## Task A-2.6.

Determine all ordered triples of integers $(a, b, c)$ for which

$$
\left\{\begin{array}{l}
a^{2}-2 a b+c^{2}=a c-13 \\
b^{2}+a c=23
\end{array}\right.
$$
## Solution.

If we add the given equations, we get:

$$
(a-b)^{2}+c^{2}=10
$$

Since $a-b$ and $c$ are integers, from the previous equality, we have that the sum of the squares of two integers is equal to 10. The only two such squares are 1 and 9 (and 9 and 1), so we conclude that $c \in\{-3,-1,1,3\}$.

If $c \in\{-3,3\}$, the expression $ac$ in the second equation is divisible by 3. Therefore, the expression $b^{2}$ must give a remainder of 2 when divided by 3 (since the number 23 also gives such a remainder). This is not possible for any square of an integer, so there is no solution in this case.

If $a-b=-3$, from the second equation we have $b^{2}-b-20=0$. The solutions to this quadratic equation are $b_{1}=-4, b_{2}=5$, which gives us candidates for solutions:

$$
(a, b, c)=(-7,-4,-1) \quad \mathrm{and} \quad(a, b, c)=(2,5,-1)
$$

If $c=1$ we again have $(a-b)^{2}=9$, i.e., $a-b= \pm 3$. If $a-b=-3$, the second equation of the system gives $b^{2}+b-26=0$ which has no integer solutions.

If $a-b=3$, we get $b^{2}+b-20=0$. The solutions to this quadratic equation are $b_{1}=-5, b_{2}=4$, which gives us new candidates for solutions:

$$
(a, b, c)=(-2,-5,1) \quad \mathrm{and} \quad(a, b, c)=(7,4,1)
$$

By direct verification, we see that all four triples

$$
(a, b, c) \in\{(-7,-4,-1),(2,5,-1),(-2,-5,1),(7,4,1)\}
$$

are indeed the solutions we are looking for.