3. (1987 13th All-Russian Mathematical Olympiad) Find all solutions to the equation $x^{2}-8[x]+7=0$. Here $[x]$ denotes the greatest integer not exceeding $x$.

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3. Let $x$ be the root of the given equation, and $[x]=n$. From the original equation, we get $x^{2}+7=8 n$. This indicates $n>0$. Using $n \leqslant x < n+1$, we have:
\[
\left\{\begin{array}{l}
n^{2}+7 \leqslant 8 n, \\
(n+1)^{2}+7 > 8 n.
\end{array}\right.
\]
Solving these, we get $1 \leqslant n < 2$ or $4 < n \leqslant 7$.

Thus, $n=1,5,6,7$.
For $n=1$, we get $x=1$,
For $n=5$, we get $x=\sqrt{33}$,
For $n=6$, we get $x=\sqrt{41}$,
For $n=7$, we get $x=7$.