Ok. When will you stop bragging about your latex?

$$ab+a+b=524$$
$$bc+b+c=146$$
$$cd+c+d=104$$

The first thing we notice is that the left side of all the equations are written almost in a quadratic format. We just need to factor it.

But when we factor, we see that we need a one on all of the equations.

To include a one on one side of the equation, we have to include a one on the other side.

Adding a one on all of the equations will give us:

$$ab+a+b+1=525$$
$$bc+b+c+1=147$$
$$cd+c+d+1=105$$

Now we can factor things out.


$$(a+1)(b+1)=525$$
$$(b+1)(c+1)=147$$
$$(c+1)(d+1)=105$$


Now this may take us a bit of guess and check work. But we can make it easier by finding the prime factorization of all the numbers so we can find common divisors.

$$525$$
$$5\cdot 105$$
$$5\cdot 5\cdot 21$$
$$5\cdot 5\cdot 3\cdot 7$$
$$5^2\cdot 3\cdot 7$$

$$147$$
$$7\cdot 21$$
$$7\cdot 3\cdot 7$$
$$7^2\cdot 3$$

$$105$$
$$5\cdot 21$$
$$5\cdot 3\cdot 7$$


At one point of all the equations, we notice that there will be a $21$.
Since 21 is the most common term, we have to make 21 be a factor of more than one equation. For example, if we make $a$ become 20, the factor will become 21 since we add one. Now we only have a 21 in one equation since $a$ doesn't appear in any other equation. But two variables, $c$ and $b$ do appear in more than one equation, so that means one of them has to be 20.


$$b=20$$
Plugging $b$ in for the first equation, we will get
$$(a+1)(20+1)=525$$
$$(a+1)21=525$$
$$\rightarrow \div 21$$
$$a+1=25$$
$$\rightarrow -1$$
$$a=24$$

So that means that $a$ must be $24$ if $b$ is $20$.

Plugging in $b$ for the second equation, we will get
$$(20+1)(c+1)=147$$
$$21(c+1)=147$$
$$\rightarrow \div 21$$
$$c+1=7$$
$$\rightarrow -1$$
$$c=6$$

And $c$ must be $6$.

Now let's plug in $c$ for the third equation.
$$(6+1)(d+1)=105$$
$$7(d+1)=105$$
$$\rightarrow \div 7$$
$$d+1=15$$
$$\rightarrow -1$$
$$d=14$$

Now let's collect all of our variables that we found.
$$a=24$$
$$b=20$$
$$c=6$$
$$d=14$$

Don't forget that there is a fourth equation: $abcd=8!$

To see if this is true, we can first plug the four variables in

$$24\cdot 20\cdot 6\cdot 14=8!$$

Let's split things up now.

$$(2\cdot 12)\cdot (2\cdot 10)\cdot (2\cdot 3)\cdot (2\cdot 7)=8!$$
$$(2\cdot 2\cdot 6)\cdot (2\cdot 2\cdot 5)\cdot (2\cdot 3)\cdot (2\cdot 7)=8!$$
$$(2\cdot 2\cdot 2\cdot 3)\cdot (2\cdot 2\cdot 5)\cdot (2\cdot 3)\cdot (2\cdot 7)=8!$$
Using communicative property, we can get:
$$(2)\cdot (3)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (7)\cdot (2\cdot 2\cdot 2)=8!$$
$$2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8=8!$$
So it's true!


Wrapping things up, we see that our four variables are:
$$\bold{\boxed{a=24;b=20;c=6;d=14}}$$