## Task 18/78

Determine the greatest common divisor of all numbers $z$ that can be represented in the form $z=n^{4 m+1}-n$ with $m ; n \in N$!
For $n=1, \mathrm{~m}$ arbitrary, $c=0$; the smallest non-trivial number $z$ is obtained for $n=2, m=1$ as $z=2^{5}-2=30$. Therefore, the greatest common divisor sought can be at most $30=2 \cdot 3 \cdot 5$. It is now necessary to investigate which of these three prime factors $2 ; 3 ; 5$ are divisors of $z$ for all $n$ and $m$. To do this, we write $z$ in the form

$$
z=n\left(n^{4 m}-1\right)
$$

Obviously, the prime factor 2 is contained in every $z$; either $n$ is even, or $n^{4 m}-1$ is even. Either $n \equiv 0(\bmod 3)$, or $n \equiv \pm 1(\bmod 3)$. In the latter case, $n^{2} \equiv 1(\bmod 3)$, thus $n^{4 m} \equiv 1(\bmod 3)$ and therefore $n^{4 m}-1 \equiv 0(\bmod 3)$. This means that in every case, the prime factor 3 is contained in $z$.

Similarly, we conclude: Either $n \equiv 0(\bmod 5)$, or $n \equiv \pm 1(\bmod 5)$ or $\equiv \pm 2(\bmod 5)$, so $n^{2} \equiv \pm 1(\bmod 5), n^{4 m} \equiv 1(\bmod 5)$ and thus $n^{4 m}-1 \equiv 0(\bmod 5)$. Therefore, the prime factor 5 is also contained in every $z$. It follows that the greatest common divisor of all numbers $z$ that can be represented in the form $z=n^{4 m+1}-n$ with $n ; m \in N$ is the number 30.