How to prove that any positive integer can be written as sum of weighted powers of another number?

Question

I am hypothesizing the following result, given two natural numbers $n$ and $x$, then there exists a unique sequence of coefficients $\{a_1,a_2...,a_n \}$ with $a_i $ in whole numbers such that:

$$ n= \sum_i a_i x^i$$

When $x=10$, this coincides with the standard form of numbers. How do I prove such a sequence exists always?

Edit: From the current answer I received and the comment, I put a further restriction $a_i <x$

A comment is that I think the reminiscent of how we prove that a vector has a unique decomposition into components given a complete basis.

This is not true. You need the set of coefficients to be bounded above strictly by $x$ for uniqueness. — Rushabh Mehta, Jun 14 '21 at 15:38
@Buraian Look at the answer provided on this question. It gives you a counterexample to uniqueness. — Rushabh Mehta, Jun 14 '21 at 15:40
Hi Don, I think I accounted for it now. I saw your comment first before the answer. — tryst with freedom, Jun 14 '21 at 15:40
One way is induction. Prove that if it's true for naturals $< x^n$ then it's true for naturals $<x^{n+1}$. — saulspatz, Jun 14 '21 at 15:47
This idea, using "radix notation" to express a large number as a polynomial evaluated at a smaller number, has an important application in setting up a General Number Field Sieve factoring project. — hardmath, Jun 14 '21 at 15:47
This got deep quite quickly haha. I will try go over the three links sent and attempt a self answer in time (if no one else writes a proof by then) — tryst with freedom, Jun 14 '21 at 15:49

score 1 · Answer 1 · answered Jun 14 '21 at 15:38

1

You need further restrictions. For example, in your example $10$ can be represented via $(10,0,0,0...,0)$ and via $(0,1,0,0,0....)$.

answered Jun 14 '21 at 15:38

Stefanos Van Dijk

129

I've edited the question now. – tryst with freedom Jun 14 '21 at 15:41

How to prove that any positive integer can be written as sum of weighted powers of another number?

1 Answers1