3. Find all quadruples $a>b>c>d$ of integers with a sum of 71, which satisfy the equation

$$
(a-b)(c-d)+(a-d)(b-c)=26 .
$$

(Josef Tkadlec)
SOLUTION. We expand and simplify the left side of the given equation:

$$
\begin{aligned}
(a-b)(c-d)+(a-d)(b-c) & =(a c-b c-a d+b d)+(a b-b d-a c+c d)= \\
& =-b c-a d+a b+c d=(a-c)(b-d) .
\end{aligned}
$$

Thus, we have the equation $(a-c)(b-d)=26=2 \cdot 13$. From the condition $a>b>c$, it follows that the integers $a>c$ differ by at least 2, i.e., $a-c \geq 2$. Similarly, from $b>c>d$ it follows that $b-d \geq 2$. Therefore, in the equation $(a-c)(b-d)=2 \cdot 13$, the factors $a-c, b-d$ must equal the numbers 2 and 13 in some order (since 13 is a prime number). We will now analyze these two possibilities.

- $a-c=2$ and $b-d=13$. Then we have $a=c+2$ and $d=b-13$. The condition $a>b>c$ thus becomes $c+2>b>c$. Since, however, the only integer between $c+2$ and $c$ is $c+1$, it must be that $b=c+1$, and thus $d=b-13=c-12$. Our quadruple $(a, b, c, d)$ thus has the form $(c+2, c+1, c, c-12)$. From the condition $a+b+c+d=71$, it follows that $4c-9=71$ or $c=20$, and therefore $(a, b, c, d)=(22,21,20,8)$. Verification is not necessary.
- $b-d=2$ and $a-c=13$. Similarly to the first case, due to $b=d+2$, the condition $b>c>d$ becomes $d+2>c>d$, according to which $c=d+1$, and thus $a=c+13=d+14$. Our quadruple $(a, b, c, d)=(d+14, d+2, d+1, d)$, when substituted into the equation $a+b+c+d=71$, gives $4d+17=71$, from which $d=27/2$, which is not an integer. This case is thus excluded.

Conclusion. The only quadruple $(a, b, c, d)$ that satisfies the problem is $(22,21,20,8)$.

For a complete solution, award 6 points. In no solution should the absence of verification be penalized. In incomplete solutions, evaluate partial steps as follows.

A1. Decomposition of the left side of the equation into the product $(a-c)(b-d)$, or possibly $(c-a)(d-b)$: 2 points.

B1. Exclusion of cases where $\{a-c, b-d\}=\{1,26\}$: 1 point

B2. Solving one or both cases where $\{a-c, b-d\}=\{2,13\}$: 1 case 2 points, 2 cases 3 points.

B3. Observation that for integers $x, y, z$, from the relations $x>y>z$ and $x-z=2$ it follows that $y=z+1$: 1 point.

B4. Guessing the answer: 1 point.

In total, then award $\mathrm{A} 1+\max (\mathrm{B} 1, \mathrm{~B} 2, \mathrm{~B} 3, \mathrm{~B} 4)$ points.

[^0]: * This observation does not need to be proven - it follows from the fact that the opposite sides $K B$ and $L C$ are equal and parallel.