## Task 24/88

The following system of equations is to be solved. Here, $n_{1}$ and $n_{2}$ are natural numbers, and $p_{i} (i=1,2, \ldots, 6)$ are prime numbers.

$$
\begin{array}{r}
n_{1}=p_{1}^{2} p_{2} p_{3} \\
n_{2}=n_{1}+1=p_{4}^{2} p_{5} p_{6} \\
p_{2}=p_{3}-p_{1}^{6} \\
p_{3}=p_{1} p_{4}^{3}+p_{6} \\
p_{5}=p_{2}+p_{1} p_{4} \\
p_{6}=p_{1}^{2}+p_{2}+p_{1} p_{4}
\end{array}
$$
From equations (1) and (2), it follows that one of the six prime numbers must be even. From (3), it follows that this can only be $p_{1}$ or $p_{2}$, and from (4) it follows that this is $p_{1}$: $p_{1}=2$. Substituting this into equations (3)...(6), we get

$$
\begin{gathered}
p_{2}=p_{3}-64 \\
p_{5}=p_{2}+2 p_{4}
\end{gathered}
$$

Using (4a), we eliminate $p_{3}$ from (3a):

$$
p_{2}=2 p_{4}^{3}+p_{6}-64
$$

Using (3b), we eliminate $p_{2}$ and $p_{6}$ from (6a); after rearrangement, we get

$$
p_{4}\left(p_{4}^{2}+1\right)=30
$$

The prime factors of 30 are 2, 3, and 5; by trial, we quickly find $p_{4}=3$, and the system takes the following form:

$$
\begin{array}{ll}
p_{2}=p_{3}-64 \\
p_{5}=p_{2}+6
\end{array}
$$

By substituting (1) into (2), we get $n_{2}=p_{1}^{2} p_{2} p_{3}+1=p_{4}^{2} p_{5} p_{6}$ (2b) with the already found values for $p_{1}$ and $p_{4}$, and by substituting (3c)...(6c) into (2b), we get the quadratic equation in $p_{2}$

$$
p_{2}^{2}-22.4 p_{2}+107.8=0
$$

with the only integer solution $p_{2}=7$. From (3c), we then get $p_{3}=71$, from (5c) $p_{5}=13$, and from (6c) or (4c) $p_{6}=17$. Finally, we get from the found 6 prime numbers in (1) and (2)

$$
n_{1}=2^{2} \cdot 7 \cdot 71=1988, \quad n_{2}=3^{2} \cdot 13 \cdot 17=1989=n_{1}+1
$$

### 2.29 Problems and Solutions 1989