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Let $x, \ r \in \mathbb{Q}$.

I need to find the conditions on $ \ x, \ r$ so that the value of $ \large (1+x)^r$ is a rational number.

Which $x, \ r$ makes $(1+x)^r$ a rational number?

Answer:

If I take $x=\frac{16}{9}$ and $r=\frac{1}{2}$, then $ (1+x)^r=(1+\frac{16}{9})^{\frac{1}{2}}=\sqrt{\frac{25}{9}}=\pm \frac{5}{3} \in \mathbb{Q}$,

If I take $x=\frac{19}{8}$ and $r=\frac{1}{3}$, then $ (1+x)^r=(1+\frac{19}{8})^{\frac{1}{3}}=\large \sqrt[3]{\frac{27}{8}}= \frac{3}{2} \in \mathbb{Q}$,

and so on $ \cdots $

How to find all $x \in \mathbb{Q}$ and $r \in \mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?

Can you give me the general form of $x$ and $r$ so that $ (1+x)^r$ becomes a rational number?

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    $x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers – Wojowu Jan 02 '19 at 11:32
  • @Wojowu, Does it cover all possible choices? How do you match this formula for $x=\frac{16}{9}, \ r=\frac{1}{2}$ ? –  Jan 02 '19 at 13:02
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    $x=(5/3)^2-1,r=1/2$. – Wojowu Jan 02 '19 at 13:12
  • @Wojowu, For convergence I need to show that $ \ |x|_p<1$. So the question is which condition in $(a/b)^n-1$ ensure that $ \ \left| (a/b)^n-1 \right|_p<1$ ? Can you help me. – MAS Mar 30 '19 at 12:17
  • @M.A.SARKAR Assuming we take $a,b$ relatively prime, this is equivalent to $a^n\equiv b^n\pmod p$. – Wojowu Mar 30 '19 at 14:06
  • @Wojowu, Sir, you mean if $a^n \equiv b^n \ (\ mod \ p)$, then the condition $ \left | \left(\frac{a}{b} \right)^n-1 \right|_p<1$ is satisfied. – MAS Mar 30 '19 at 14:15
  • @M.A.SARKAR I mean that the conditions are equivalent - $a^n\equiv b^n\pmod p$ iff $|(a/b)^n-1|_p<1$ for $a,b$ relatively prime. – Wojowu Mar 30 '19 at 14:29
  • @Wojowu, $ |(a/b)^n-1|_p<1 \Rightarrow |a^n-b^n|_p<b^n$, then how $a^n \equiv b^n \ (mod \ p)$ ? How is it possible? Would you show a little work? – MAS Mar 30 '19 at 14:53
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    For $b\neq 0$, $|(a/b)^n-1|_p<1\iff |a^n-b^n|_p<|b^n|_p$. If $b$ is not divisible by $p$, $|b^n|_p=1$ so $|a^n-b^n|_p<1$, meaning $p\mid a^n-b^n$. If $p$ divides $b$ but doesn't divide $a$, then $|a^n-b^n|_p<|b^n|_p$ is easily seen to never hold – Wojowu Mar 30 '19 at 14:56
  • @Wojowu, thank you very much sir – MAS Mar 30 '19 at 15:02

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