## Task 1 - 200921

Determine the largest prime number $p$ for which there exists a triple $(a, b, c)$ of natural numbers $a, b$, and $c$ such that $(a+p)(b+p)(c+p)=1980$ holds!

Determine all different triples $(a, b, c)$ corresponding to this prime number $p$ with the mentioned property!
Remark: In this solution, the set of natural numbers is assumed to be $\mathbb{N}=\{0,1,2, \ldots\}$, as the problem likely presupposes.

Let o.B.d.A. $a \leq b \leq c$ and thus also $a+p \leq b+p \leq c+p$.

$p$ is maximized when $a+p$ also takes its largest possible value. Since $13^{3}=169 \cdot 13=2197>1980$, it follows that $a+p<13$, so $a+p \leq 12$ must hold. We have $1980=2^{2} \cdot 3^{2} \cdot 5 \cdot 11$, so for $a+p=12=2^{2} \cdot 3$, we also have $(b+p)(c+p)=3 \cdot 5 \cdot 11$. However, any factorization of this product into two factors always has at least one factor that is at most 11, which contradicts the fact that $a+p$ is the smallest of the three factors $a+p, b+p, c+p$. Therefore, $a+p \leq 11$ must hold.

If $a+p=11$, then $(b+p)(c+p)=2^{2} \cdot 3^{2} \cdot 5$, which can be factored as $b+p=2^{2} \cdot 3=12 \geq a+p$ and $c+p=3 \cdot 5=15 \geq a+p$, allowing $p$ to be maximized at 11.

If $p$ is chosen to be maximal, i.e., $p=11$, then, under the continued assumption of the ordering of the variables $a, b$, and $c-a=0$ and $(b+p)(c+p)=2^{2} \cdot 3^{2} \cdot 5$. Then exactly one of the two factors $b+p$ or $c+p$ must be divisible by 5.

If this factor is also divisible by 3, then the remaining factor can have at most the factors $2^{2} \cdot 3=12$. Therefore, the factor divisible by 5 cannot contain any additional factors (and also no factor of 2), as otherwise the other factor would be a proper divisor of 12, and thus at most 6, which contradicts $a+p \leq b+p \leq c+p$.

Therefore, in this case, there is only one factorization into $12 \cdot 15$, so due to the ordering, $b+p=12$ and $c+p=15$, which means $(a, b, c)=(0,1,4)$.

If the factor divisible by 5 is not divisible by 3 but by 2, then it must also be divisible by the second factor 2 because $5 \cdot 2=10<11$. However, this results in the other factor being at most $3^{2}=9$, which is smaller than $a+p=11$, leading to a contradiction. A similar contradiction arises if the factor divisible by 5 is neither divisible by 2 nor 3, as it would then be exactly 5, and $5<11$.

Therefore, under the additional assumption $a \leq b \leq c$, there is exactly one triplet $(a, b, c)$ for the maximal possible $p=11$. If the additional assumption is dropped, all 6 possible permutations of this one triplet are solutions:

$$
(a, b, c) \in\{(0,1,4) ; \quad(0,4,1) ;(1,0,4) ;(1,4,0) ;(4,0,1) ;(4,1,0)\}
$$