# A problem regarding rational and irrational numbers.

Let $n$ be a positive integer greater or equal to $2$. Prove that there are infinitely many irrational numbers $a$ such that $a+a^2+\cdots+a^n$ is rational.

Well, let $p$ be a prime. We consider the equation: $x^n + x^{n-1} +\cdots+ x=1/p$, which is equivalent to: $px^n+px^{n-1}+\cdots+px-1=0$ We can prove that the equation has a solution $a$ in the interval $I=(0, 1/p)$ using continuity and that the solution is not rational. But, is there another solution without using continuity? Thanks!

• Did you have something specific in mind? Most ways of proving that polynomial equations have irrational solutions boil down to continuity in one way or another... – Micah Sep 20 '13 at 17:49
• No. I really don't have something specific in mind. I'm thinking about it now. :) – user85046 Sep 20 '13 at 18:00
• Let m be a rational number such that 1 - 4*m >= 0, n = 2. Then there are infinitely many values of a = (-1 +- sqrt(1 - 4*m)) / 2 that are irrational but a^2 + a = m is rational. For n>2 it's not so good. – user85046 Sep 20 '13 at 19:35
• Even that might involve continuity if you look at the details. Certainly the easiest way to prove that real numbers have square roots is via exactly the same kind of continuity argument you're making in the question statement, though there are others (e.g., explicitly constructing a Cauchy sequence whose square converges to a given rational). – Micah Sep 20 '13 at 20:16

Take your equation $px^n+\cdots+px-1=0$, and set $y=1/x$, giving you $-py^{-n}-\cdots-p/y+1=0$, thus $y^n-py^{n-1}-\cdots-p=0$, Eisenstein at $p$, so that any of its roots is irrational, and there’s even at least one real one. So $x=1/y$ is also irrational.
• Of course the remark that one of its roots is real is justified only on topological (continuity) grounds. But within the algebraically closed field of all algebraic numbers, there are $n$ roots... – Lubin Sep 21 '13 at 22:00
Let $p$ be a prime. Then the only possible rational solutions of $x^n+x^{n-1}+\cdots+x=p$ are $x=\pm1$ and $x=\pm p$. We can rule out $x=\pm1$ by taking $p\gt n$, and it's also clear that $x=\pm p$ can't work since the left side far exceeds the right. But the equation has a real solution, so there is an irrational $a$ such that $a^n+a^{n-1}+\cdots+a=p$. And there are infinitely many primes, so, infinitely many $a$.
• +1 Nice answer! But how do you grantee that there is a real solution if $n$ is even? – Ovi Sep 21 '13 at 13:17
• Well, the easiest way is by continuity (the left side is zero at zero, and exceeds $p$ at $p$), but I guess that's forbidden. But as has been suggested in the comments, how do you ever prove a polynomial equation has a solution, other than by invoking continuity? I think what I've written makes the invocation of continuity as painless as possible. – Gerry Myerson Sep 21 '13 at 13:23