## Task 3 - 260923

Determine all triples $(a ; b ; c)$ of natural numbers $a, b, c$ that satisfy the following conditions (1) to (5)!

(1) It holds that $b<c$.

(2) $b$ and $c$ are coprime.

(3) $a$ is different from each of the numbers 4, 9, 12.

(4) $b$ and $c$ are different from each of the numbers 13, 16, 21.

(5) Every number that is the sum of two different numbers from the set $A=\{4 ; 9 ; 12 ; a\}$ is contained in the set $B=\{13 ; 16 ; 21 ; b ; c\}$.
After (5), the sums $4+9=13, 4+12=16, 9+12=21, 4+a, 9+a$ and $12+a$ are all contained in $B$. Since according to (3) $a$ is different from 4, 9, and 12, there is a duplicate in this list of six addition problems. Since $B$ contains at most five elements, at least two must have the same result. According to (4), $b$ and $c$ are different from the other elements of $B$ and, according to (1), also from each other, so $B$ actually has exactly five elements, and thus one of the three above sums, which contain $a$ as a summand, must take an already existing value of 13, 16, or 21.

Case 1: $4+a \in \{13; 16; 21\}$. Then $4+a$ cannot be 13 or 16, as otherwise $a=9$ or $a=12$ would follow; in contradiction to (3). Therefore, $a+4=21$ and thus $a=17$ must hold. Then $a+9=26=b<a+12=29=c$, which also satisfies (2), so a first solution triplet $(a, b, c)=(17,26,29)$ is obtained.

Case 2: $9+a \in \{13; 16; 21\}$. Then $9+a$ cannot be 13 or 21, as otherwise $a=4$ or $a=12$ would follow; in contradiction to (3). Therefore, $a+9=16$ and thus $a=7$ must hold. Then $a+4=13=b<a+12=19=c$, which also satisfies (2), so a second solution triplet $(a, b, c)=(7,13,19)$ is obtained.

Case 3: $12+a \in \{13; 16; 21\}$. Then $12+a$ cannot be 16 or 21, as otherwise $a=4$ or $a=9$ would follow; in contradiction to (3). Therefore, $a+12=13$ and thus $a=1$ must hold. Then $a+4=5=b<a+9=10=c$. This contradicts (2), so there is no solution triplet in this case.

In summary, there are exactly two solution triplets that satisfy the problem statement, namely $(7,13,19)$ and $(17,26,29)$.