5. (Austria-Poland Olympiad 2000) Find all positive integers $N$, such that $N$ has only the prime factors 2 and 5, and $N+25$ is a perfect square.
5. Let $N+25=(x+5)^{2}$ where $x \in \mathbf{N}^{*}$, then $N=x(x+10)$. According to the problem, we can set
$$
\left\{\begin{array}{l}
x=2^{a_{1}} \cdot 5^{\beta_{1}} \\
x+10=2^{a_{2}} \cdot 5^{\beta_{2}}
\end{array},\right.
$$

where $\alpha_{i}, \beta_{i}$ are non-negative integers. Thus, $2^{\alpha_{2}} \cdot 5^{\beta_{2}}-2^{\alpha_{1}} \cdot 5^{\beta_{1}}=10$.
If $\alpha_{1}=\alpha_{2}=0$ then $5 \beta_{2}-5 \beta_{1}=10$ is impossible. Therefore, at least one of $\alpha_{1}, \alpha_{2}$ is greater than 0. Hence, $x$ is even. Thus, $\alpha_{1}$, $\alpha_{2} \in \mathbf{N}^{*}$, and when $\alpha_{1} \geqslant 2$, we know: $\alpha_{2}=1$; when $\alpha_{2} \geqslant 2$, $\alpha_{1}=1$. Therefore, at least one of $\alpha_{1}, \alpha_{2}$ is 1.
Similarly, we can conclude: $\beta_{1}, \beta_{2} \in \mathbf{N}^{*}$ and at least one of $\beta_{1}, \beta_{2}$ is 1.
Case 1: $\alpha_{1}=1$, in this case, if $\beta_{1}=1$ then $x=10 . x+10=20=2^{2} \times 5$ yielding a valid $N=200$. If $\beta_{2}=1$ then from (1) we get $2^{\alpha_{2}-1}-5^{\beta_{1}-1}=1$ which requires $\alpha_{2}-1 \geqslant 3$. Modulo 4 on both sides, we get $-1 \equiv 1(\bmod 4)$, a contradiction.
Case 2: $\alpha_{2}=1$, in this case $\beta_{2}>1$, so $\beta_{1}=1$. Thus from (1) we get: $5^{\beta_{2}-1}-2^{a_{1}-1}=1$.
Modulo 5 on both sides, we find: $\alpha_{1}-1$ is even. Let $\alpha_{1}-1=2 m$
Then for $m=1$, $\alpha_{1}=3, \beta_{2}=2, N=2000$.
When $m>1$, modulo 8 on both sides of (2) shows: $\beta_{2}-1$ is even. Let $\beta_{2}-1=2 n$ leading to $\left(5^{n}-5^{m}\right)\left(5^{n}+5^{m}\right)=1$. The left side is clearly greater than the right side, a contradiction.
In summary, the valid $N$ values are 200 or 2000.