Power n divisible by n powers

Warning: Gigantic solution ahead.

Well, in order for us to have $n^{3} + n^{2} + n + 1|10^{n}$ , $n^{3} + n^{2} + n + 1$ must be of the form $2^{a}5^{b}$ , $a$ and $b$ being non-negative integers.

Let's consider two main cases:

Case $I$ : $a = 0$ , $b = 0$

This gives us: $n^{3} + n^{2} + n + 1 = 1$ , or $n^{3} + n^{2} + n = 0$ ; the only real solution for this equation is n = 0, which doesn't suit our purposes.

Case $II$ : At least one of $a, b$ is different from 0.

Now, we must consider a few subcases for the values of $n$ itself. Notice that, for all $a > 0$ , the values of $x$ such that $2^{a} \equiv x \pmod{10}$ cycle through the set $2, 4, 8, 6$ in this particular order; also, for all $b > 0$ , $5^{b} \equiv 5 \pmod{10}$ .

Further observation tells us that $2^{a}5^{b} \equiv 0 \pmod{10}$ when both $a$ and $b$ are greater than 0 at the same time, and thus $2^{a}5^{b}$ can only be non-congruent to $0 \pmod{10}$ iff one of them is $0$ ; thus, the set of possible congruences modulo $10$ for $n^{3} + n^{2} + n + 1$ is $(0, 2, 4, 5, 6, 8)$ .

Then, let's observe what happens to $n^{3} + n^{2} + n + 1$ when we cycle for all possible values of $z$ such that $0 \leq z \leq 9$ and $n \equiv z \pmod{10}$ .

$n \equiv 0 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 1 \pmod{10}$

$n \equiv 1 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 4 \pmod{10}$

$n \equiv 2 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 5 \pmod{10}$

$n \equiv 3 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 0 \pmod{10}$

$n \equiv 4 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 5 \pmod{10}$

$n \equiv 5 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 6 \pmod{10}$

$n \equiv 6 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 9 \pmod{10}$

$n \equiv 7 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 0 \pmod{10}$

$n \equiv 8 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 5 \pmod{10}$

$n \equiv 9 \pmod{10} \rightarrow n^{3} + n^{2} + n + 1 \equiv 0 \pmod{10}$

From this we derive that $n$ cannot be congruent to either $0$ or $6$ when divided by $10$ ; this is the same as stating that the last digit of n cannot be $0$ or $6$ . Also, we can narrow the set of possible congruences modulo $10$ for $n^{3} + n^{2} + n + 1$ down to $(0, 4, 5, 6)$ .

The first case work is done; we move into a new case analysis.

Case $1$ : $n^{3} + n^{2} + n + 1 \equiv 4 \pmod{10}$

What this means is that $n^{3} + n^{2} + n + 1$ is a number of the form $2^{a}$ such that $a \equiv 2 \pmod{4}$ .

Thus, we have: $(n + 1)(n^{2} + 1) = 2^{a}$ . We've ruled out the possibility $n = 1$ back at case $I$ ; besides, we must have:

$\begin{cases} n + 1 = 2^{a_{1}} \\ n^{2} + 1 = 2^{a_{2}} \end{cases}$

Since $n^{2} + 1$ is a power of two, as well as $n + 1$ , and since $n^{2} + 1 > n + 1$ for all $n > 1$ , then $n + 1|n^2 + 1$ .

Performing the polynomial division yields: $n^{2} + 1 = (n + 1)*(n - 1) + 2$ ; thus, $n + 1$ must divide 2 so that $n + 1|n^2 + 1$ . This could only happen if either $n = 0$ or $n = 1$ ; both possibilities have been ruled out, and thus no numbers fit this case.

Case $2$ : $n^{3} + n^{2} + n + 1 \equiv 6 \pmod{10}$

Like the previous case, $n^{3} + n^{2} + n + 1$ is a number of the form $2^{a}$ such that $a \equiv 0 \pmod{4}$ . By reasoning similar to the previous case, we'd arrive at the same result, which is that no numbers fit this case.

Case $3$ : $n^{3} + n^{2} + n + 1 \equiv 5 \pmod{10}$

This time, $n^{3} + n^{2} + n + 1$ is a number of the form $5^{b}$ ; however, by reasonings similar to the previous cases, we'd arrive once more at the same result.

Case $4$ : $n^{3} + n^{2} + n + 1 \equiv 0 \pmod{10}$

Finally, the last case. Not only this means that $n^{3} + n^{2} + n + 1$ is a number of the form $2^{a}5^{b}$ , but also that either $n \equiv 3 \pmod{10}$ , $n \equiv 7 \pmod{10}$ , or $n \equiv 9 \pmod{10}$ .

$(n + 1)*(n^2 + 1) = 2^{a}5^{b}$

Subcase $4.1$ : $n \equiv 3 \pmod{10}$

This means that $n + 1 \equiv 4 \pmod{10}$ , and $n^2 + 1 \equiv 0 \pmod{10}$ ; thus, $n + 1$ is a power of two and $n^2 + 1$ is the product of a power of two by $5^{b}$ .

$\begin{cases} n + 1 = 2^{a_{1}} \\ n^{2} + 1 = 2^{a_{2}}5^{b} \end{cases}$

Thus: $(2^{a_{1}} - 1)^{2} + 1 = 2^{a_{2}}5^{b}$

$2^{2a_{1}} - 2^{a_{1} + 1} + 2 = 2^{a_{2}}5^{b}$

$2*(2^{2a_{1} - 1} - 2^{a_{1}} + 1) = 2^{a_{2}}5^{b}$

Since $a_{1} > 1$ (Otherwise we'd have $n = 0$ or $n = 1$ ), we can write:

$\begin{cases} 2^{a_{2}} = 2 \\ 2^{2a_{1} - 1} - 2^{a_{1}} + 1 = 5^{b} \end{cases}$

Thus, $a_{2} = 1$ . As for the other equation, let $2^{a_{1}} = x$ and $5^b = y$ .

We have: $\frac{x^{2}}{2} - x + 1 = y \rightarrow x^{2} - 2x + 2(1 - y) = 0$ .

Solving for $x$ : $x = 1 \pm \sqrt{2y -1}$

Now, $2y - 1$ must be a square number; this means $2*5^{b} - 1 = z^{2}$

The only non-negative integer solutions $(b, z)$ for this equation are: $(0, 1)$ , $(1, 3)$ and $(2, 7)$ . To see why this holds, let's examine this lemma:

Lemma $I$ : The only pairs $(b, z)$ of integer solutions for $2*5^{b} - 1 = z^{2}$ are $(0, 1)$ , $(1, 3)$ and $(2, 7)$ .

To prove this, first note that, since $5^{b}$ is odd for all integers $b$ , then so is $2*5^{b} - 1$ . This implies that $z^{2}$ is odd, which means that $z$ must be odd as well. Now, if there are two solutions $(b_{1}, z_{1})$ and $(b_{2}, z_{2})$ , with $b_{1} \neq b_{2}$ (which imply in $z_{1} \neq z_{2}$ due to the fact that the exponential function is injective), then the difference between $z_{1}$ and $z_{2}$ is even; in other words, $z_{2} - z_{1} = 2p$ .

Now, notice that for $z = 5$ , there are no solutions for $b$ . Now, I'll prove that if any odd value of $z > 7$ is not a solution, then neither $z + 4$ can produce a solution; this will eventually assure that we'll cover all of the odd integers.

Firstly, let's assume that there is a value $z_{1} > 7$ such that there is a corresponding $b_{1}$ , and also that there is a second value, $z_{2}$ , such that $z_{2} = z_{1} + 4$ and there is a value $b_{2}$ ;

Thus:

$\begin{cases} 2*5^{b_{1}} - 1 = z_{1}^{2} \\ 2*5^{b_{2}} - 1 = z_{2}^{2} \end{cases}$

Now, subtracting the first equation from the second, and using the fact that $z_{2} = z_{1} + 4$ , we have:

$2*(5^{b_{2}} - 5^{b_{1}}) = 8*(z_{1} + 2)$

$5^{b_{1}}*(5^{b_{2} - b_{1}} - 1) = 4*(z_{1} + 2)$

Now, since $z_{1}$ is odd, so is $z_{1} + 2$ ; also, on the LHS the only odd factor is $5^{b_{1}}$ ; then, we can conclude that $5^{b_{1}} = z_{1} + 2$ .

What this means is that: $2*(z_{1} + 2) - 1 = z_{1}^{2}$ ; from here:

$z_{1}^{2} - 2z_{1} - 3 = 0$ .

The only positive solution for $z_{1}$ is $z_{1} = 3$ ; however, this contradicts our earlier assumption that $z_{1} > 7$ ; thus, by contradiction, if $z_{1}$ is not a solution of the equation, neither $z_{1} + 4$ will be. Note that $z = 9$ does not produce a solution (we'd have $5^{b} = 41$ ), and neither does $z = 11$ (we'd have $5^{b} = 60$ ). Then, we get all of the odd integers covered, meaning that our lemma is valid; thus, the only solution is given by $b = 1$ and $z = 3$ ( $b = 0$ produces the solution $n = 1$ , which is not valid, and $b = 2$ produces the only solution for the next subcase, as you will see).

Finally, verifying that $n = 3$ gives us a solution: $\frac{10^{3}}{3^{3} + 3^{2} + 3 + 1} = \frac{1000}{40} = 25$ .

Subcase $4.2$ : $n \equiv 7 \pmod{10}$

This means that $n + 1 \equiv 8 \pmod{10}$ , and $n^2 + 1 \equiv 0 \pmod{10}$ ; thus, $n + 1$ is a power of two and $n^2 + 1$ is the product of a power of two by $5^{b}$ .

By the reasoning shown in the previous subcase, we arrive at the fact that the only solution is $n = 7$ ; verifying: $\frac{10^{7}}{7^{3} + 7^{2} + 7 + 1} = \frac{10^7}{400} = 25000$ .

Subcase $4.3$ : $n \equiv 9 \pmod{10}$ This means that $n + 1 \equiv 0 \pmod{10}$ , and $n^2 + 1 \equiv 2 \pmod{10}$ ; thus, $n^2 + 1$ is a power of two and $n + 1$ is the product of a power of two by $5^{b}$ .

$\begin{cases} n + 1 = 2^{a_{1}}5^{b} \\ n^{2} + 1 = 2^{a_{2}} \end{cases}$

$n = 2^{a_{1}}5^{b} - 1 \rightarrow (2^{a_{1}}5^{b} - 1)^{2} + 1 = 2^{a_{2}}$

$2^{2a_{1}}5^{2b} - 2^{a_{1} + 1}5^{b} + 2 = 2^{a_{2}}$

$\begin{cases} 2^{a_{2}} = 2 \\ 2^{2a_{1} - 1}5^{2b} - 2^{a_{1}}5^{b} + 1 = 1 \end{cases}$

This means that $a_{2} = 1$ , and thus $n^2 = 1 \rightarrow n = 1$ , which we've proven is not a valid solution, and therefore no solutions exist here.

Thus, the only valid numbers that are answers to this problem are $3$ and $7$ ; their sum is $10$ , which leaves a remainder of $\boxed{0}$ when divided by $10$ , and this is the solution sought.