Five, if positive integers $a, b, c$ satisfy $a^{2}+b^{2}=c^{2}$, then $(a, b, c)$ is called a Pythagorean triple. Find all Pythagorean triples that contain 30.
(Yang Chunhong, contributor)
Five, Solution 1 Since $30=2 \times 3 \times 5$, therefore, 30 has four even factors: $2,6,10,30$ and three odd factors: $3,5,15$.
(1) Let $30=2 \mathrm{mn}$.
Then $m n=15=1 \times 15=3 \times 5$.
By $m=15, n=1$, we do not get a basic group;
By $m=5, n=3$, we also do not get a basic group.
(2) Let $10=2 \mathrm{mn}$.
Then $m n=5=1 \times 5$.
By $m=5, n=1$, we know that $m$ and $n$ are both odd, so we do not get a basic group.
(3) Let $6=2 \mathrm{mn}$.
Then $m n=3=1 \times 3$.
By $m=3, n=1$, we know that $m$ and $n$ are both odd, so we do not get a basic group.
(4) Let $2=2 \mathrm{mn}$.
Then $m n=1=1 \times 1$.
By $m=n=1$, we do not get a basic group.
(5) Let $15=m^{2}+n^{2}$. Inspection modulo 4 shows no solution.
Again, let $15=m^{2}-n^{2}$. Then by
$m^{2}-n^{2}=15=15 \times 1$
$\Rightarrow m=8, n=7$, we get $(15,112,113)$.
Thus, the Pythagorean triple is $(30,224,226)$.
Again, by $m^{2}-n^{2}=15=5 \times 3$
$\Rightarrow m=4, n=1$, we get $(15,18,17)$.
Thus, the Pythagorean triple is $(30,16,34)$.
(6) Let $5=m^{2}+n^{2}$.
Then $m=2, n=1$, we get $(5,4,3)$.
Thus, the Pythagorean triple is $(30,24,18)$.
Again, by $m^{2}-n^{2}=5=5 \times 1$
$\Rightarrow m=3, n=2$, we get $(5,12,13)$.
Thus, the Pythagorean triple is $(30,72,78)$.
(7) Let $3=m^{2}+n^{2}$. Inspection modulo 4 shows no solution.
Again, let $3=m^{2}-n^{2}$, then by
$m^{2}-n^{2}=3=3 \times 1$
$\Rightarrow m=2, n=1$, we get $(3,4,5)$.
Thus, the Pythagorean triple is $(30,40,50)$.
In summary, there are 5 Pythagorean triples that meet the requirements:
$(30,224,226),(30,16,34),(30,24,18)$,
$(30,72,78),(30,40,50)$.
Solution 2 Let the other two sides of the right-angled triangle be $a 、 b(a>b)$. Then
(1) $a^{2}+b^{2}=30^{2}$;
(2) $a^{2}-b^{2}=30^{2}$.
It is easy to see that (1) has only one solution $(a, b)=(24,18)$.
For (2), we have
$(a+b)(a-b)=2^{2} \times 3^{2} \times 5^{2}$.
Since $a+b$ and $a-b$ are both even, and $a+b > a-b$, therefore,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2 \times 3^{2} \times 5^{2}, \\
a-b=2 ;
\end{array}\right. \\
\left\{\begin{array}{l}
a+b=2 \times 3 \times 5^{2}, \\
a-b=2 \times 3 ;
\end{array}\right. \\
\left\{\begin{array}{l}
a+b=2 \times 3^{2} \times 5, \\
a-b=2 \times 5 ;
\end{array}\right. \\
\left\{\begin{array}{l}
a+b=2 \times 5^{2}, \\
a-b=2 \times 3^{2} .
\end{array}\right.
\end{array}
$$

And each of them has one solution, hence there are 5 Pythagorean triples that meet the requirements:
$$
\begin{array}{l}
(30,224,226),(30,72,78),(30,40,50), \\
(30,16,34),(30,24,18) .
\end{array}
$$