## Task 17/85

Determine all triples $(x ; y ; z)$ of natural numbers $x ; y ; z$ (where $0 \in N$), that satisfy the equation $5 x!=y!+z!+1$.
1. If $x=0$ or $x=1$, then $x!=1$ and the equation reduces to the equation $4=y!+z!$. This equation can be solved very quickly by trial: $y=z=2$. Thus, two triples are found: $\left(x_{1} ; y_{1} ; z_{1}\right)=(0 ; 2 ; 2)$ and $\left(x_{2} ; y_{2} ; z_{2}\right)=(1 ; 2 ; 2)$.
2. If $x>1$, then $x!$ contains the factor 2; thus, $5 x!\equiv 0(\bmod 10)$, and therefore $y!+z!\equiv 9(\bmod 10)$.

It is certain, therefore, that exactly one of the two summands $y!$ and $z!$ is odd. Due to the symmetry of the equation or the congruence in $y$ and $z$, we can assume without loss of generality that it is $y!$. Since $y!$ for $y>1$ contains the factor 2, it follows that $y!=1$ and thus $z!\equiv 8(\bmod 10)$.

However, this is not possible; because $2!=2,3!=6,4!=24$, and $x!\equiv 0(\bmod 10)$ for $x \geq 5$. From this, it follows that there are no further triples of the required type.