6. Determine all possible values of the sum $a+b+c+d$, where $a, b, c, d$ are natural numbers satisfying the equation

$$
\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)+\left(b^{2}-d^{2}\right)\left(c^{2}-a^{2}\right)=2021 .
$$

(Mária Dományová, Patrik Bak)
SOLUTION. Let's expand the left side of the given equation and further simplify:

$$
\begin{aligned}
\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right) & +\left(b^{2}-d^{2}\right)\left(c^{2}-a^{2}\right)= \\
& =\left(a^{2} c^{2}-a^{2} d^{2}-b^{2} c^{2}+b^{2} d^{2}\right)+\left(b^{2} c^{2}-a^{2} b^{2}-c^{2} d^{2}+a^{2} d^{2}\right)= \\
& =a^{2} c^{2}+b^{2} d^{2}-a^{2} b^{2}-c^{2} d^{2}=a^{2}\left(c^{2}-b^{2}\right)+d^{2}\left(b^{2}-c^{2}\right)= \\
& =\left(a^{2}-d^{2}\right)\left(c^{2}-b^{2}\right)=(a-d)(a+d)(c-b)(c+b) .
\end{aligned}
$$

The equation from the problem is thus equivalent to the equation

$$
(a-d)(a+d)(c-b)(c+b)=2021=43 \cdot 47 .
$$

The numbers $a, b, c, d$ are natural, so both factors $a+d$ and $c+b$ on the left side of the equation are integers greater than 1. However, since 43 and 47 are prime numbers, their product is divisible by the product of two numbers greater than 1 only if these numbers are 43 and 47 themselves. Therefore, our equation can only be satisfied in one way:

$$
\{a+d, c+b\}=\{43,47\} \quad \text { and } \quad(a-d)(c-b)=1 \text {. }
$$

Thus, it must be true that $a+b+c+d=90$. If we can show that the numbers $a, b, c, d$ satisfying the problem's conditions exist, we will be done with the solution.

Let's try to satisfy the equation as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_29a3dddc7a9db375e695g-16.jpg?height=127&width=778&top_left_y=1638&top_left_x=639)

We easily find that the (only) quadruple $(a, b, c, d)=(22,23,24,21)$ satisfies the conditions and is indeed composed of natural numbers.

Conclusion. The sum $a+b+c+d$ has only one possible value equal to the number 90.

Note. We can supplement our approach by finding all quadruples $(a, b, c, d)$ that satisfy the problem's conditions. Since the second part of the condition (1) means either $a-d=c-b=1$ or $a-d=c-b=-1$, we are looking for all solutions to four systems of equations:

$$
\begin{aligned}
& a+d=43 \quad a+d=47 \quad a+d=43 \quad a+d=47 \\
& a-d=1 \quad a-d=1 \quad a-d=-1 \quad a-d=-1 \\
& c+b=47 \quad c+b=43 \quad c+b=47 \quad c+b=43 \\
& c-b=1 \quad c-b=1 \quad c-b=-1 \quad c-b=-1
\end{aligned}
$$

Each of these systems has a unique solution $(a, b, c, d)$, which are the quadruples

$$
(22,23,24,21), \quad(24,21,22,23), \quad(21,24,23,22), \quad(23,22,21,24) .
$$

## ADDITIONAL AND COMPLEMENTARY PROBLEMS:

N1. For natural numbers $a, b, c, d$, it holds that $a b+b c+c d+d a=77$. Determine all possible values of their sum. [The only value is 18. We have $a b+b c+c d+d a=b(a+c)+d(a+c)=(a+c)(b+d)$. Since $77=7 \cdot 11, a+c>1, b+d>1$, it must be $\{a+c, b+d\}=\{7,11\}$, thus $a+b+c+d=18$, and the quadruple $(a, b, c, d)=(1,1,6,10)$ satisfies the conditions.]

N2. For natural numbers $a, b, c$, it holds that $a(a+b+c)+b c=143$. Determine all possible values of $|b-c|$. [The only value is 2. By expanding the left side of the equation, we get $a^{2}+a b+a c+b c$, which we can rearrange as follows: $a(a+b)+c(a+b)=(a+b)(a+c)$. Since $143=11 \cdot 13, a+b>1$ and $a+c>1$, it must be $\{a+b, a+c\}=\{11,13\}$, so $|b-c|=|(a+b)-(a+c)|=|11-13|=2$, and the triple $(a, b, c)=(1,10,12)$ satisfies the conditions.]

D1. In each cell of a $2 \times 2$ table, a natural number is written. If we add the product of the numbers in the first column, the product of the numbers in the second column, the product of the numbers in the first row, and the product of the numbers in the second row, we get the result 2021. a) Determine all possible values of the sum of all four numbers in the table. b) Find the number of tables satisfying the conditions that contain four distinct numbers. [a) The only value is 90, b) 3528. a) The numbers in the table can be denoted as $a, b, c, d$ such that $a b+c d+a c+b d=(a+d)(b+c)=2021=43 \cdot 47$. Therefore, $\{a+d, b+c\}=\{43,47\}$, so $a+b+c+d=43+47=90$. b) According to the solution in a), we distinguish two cases: $a+d=43$ and $b+c=47$, or $a+d=47$ and $b+c=43$. In the first case, the pair $(a, d)$ can be chosen in 42 ways and the pair $(b, c)$ in 46 ways, and in the second case, these numbers are reversed. Thus, there are $2 \cdot 42 \cdot 46$ different satisfying tables, including those with some numbers equal. We need to determine the number of such tables to subtract it from the total number. Since 43 and 47 are odd numbers, in each satisfying table, $a \neq d$ and $b \neq c$. In each table with some numbers equal, at least one of the equalities $a=b, a=c, d=b, d=c$ must hold, and due to $a+d \neq b+c$, exactly one of them (verify by considering all possible selections of two equalities). Each of these four equalities is satisfied by 42 satisfying tables in each of the two distinguished cases, so their total number is $2 \cdot 4 \cdot 42=8 \cdot 42$. The desired number is therefore $2 \cdot 42 \cdot 46-8 \cdot 42=3528$. (CPSJ 2021)]

D2. On each face of a cube, a natural number is written. For each of its vertices, the product of the three numbers on adjacent faces is written. The sum of the eight numbers at the vertices is 1001. Determine all possible values of the sum of the numbers on the faces. [31. If $(a, b)$ is a pair of numbers on the front and back faces, $(c, d)$ is a pair of numbers on the top and bottom faces,