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Let $0<q<1$ be rational. I am suspecting that $\sqrt[n]{(1+q^n)^2}$ is irrational. Can someone please help me to prove or to disprove this?

  • $n=1$ and $n=2$ are simple cases. I am interested in $n>2$.
Bart Michels
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Optima
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  • Hint: use power series – Aman Rajput Jun 27 '16 at 11:54
  • @AmanRajput Thanks man! My rusty math skills helped me to use power series (as you suggested) to expand this function. So if the power series is convergent but the number of terms is infinite, does this imply that the limit value is irrational? Could you guide me to a reference to read about when the limit value of power series is irrational? – Optima Jun 27 '16 at 14:40
  • Check on math world wolfram.. or download any PDF regarding general series. There are three kinds of series.. power series, Dirichlet or q-series and Fourier series – Aman Rajput Jun 27 '16 at 14:43
  • To be more specific, power series includes two kinds of series.. Laurent and piuseux series – Aman Rajput Jun 27 '16 at 14:44
  • @Optima This question perfectly fits the [tag:rationality-testing] tag. – Bart Michels Jul 19 '16 at 11:42

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Suppose $(1+q^n)^2=p^n$, $p\in\mathbb Q$. Let $q=\frac ab$, $p=\frac cd$ with $\gcd(a,b)=\gcd(c,d)=1$. We have

$$d^n(a^n+b^n)^2=b^{2n}c^n.$$

If $n$ is odd, then $c$ and $d$ are squares, so we get $x^n+y^n=z^n$ for some $x,y,z\in\mathbb Z$. If $n>1$ Fermat's Last Theorem says $q=0$.
If $n=2m$ is even, then $(da^2)^m+(db^2)^m=(b^2c)^m$. If $n>4$, Fermat's Last Theorem says $q=0$.

As for $n=4$, note that $d\mid b^2$, say $b^2=kd$, to get $a^4+b^4=(kc)^2$. Fermat proved that this has no non-trivial solutions. See e.g. this post:
${\;\;\;\;\;\;\;}$ $x^4 + y^4 = z^2$

Remark: Unlike some funny applications of Fermat's Last Theorem to similar problems (such as in this proof that $\sqrt[3]2$ is irrational) here it is really necessary to invoke it. For suppose $x^n+y^n=z^n$ is a non-trivial counterexample to FLT, then $\sqrt[n]{(1+\frac xy)^2}=(\frac zy)^2$.

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Bart Michels
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  • FLT solves it. But if using FLT is the only approach, then this problem is really as difficult as the FLT and there is no "elementary" type solution to it. Right? – Optima Jul 20 '16 at 12:27
  • @Optima Exactly; the last paragraph says that the question is equivalent to FLT. – Bart Michels Jul 20 '16 at 12:28