Find $n \in \mathbb{N}$ which makes $x_1, x_2, ..., x_{26} \in \mathbb{N}$ exist which satisfies $\sum_{k=1}^{25} n^{x_k}=n^{x_{26}}$.

Question

Find $n \in \mathbb{N}$ which makes $x_1, x_2, ..., x_{26} \in \mathbb{N}$ exist which satisfies $\displaystyle \sum_{k=1}^{25} n^{x_k}=n^{x_{26}}$.

My attempt:

\begin{align} &\sum_{k=1}^{25} n^{x_k} \equiv 25 (\text{mod } {n-1}). \\ &n^{x_{26}} \equiv 1 (\text{mod } {n-1}). \\ \ \\ &\therefore n-1 \Big| 24. \longrightarrow n=2, 3, 4, 5, 7, 9, 13, 25. \end{align}

I can't proceed more from here.

P.S. I am looking for simple solution, not just putting each into the equation.

Answer 1

OP has showed that a necessary condition for the equation is $n-1 \mid 24$.

Conversely, if $ n-1 \mid 24$ , let $ 24 = k(n-1)$, then find an equation which shows that this is also sufficient.

Hints:

• We can replace $n^a$ with $ n \times n^{a-1}$. This increases the number of terms by $ n-1$. How many times do we need to do this?
• Try $ n = 3$, can you find 25 such terms?
• Try $ n = 13$, can you find 25 such terms?

$ n^k = (n-1) \times n^{k-1} + (n-1) \times n^{k-2} + \ldots + (n-1) \times n^1 + n \times n^0 $

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Notes:

• Arguably, showing that $n -1 \mid 24$ is a necessary condition is the harder part of this question. OP has a very slick way of doing this.
  • EG The naive "bound LHS with $ n^{x_{26}} = RHS = LHS \leq 25 \times n^{\max x_k} \leq 25 \times n ^ { x_{26} - 1 } $ " tells us that $ n \leq 25$. However, this admits many more solutions to test.
• One can arrive at that answer by realizing that essentially, we can only replace $n^a$ by $(n^b) \times n^{a-b}$, which increases the number of terms by $n^b-1$ which is a multiple of $n-1$. Hence, this is a necessary and sufficient condition.
• Starting with $n^{x_{26}}$, we just need to to "lower the degree $k$ times" which will result in $1+k(n-1) = 25$ terms.
  • All that we needed was to start off with a high enough $x_{26}$ that will allow for $ k$ lowerings.
  • EG I lowered the degree $k$ times, once each from $ n^i$ to $n^{i-1}$ (resulting in $n$ $n^{i-1}$ each time).

Answer 2

Continuing from your conclusion, we can show that all of these are valid answers for $n$. This is a recipe for specifying $x_j$s for any such valid $n$. Let us define two parameters. Suppose $p=\frac{24}{n-1}$. By OP's proof this is a valid natural number. Now let $\beta$ be the maximum integer for which $p\geq1+n+n^2+...+n^{\beta}$

As examples,

• for $n=2$, we have $p=\frac{24}{1} = 24$ and $\beta=3$ since $24\geq2^0 + 2^1 + 2^2 + 2^3$ (cannot add $2^4$)
• for $n=7$, we have $p=\frac{24}{6} = 4$ and $\beta=0$ since $4\geq7^0$ (cannot add $7^1$)

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We propose that the following choice of $x_j$'s suffices:

$x_{26} = \beta+3$ and for other $x_j$'s

$x_j = 1 \qquad \text{if} \qquad j\leq n \cdot \big(p-\sum_{j=0}^{\beta} n^j\big) \qquad (\text{see additional proof at end})$

$x_j = 2 \qquad $otherwise

The proof is simple. The summation becomes:

$\sum_{k=1}^{25}n^{x_k} = n \cdot \big(p-\sum_{j=0}^{\beta} n^j\big) \times n^1 + \bigg(25 - n\cdot \big(p-\sum_{j=0}^{\beta} n^j\big) \bigg)\times n^2$

$ \qquad \qquad = n^2\bigg(25-(n-1)\cdot\big(p-\sum_{j=0}^{\beta} n^j\big)\bigg)$

$\qquad \qquad = n^2\bigg(25 - (n-1)p \quad + \quad (n-1)\cdot\big(\sum_{j=0}^{\beta} ​n^j\big)\bigg)$

Using the well known formula for the sum of GP and the definition of $p$, we get:

$\qquad \qquad = n^2\bigg(25 - 24 \quad + \quad n^{\beta+1}-1\bigg)=n^{\beta+3} = n^{x_{26}}$

Of course in either case for a given $n$, we can increase all $x_j$'s by the same amount and still get valid solutions, so for each of these $n$ values, we have infinite set of $x_j$ solutions

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Additional Proof

We also show that the number of terms set equal to $1$ are necessarily less than $25$ so that this is a valid prescription. Note that by definition of $\beta$, we have:

$\qquad \quad p<n^{\beta+1} + \sum_{j=0}^{\beta} n^j$

$\iff 25 + \big(p - \sum_{j=0}^{\beta} n^j\big) - n^{\beta+1} < 25$

$\iff 24+p+1 - \sum_{j=0}^{\beta} n^j - n^{\beta+1} < 25$

$\iff (n-1)p + p - \sum_{j=1}^{\beta+1} n^j < 25$

$\iff np - n\cdot \sum_{j=0}^{\beta} n^j < 25$

$\iff n \cdot \big(p-\sum_{j=0}^{\beta} n^j\big) < 25$ $$\tag*{$\blacksquare$}$$