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Let $f(x)$ be a polynomial of $n^{th}$ degree with integer coefficients and let the leading coefficient be 1. Is it true that $f(x) = k^n$ for infinitely many integers $k$ and $x$ if and only if all roots of $f(x)$ are equal?

One side of the proof is fairly simple. If all roots of $f(x)$ are equal, then $f(x) = (x-x_1)^n$, where $x_1$ is the root of $f(x)$

Now I have problems with the other side. I don't know how to start. I started looking at special cases and I proved it for $n=2$. Let $f(x) = x^2 + ax + b$. Then the roots are $x_1 = \frac{-a + \sqrt{D}}{2}$ and $x_2 = \frac{ -a - \sqrt{D}}{2}$

Then $f(x) = (x-x_1)(x-x_2)$. Now set $f(x)$ to $k^2$ and we have:

$$(2x + a + \sqrt{D})(2x + a - \sqrt{D}) = (2k)^2$$ $$(2x + a)^2 - D = (2k)^2$$

Since $D$ is some fixed number for some large enough $x$ we will have:

$$(2x+a - 1)^2 < (2x+a)^2 - D < (2x+a+1)^2$$

Hence $(2x+a)^2 - D = (2x +a)^2 \implies D=0 \implies x_1 = x_2$

But i'm stuck for $n=3$ and more generally you can't use this method when the degree of $f(x)$ is bigger than 4, since there's no explicit formula for the roots of $f(x)$

Stefan4024
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  • I think something about the question is strange. If the leading coefficient of a polynomial of $n^{th}$ degree is positive then the $lim_{x\rightarrow\infty}=+\infty$ and $f(x)$ is continuous so $\exists N$ s.t. for all integer $j>N$ $\exists x$ s.t. $f(x)=j^n$. That's just continuity. Is there something I'm not seeing? – DRF Mar 12 '15 at 18:36
  • Reading over the proof you wrote,before that last implication, I'm not quite seeing why you are getting that $(2x+a)^2+D=(2x+a)^2$ can you elaborate? – DRF Mar 12 '15 at 18:51
  • @DRF We need to prove that if $f(x) = k^n$ then all roots of $f(x)$ are equal, i.e. $f(x) = (x-x_1)^n$, where $x_1$ is it's root. And to be fair I don't know what you're talking about. BTW Look at $f(x) = (x+1)^3 + 3$ it's not cube for any $x$, although as $x\to \infty, f(x) \to \infty$ and it's continuous function. – Stefan4024 Mar 12 '15 at 18:54
  • @DRF It should be $(2x + a)^2 - D = (2x+a)^2$. Now since for big enough $x$ it's between $(2x+a-1)^2$ and $(2x+a+1)^2$ and since we want it to be full square it has to be equal to $(2x+a)^2$ since that's the only square in the given range. – Stefan4024 Mar 12 '15 at 18:56
  • $(x+1)^3+3$ is cube for infinitely many $x$. Particularly for each $x$ such that, $x=\sqrt[3]{k^3-3}-1$ where $k$ is an integer. – DRF Mar 12 '15 at 18:57
  • @DRF $x$ needs to be an integer. – Stefan4024 Mar 12 '15 at 19:06
  • That's the bit that was missing then.:) – DRF Mar 12 '15 at 19:08
  • This is not true if we can vary number $k$ and roots of polynomial at the same time.

    Let $k$ be a product of $n$ numbers: $k = p_1p_2...p_n$. Then for arbitrary $x$ build $x_i = x - p^n_i$. Then $f(x) = (x - x_1)...(x - x_n) = k^n$

    But I think you are asking if roots are fixed and equality is hold for infinitely many $x$ and $k$.

     – hOff Mar 12 '15 at 21:43
  • @hOff Yes, $f(x)$ is a fixed polynomial. Your way varies $f(x)$ and it's roots. – Stefan4024 Mar 12 '15 at 22:01

1 Answers1

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Answer : The only polynomials $f$ for which the $f(x)$ is a k-th power for infinitely many integers are the polynomials of the form $f(x)=(x+k)^n$.

Assume that, $f(x)=x^n+\sum_{i=1}^N f_i x^i, f_0=1$

First, There exists two positive integers $a$ and $b$ such that for larger $x>A$ we have $(x-b)^{n}\leq f(x)\leq (x+a)^{n}$ (to prove this it suffices to pass through the coefficients and take an integer $a$ which makes $\dbinom n k a^{n-k}$ greater then the coefficients and the same thing for $b$)

So let $N$ be a very larger integer such that $N>\max(2^{n+1}(\max(a,b))^n,A,2|f_i|)$ and $f(N)$ is a $n$th power we have : $$f(N)=(N+k)^n $$ for some $-b\leq k\leq a$, and we know that every integer have a unique writing of the form: $$a_kN^{k}+a_{k-1}N^{k-1}+\cdots+a_1N^1+a_{1} $$

where $|a_i|\leq N/2$ But: $$f(N)=\sum_{i=0}^{n}f_iN^n=\sum_{i=0}^{N}\dbinom n i k^{n-i}N^{i}$$ $|f_i|\leq N/2$ and $|\dbinom n i k^{n-i}|\leq \max(a,b)^n2^n\leq N/2$

we conclude that $f_i=\dbinom n i k^{n-i}$ and $f(x)=(x+k)^n$

Note that this method is well known for this algorithm:Polynomial determined by two inputs, this problem have a lot of generalization which can not be solved using this method, see my question here.

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Elaqqad
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