China Mathematical Olympiad

We will discuss the following three cases for $n$ and $m$.

(i) In the case when $n<m$, from $a^n+203\ge a^m+1$, we have $$202\ge a^m-a^n\ge a^n(a-1)\ge a(a-1).$$ Therefore, $2\le a\le 14$.

When $a=2$, we can take $n$ to be $1,2,\dots ,7$. When $a=3$, we can take $n$ to be $1,2,3$ and $4$. When $a=4$, we can take $n$ to be $1,2$ and $3$. When $5\le a\le 6$, we can take $n$ to be $1$ and $2$. When $7\le a\le 14$, $n=1$. From $a^m+1\mid a^n+203$, we can deduce that the solutions are $(2,2,1),(2,3,2)$ and $(5,2,1)$.

(ii) In the case when $n=m$, $a^m+1\mid 202$. Since 202 has only four divisors 1, 2, 101 and 202, but $a\ge 2$, $m\ge 2$ and $a^m+1\ge 5$, so $a^m=100$ or 201. Since $m\ge 2$, therefore the solution is $(10,2,2)$.

(iii) In the case when $n>m$, from $a^m+1\mid 203(a^m+1)$, we have $$a^m+1\mid a^n+203-(203a^m+203),$$ that is, $$a^m+1\mid a^m(a^{n-m}-203).$$ Since $(a^m+1,a^m)=1$, so $$a^m+1\mid a^{n-m}-203.$$

① If $a^{n-m}<203$, then set $n-m=s\ge 1$, we have $a^m+1\mid 203-a^s$. Hence $$\begin{gathered} 203-a^s\ge a^m+1, \\ 202\ge a^s+a^m\ge a^m+a \\ =a(a^{m-1}+1)\ge a(a+1), \end{gathered}$$ thus $$2\le a\le 13.$$ Using the same argument as in Case (i), we show that the solutions for $(a,m,s)$ are: $(2,2,3),(2,6,3),(2,4,4),(2,3,5),(2,2,7),(3,2,1),(4,2,2),(5,2,3)$ and $(8,2,1)$. Hence, $(a,m,n)$ are $(2,2,5),(2,6,9),(2,4,8),(2,3,8),(2,2,9),(3,2,3),(4,2,4),(5,2,5)$ and $(8,2,3)$.

② If $a^{n-m}=203$, then $a=203$ and $n-m=1$, and the solution is $(203,m,m+1)$, $m\ge 2$.

③ If $a^{n-m}>203$, set $n-m=s\ge 1$, then $a^m+1\mid a^s-203$. Since $a^s-203\ge a^m+1$, so $s>m$. By $$\begin{gathered} a^m+1\mid a^s+203a^m=(a^{s-m}+203)a^m \\ =(a^{n-2m}+203)a^m, \end{gathered}$$ and $(a^m+1,a^m)=1,$ we have $a^m+1\mid a^{n-2m}+203.$ Now $s>m\iff n-m>m\iff n>2m\iff n-2m>0$. In this case, the solutions can only be derived from the preceding solutions, that is, from $(a,m,n)\to (a,m,n+2m)\to \cdots \to (a,m,n+2km)$, and each solution derived also satisfies $a^m+1\mid a^n+203$.

Summarizing what described above, we obtain all solutions $(a,m,n)$ to be: $(2,2,4k+1)$, $(2,3,6k+2)$, $(2,4,8k+8)$, $(2,6,12k+9)$, $(3,2,4k+3)$, $(4,2,4k+4)$, $(5,2,4k+1)$, $(8,2,4k+3)$, $(10,2,4k+2)$ and $(203,m,(2k+1)m+1)$, where $k$ is any nonnegative integer, and $m\ge 2$ is an integer.