, 2009

Solution:
We will use the following fact:

Lemma: If $x, y, z$ are integers such that
$$x+y+z=0$$
then
$$x^{3}+y^{3}+z^{3}=3 x y z$$
Proof: Let
$$x+y+z=0.$$
Then we have
$$x^{3}+y^{3}+z^{3}=x^{3}+y^{3}+(-x-y)^{3}=x^{3}+y^{3}-x^{3}-y^{3}-3 x y(x+y)=3 x y z$$
Now, from
$$a+b+c=15$$
we obtain:
$$(a-3)+(b-5)+(c-7)=0$$
Using the lemma and the given equations, we get:
$$540=(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=3(a-3)(b-5)(c-7)$$
Now,
$$(a-3)(b-5)(c-7)=180=2 \times 2 \times 3 \times 3 \times 5$$
Since
$$(a-3)+(b-5)+(c-7)=0$$
only possibility for the product $(a-3)(b-5)(c-7)$ is $(-4) \times(-5) \times 9$. Finally, we obtain the following systems of equations:
$$\left\{\begin{array} { l } { a - 3 = - 4 } \\ { b - 5 = - 5 } \\ { c - 7 = 9 , } \end{array} \quad \left\{\begin{array} { l } { a - 3 = - 5 } \\ { b - 5 = - 4 } \\ { c - 7 = 9 , } \end{array} \quad \left\{\begin{array} { l } { a - 3 = - 4 } \\ { b - 5 = - 5 } \\ { c - 7 = 9 , } \end{array} \quad \left\{\begin{array}{l} a-3=-5 \\ b-5=-4 \\ c-7=9 \end{array}\right.\right.\right.\right.$$
From here we get:
$$(a, b, c) \in\{(-1,0,16),(-2,1,16),(7,10,-2),(8,9,-2)\}$$

Solution II:
We use the substitution $a-3=x, b-5=y, c-7=z$.
Now, equations are transformed to:
$$\begin{aligned} x+y+z & =0 \\ x^{3}+y^{3}+z^{3} & =540 . \end{aligned}$$
Substituting $z=-x-y$ in second equation, we get:
$$-3 x y^{2}-3 x^{2} y=540$$
or
$$x y(x+y)=-180$$
or
$$x y z=180.$$
Returning to starting problem we have:
$$(a-3)(b-5)(c-7)=180$$
Solution proceeds as the previous one.