Find all quadruples $(a, b, c, d)$ of positive integers which satisfy

$$
\begin{aligned}
a b+2 a-b & =58 \\
b c+4 b+2 c & =300 \\
c d-6 c+4 d & =101
\end{aligned}
$$
Consider the first equation, $a b+2 a-b=58$.

Factoring $a$ out of the first two terms, we get $a(b+2)-b=58$.

If the left side of this equation were instead $a(b+2)-b-2$, we could further factor the left side of the equation to $(a-1)(b+2)$.

To achieve this, we add -2 to both sides of the first equation to get

$$
(a-1)(b+2)=58-2=56=2^{3} \times 7 .
$$

In a similar way, we can add 8 and -24 to both sides of the second and third equations, respectively and factor to get

$$
\begin{aligned}
& (a-1)(b+2)=56 \\
& (b+2)(c+4)=308 \\
& (c+4)(d-6)=77
\end{aligned}
$$

We are given that $c$ and $d$ are positive integers, which means $c+4$ is a positive integer and $d-6$ is an integer.

From (3) and since $c$ is positive, we can see that $c+4$ is a positive divisor of $77=7 \times 11$.

This means that $c+4$ is one of $1,7,11$, and 77 .

We cannot have $c+4=1$ since this would mean $c=-3$, which is negative.

If $c+4=7$, then $c=7-4=3$ and rearranging (2) gives

$$
b+2=\frac{308}{c+4}=\frac{308}{7}=44
$$

Rearranging (1), this leads to

$$
a-1=\frac{56}{b+2}=\frac{56}{44}=\frac{14}{11}
$$

which is not an integer, but $a-1$ must be an integer because $a$ is an integer.

Therefore, we cannot have $c+4=7$.

If $c+4=11$, then (3) implies

$$
d-6=\frac{77}{c+4}=\frac{77}{11}=7
$$

which means $d=7+6=13$.

From (2),

$$
b+2=\frac{308}{c+4}=\frac{308}{11}=28
$$

so $b=28-2=26$.

Finally, using (1), we get

$$
a-1=\frac{56}{b+2}=\frac{56}{28}=2,
$$

so $a=1+2=3$.

Therefore, $(a, b, c, d)=(3,26,7,13)$ is one possible solution.

If $c+4=77$, then similar calculations to those above lead to $d-6=\frac{77}{77}=1, b+2=\frac{308}{77}=4$, and $a-1=\frac{56}{4}=14$.

Therefore, $c=77-4=73, d=1+6=7, b=4-2=2$, and $a=14+1=15$.

Thus, the only other solution is $(a, b, c, d)=(15,2,73,7)$.

ANSWER: $(3,26,7,13),(15,2,73,7)$

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