \cdot \frac{197 \cdot 201}{199^{2}}
\end{aligned}
\]

In the numerators, the first factors run through the sequence of odd numbers from 1 to 197, and the second factors run through the sequence of odd numbers from 5 to 201. Thus,

\[
z=\frac{-1 \cdot 3^{2} \cdot 5^{2} \cdot \ldots \cdot 197^{2} \cdot 199 \cdot 201}{1 \cdot 3^{2} \cdot 5^{2} \cdot \ldots \cdot 195^{2} \cdot 197^{2} \cdot 199^{2}}
\]

and thus \(z=-\frac{201}{199}\).

Taken from \([2]\)