# Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots

Let $\overline{a_n \ldots a_1a_0}$ be the decimal representation of $65^k$ for some $k \geq 2$. Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots.

I thought about using the rational root theorem. We know that $a_0 = 5$ since $65$ ends in a $5$, so any rational root must be of the form $-\dfrac{5}{a}$ or $-\dfrac{1}{a}$ where $a$ is an integer factor of $a_n$. How can we continue?

• Potentially useful: a quick computer experiment shows that all the digits $\{1, \dots, 9\}$ occur for $a_n$. Mar 31, 2017 at 23:06
• That follows from $\log_{10} 65$ being irrational, @erfink Mar 31, 2017 at 23:09
• you have $a_1=2$ too, though that probably won't be helpful. Mar 31, 2017 at 23:31
• Similar to @dezdichado comment, for $k\geq 6$, we seem to have that all the decimal representations end with $\dots 890625$. For $k \geq 7$, we seem to have endings of $\dots 2890625$ or $\dots 7890625$. Mar 31, 2017 at 23:42
• @астонвіллаолофмэллбэрг What do you mean by reduced it to 1 case? Apr 1, 2017 at 0:14

An incomplete answer, but I hope it may help in getting the final answer itself.

Let $\frac pq$ be a rational root of $f$. Since $f-f(10)$ has $10$ as a root, it follows that $f-f(10) = (x-10)g(x)$ for some $g$. Now, let $x = \frac pq$: $$\left(\frac pq - 10\right) g\left(\frac pq\right) = -f(10) \neq 0$$ Multiplying by $q^n$ : $$\left(p-10q\right)q^{n-1}g\left(\frac pq\right) = -f(10)q^n$$ From here, we know that $\deg g = n-1$, so that $q^{n-1}g\left(\frac pq\right)$ is an integer. Hence, $p-10q | q^n f(10)$, and by co-primality, $p-10q | f(10) = 65^k$.

We have that $p|5$ and $q | a_n$. So, $p = 1/5$, and $q$ must satisfy $p-10q | 65^k$, where $-9 \leq q \leq -1$ (the polynomial does not have positive roots by Descartes' rule of signs).

Note that if $p=1$, then for the above $q$, not case matches. However, if $p=5$, then $q=-6$ fits the criteria. So all we have to do is to check that $\frac{-5}{6}$ is not a rational root, then we are done. In fact, this shows that if the leading coefficient $a_n \neq 6$, then we are done.

• have not written down it rigorously, but if you let $f(x)=(6x+5)h(x)$ and manage to prove $h(x)\in\mathbb{Z}[x]$, then $h(10) = 65^{k-1}$ and $h$ is monic, which is a contradiction I think. Apr 1, 2017 at 3:16
• Can't you also have $q=-2$? Apr 1, 2017 at 4:39
• Let $f_n(x)=(6x+5)f_{n-1}(x)+(x-10)R(x)$. Then $f_n$ has $-\frac{5}{6}$ as a root iff $R$ does. But if $n \geq 2$ then $f_{n-1}(0)=f_n(0)=5$, which means we must have $R(0)=2$. So $R(x)$ can't possibly have $-\frac{5}{6}$ as a root. I still don't see any way to rule out $-\frac{5}{2}$ (which your argument does not do; if $p=5$ and $q=-2$ then $p-10q=25$ is a factor of $65^k$ for $k \geq 2$). Apr 1, 2017 at 6:26
• The argument with the Descartes-sign rule is correct, but it is obvious without this rule that a polynomial with positive coefficients cannot have non-negative roots. Apr 1, 2017 at 8:21
• Oh, I forgot about $q=-2$! All right, thank you all for reminding me of that. However, my feeling is that we can get a trivial bound, from the fact that all coefficients are bounded by $10$, and the first and last are $6$ and $5$ respectively. I am not sure, but I'll look for some bounds and tell you. Apr 1, 2017 at 10:44

Let $f_n$ be the polynomial obtained from the digits of $65^n$. The other answer shows that if $p/q$ is a root of this polynomial in lowest terms, then $p-10q|65^n$; along with the bounds on $f_n$, this implies that we must have $p=-5$ and either $q=2$ or $q=6$. That is, the only possible rational roots are $-\frac{5}{6}$ and $-\frac{5}{2}$.

Note that $(6x+5)^n$ is also a polynomial which evaluates to $65^n$ at $x=10$. So we can write $$f_n(x)=(6x+5)^n+(x-10)R(x)$$ for some polynomial $R$.

Setting $x=0$ in this expression, we have $5=5^n-10R(0)$, and so $R(0)=\frac{5^{n-1}-1}{2}$, which is not a multiple of $5$.

Now, suppose $n \geq 3$ and $x$ is either $-\frac{5}{6}$ or $-\frac{5}{2}$. Then $(6x+5)^n$ has $5$-adic valuation at least $3$ (that is, its numerator is a multiple of $5^3$ when it is written in lowest terms). In contrast, the $5$-adic valuation of $R(x)$ is $0$ (because the $5$-adic valuation of $R(0)$ is $0$) and $(x-10)$ is either $-\frac{5}{6}-10=-\frac{65}{6}$ or $-\frac{5}{2}-10=-\frac{25}{2}$, neither of which has $5$-adic valuation greater than $2$. So $(6x+5)^n$ and $(x-10)R(x)$ have unequal $5$-adic valuations, which means that $f_n(x)=(6x+5)^n+(x-10)R(x)$ cannot possibly be $0$.

It only remains to check that $f_2$ has no rational roots, which is an easy calculation.