, 2015

For integers $a$ and $b$ and for a positive integer $c$, let us write $a \equiv b \pmod c$ if $a-b$ is divisible by $c$. We shall show that $n=3, 7$ are the answer we seek for the problem. Since
$$\frac{10^3}{3^3+3^2+3+1} = 25, \quad \frac{10^7}{7^3+7^2+7+1} = 25000,$$
we see that $n=3, 7$ satisfy the condition of the problem. We shall show in the sequel that there are no other values of $n$ satisfying the condition of the problem.
Since $n^3 + n^2 + n + 1 = (n+1)(n^2 + 1)$, $n+1$ and $n^2 + 1$ do not have prime factors other than 2 and 5 if $n$ satisfies the condition of the problem. Also from the fact that $(n^2 + 1) - (n+1)(n-1) = 2$ it follows that the greatest common divisor $(\text{gcd})$ of $n^2 + 1$ and $n + 1$ is either 1 or 2. If $n$ is even, then both $n+1$ and $n^2 + 1$ are odd, therefore must be divisible by 5, which contradicts the fact that the $(\text{gcd}\{n+1, n^2+1\}) = 1$ or 2. Hence we can assume in the sequel that $n$ is odd. Then, $n+1$ and $n^2 + 1$ are both even. Furthermore, since $n^2 + 1 \equiv 2 \pmod 4$, $n^2 + 1$ is not divisible by 4. In the sequel, we consider various cases separately.

(1) When $n^2 + 1$ is not divisible by 5:
$n^2 + 1$ is a power of 2, but since it is not divisible by 4, it must be the case that $n^2 + 1 = 2$, i.e., $n=1$. But since $\frac{10^1}{1^3+1^2+1+1} = \frac{5}{2}$ is not an integer, we see that $n=1$ does not satisfy the requirement of the problem.

(2) When $n^2 + 1$ is divisible by 5:
Since $n > 1$ and $n^2 + 1$ is not divisible by 4, we can represent $n+1 = 2^k$, $n^2 + 1 = 2 \cdot 5^l$, where $k, l$ are positive integers and $k \ge 2$. If $k=2$, then we must have $n=3$. Let us assume in the sequel that $k \ge 3$ holds. From $2 \cdot 5^l = (2^k - 1)^2 + 1$ it follows that $5^l - 1 = 2^k(2^{k-1} - 1)$. Consequently, $5^l - 1$ is a multiple of 8. If $l$ is even, we have $5^l \equiv 1 \pmod 8$, while if $l$ is odd, $5^l \equiv 5 \pmod 8$; therefore, $l$ must be even. If we put $l=2m$, where $m$ is a positive integer, we get $(5^m - 1)(5^m + 1) = 2^k(2^{k-1} - 1)$. Since $5^m + 1 \equiv 2 \pmod 4$, we can write $5^m - 1 = 2^{k-1}a$, where $a$ is a positive odd number. Then, from $2^{k-1}a(2^{k-1}a + 2) = 2^k(2^{k-1} - 1)$, it follows that $a(2^{k-2}a + 1) = 2^{k-1} - 1$ holds. If we let $a \ge 3$, then we get $a(2^{k-2}a + 1) > 2k - 2 \cdot 2 + 1 > 2^{k-1} - 1$, which is a contradiction. Thus, we must have $a=1$. In this case, we get $2^{k-1}(2^{k-1} + 2) = 2^k(2^{k-1} - 1)$, which implies that $2^{k-1} = 4$ so that $k=3$. Consequently, we have $n=7$.

In summarizing above, we conclude that $n=3, 7$ are the numbers that satisfy the condition of the problem.