$f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$ is a polynomial of degree $n$ with positive integer coefficients.

Primary problem statement: Is the Exponential Diophantine Equation $f(f(a) + 1) = y^m$ solvable in integers $y, m \geq 2, a$?

Background: This problem arises in data encoding and representation for compression. Given $n + 1$ data values (think bytes with values ranging 0 to 255), we represent them as integer coefficients of $f(x)$. We require the data values to all be positive integers $\ge 0$. Under this condition, $f(1)$ is just the sum of the coefficients. Let $b = f(1) + 1$ and $c = f(b)$. Given just the values $c$ and $b$, we can recover the coefficients $a_0, a_1, \dots, a_n$ through repeated division of $c$ by $b$. i.e., the base-$b$ representation of $c$ has the coefficients of $f(x)$.

This base-$b$ representation and recovery works for any choice of $b$ that is greater than the height of the polynomial $f(x)$. $f(a) + 1$ is guaranteed to be greater than the height of the polynomial.

Alternate Problem Statement: Does the Exponential Diophantine Equation $f(b) = y^m$ have integer solutions $b, y, m \geq 2$, where $b > \max(a_0, a_1, a_2, \dots, a_n)$.

The compression arises from the fact that we are using perfect powers to encode a set of values. The hope is $b, y, m$ are small and require fewer bits to represent than the original data.

An acceptable answer may solve either the Primary Problem Statement or the Alternate Problem Statement.


  • $GCD(a_0, a_1, \dots, a_n) = 1$ can be considered as a condition


Richmond, B. On a Perplexing Polynomial Puzzle.

Shorey, T. N. Perfect powers in values of certain polynomials at integer points

  • $\begingroup$ +1 Interesting problem - well articulated/presented. Personally, way out of my league. $\endgroup$ Sep 24 '20 at 15:10
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    $\begingroup$ I assume you want $m\geq 2$? $\endgroup$
    – QC_QAOA
    Sep 25 '20 at 19:27
  • $\begingroup$ @QC_QADA: Yes. $m>=2$ desired. Good catch. $\endgroup$
    – vvg
    Sep 25 '20 at 19:30
  • $\begingroup$ You appear, since a is a constant, to be given information about y at a single point, a+ 1. That is not enough information to determine the function y for all x. $\endgroup$
    – user247327
    Sep 25 '20 at 20:13
  • $\begingroup$ Please take a look at the linked article titled 'On a Perplexing Polynomial Puzzle'. It talks about how we can encode information about the coefficients of any polynomial and recover those coefficients with just two evaluations of the polynomial. For eg: $f(x) = 2x^2 + 3x + 1$. $f(1) = 2.1^2 + 3.1 + 1 = 2 + 3 + 1 = 6.$ $b = f(1) + 1 = 6 + 1 = 7$ $f(b) = 2.7^2 + 3.7 + 1 = 4802 + 21 + 1 = 4824$ The base-$b$ (i.e. base-7 representation of 4824 is exactly $2.7^2 + 3.7 + 1$, the coefficients of $f(x)$. $\endgroup$
    – vvg
    Sep 25 '20 at 20:23

Consider the simplest case where $f(x)=a_0+a_1x$ and $(a_0,a_1)=1$. Then $$y^m=f(f(x)+1)=(a_0+a_1+a_0a_1)+a_1^2x$$ and write $y=a_1^2n+k$ for some integer $k$ such that $(a_1,k)=1$. We obtain $$k^m\equiv a_0+a_1+a_0a_1\pmod{a_1^2}$$ and choosing $m=2$ means that $a_0+a_1+a_0a_1$ is a quadratic residue of $a_1^2$. This means that $$\left(\frac{a_0+a_1+a_0a_1}{a_1^2}\right)=\left(\frac{a_1+1}{a_1^2}\right)\left(\frac{a_0+a_1}{a_1^2}\right)=1$$ using Legendre symbol notation. We can evaluate the first term as follows $$\left(\frac{a_1+1}{a_1^2}\right)=\begin{align}\begin{cases}1&\text{if}\,a_1\,\text{is odd or}\, \sqrt{a_1+1}\,\text{is an integer}\\-1&\text{otherwise}\end{cases}\end{align}$$ by noting that $(a_1^2-a_1-2)^2/4\equiv a_1+1\pmod{a_1^2}$ when $a_1$ is odd.

In the first case, choosing $a_0=a_1^2-3a_1+1$ suffices for all $a_1>3$ as $a_0+a_1$ is a perfect square and $(a_0,a_1)=1$. This is enough to generate not one, but two infinite sets of solutions.

For this value of $a_0$, the congruence $k^m\equiv a_0+a_1+a_0a_1\pmod{a_1^2}$ reduces to $$k^2\equiv-a_1+1\pmod{a_1^2}\implies k=\frac{a_1^2\pm(a_1-2)}2$$ after removing higher-order terms.

As the "generating" function is $f(x)=r^2-3r+1+r^2x$ on replacing $r:=a_1$, substituting this back into $y^2=f(f(x)+1)$ yields the two families $$(x,y)=\left(r^2n^2+(r^2\pm(r-2))n+\frac{(r-1)^2}2\pm(r-2),r^2n+\frac{r^2\pm(r-2)}2\right)$$ for all positive integers $n$ and odd $r\ge5$.

One can obtain further solution sets by choosing $a_0=a_1^2-(2c+1)a_1+c^2$ for a positive integer $c>1$ provided that $(a_0,a_1)=1$ and then repeating the process above.


I have found a polynomial without any solutions. Let $n=1$ and




This implies $y$ is even. But then $y=2k$ gives

$$2^m k^m=2(7+8r)$$


The left side is always even (since $m\geq 2$) while the right side is always odd. We conclude there is no $y$ and $m\geq 2$ which solves the exponential diophantine equation.

However, there is still a lot of interesting stuff that might be done with this problem. I think that in general, adding the condition that


might succeed in guaranteeing a solution. However, I have not proved this. For example, the same sort of problem arises when you consider the following polynomials





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    $\begingroup$ Interesting. So, there are potentially a class of polynomials where $GCD(a0, a1, .. an) \neq 1$ that are likely (not always) to fail generating perfect power values. So, we can always consider $f(x) = x.g(x) + c$ as the polynomial where $a_0, a_1, .. a_{n-1}$ are the coefficients of $g(x)$ and we force a constant $c$ that ensures $GCD(a0, a1, .., a_{n-1}, c) = 1$. I think adding the GCD condition is reasonable and if we can establish a solution can always be generated, we are making progress! $\endgroup$
    – vvg
    Sep 25 '20 at 21:38

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