
A1 Determine all integers $a, b, c$ satisfying the identities:

$$
\begin{gathered}
a+b+c=15 \\
(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540
\end{gathered}
$$


Solution I: 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

$$
x+y+z=0 \text {. }
$$

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:

$$
(a, b, c) \in\{(-1,0,16),(-2,1,16),(7,10,-2),(8,9,-2)\}
$$
