Sum of Integers

$n^3+n^2+n+1=(n+1)(n^2+1)$ Note that if $n$ be an even number, then $n+1$ and $n^2+1$ will be two odd co-prime numbers, but $10^n$ contains only one odd prime and $(n+1)(n^2+1) \nmid 10^n$ . So $n$ must be an odd number. $n=1$ is not an answer, suppose $n>1$ .

With $n$ odd, $n^2+1$ has an odd divisor since if $n=2k+1$ then $n^2+1=2(2k^2+2k+1)$ and $\gcd(n+1, n^2+1)=2$ , It follows that $n+1=2^a$ and $n^2+1=2 \times 5^b$ . Combine to get $2^a(2^{a-1}-1)=5^b-1 \ \ \ \star$ So $2^a |5^b-1$ and power of $2$ in $5^b-1$ is $a$ , so using LTE lemma we get $2+v_2(b)=v_2(5^b-1)=a \\ v_2(b)=a-2$ Let $b=2^{a-2}(2c+1); c\ge0$ , equation $\star$ turns into $2^a(2^{a-1}-1)=5^{2^{a-2}(2c+1)}-1$ But for $a\ge 4$ $\displaystyle \begin{array}{c}\\ 5^{2^{a-2}(2c+1)}-1 &=(5^{2^{a-3}(2c+1)}-1)(5^{2^{a-3}(2c+1)}+1) \\ &> 8\times 2^{a-3}(2c+1)(1+8\times 2^{a-3}(2c+1)+1) \\ &=2^a(2c+1)( 2^a(2c+1)+2) \\ &\ge2^a( 2^a+2) \\ &>2^a( 2^{a-1}-1)\end{array}$ using the fact that $5^z > 1+8z$ for $z>1$ . So $a<4$ and trying values gives $a=2, 3$ and $n=2^a-1=3, 7$ as answers. For $n=3$ and $n=7$ the expression $(n+1)(n^2+1)$ indeed divides $10^n$ .

If f(n) is to be an integer, $(n^3+n^2+n+1) \ must\ be\ divisible\ by \ 2\ Or \ 5\ only.\\ n^3+n^2+n =n(n^2+n+1)\ must\ end\ in \ 9. \ \ So\ n\ and\ (n^2+n+1)\ must\ be\ odd.\\ For\ n=1,\ 5,\ 9\ \ n^3+n^2+n \ will\ not\ end\ in\ 9.\\ n=3\ and\ 7 \ makes \ f(n)\ an \ integer.$
I will complete this shortly.