My teacher taught me a shortcut for finding number of rational terms in $\left(a^{\frac{1}{p}}+b^{\frac{1}{q}}\right)^n$.

For example, find the number of rational terms in $\left(5^{\frac{1}{6}}+2^{\frac{1}{8}}\right)^{100}$.


  1. Find LCM of $(p,q)$. In the above example, its $24$.
  2. Divide $n$ by the LCM obtained. Let quotient be $Q$ and remainder be $R$.
  3. If $R=0$, number of rational terms is $Q+1$. Else its $Q$.

In the above example, $R\neq 0$. So number of rational terms is $4$.

How did he derive this shortcut?

  • $\begingroup$ What are $a,b$? What is counted as a rational term? Example: $a,b \in \mathrm{Q}_{>0}, p=q=n=2$. That is you ask for the number of rational terms in $(\sqrt{a}+\sqrt{b})^2=a + 2\sqrt{ab}+b.$ IMO this number is either $2$ or $3,\,$ but you formula predicts always $2.$ $\endgroup$ – gammatester Mar 9 '16 at 8:15
  • $\begingroup$ @gammatester I haven't discovered it's flaws yet. a,b are two positive integers. But he did specify a rule. If it's $(2+3^{\frac{1}{4}})^6$ you have to express the $2$ as $4^{\frac{1}{2}}$ $\endgroup$ – Aditya Dev Mar 9 '16 at 8:55
  • $\begingroup$ Interesting. Since $R=0$, then that makes $\sqrt{ab}$ rational, thus leading to $Q+1$. Hm... $\endgroup$ – Simply Beautiful Art Mar 9 '16 at 19:44
  • $\begingroup$ p,q must be prime numbers. $\endgroup$ – Sujith Sizon Mar 10 '16 at 9:02
  • $\begingroup$ @SujithSizon no. In the example I mentioned, p,q are not prime $\endgroup$ – Aditya Dev Mar 10 '16 at 13:05

Since the binomial expansion of this expression is $$ \left(a^{1/p} + b^{1/q}\right)^n=\sum_{i=0}^{n}{{n}\choose{i}}a^{i/p}b^{(n-i)/q}, $$ the $i$-th term is certainly rational (indeed, an integer) when $i\equiv 0$ (mod $p$) and $i\equiv n$ (mod $q$). By the Chinese remainder theorem, all solutions to these two equations are equal modulo ${\text{lcm}}(p, q)$; i.e., we get one solution every ${\text{lcm}}(p, q)$ steps. Therefore we get $Q$ or $Q+1$ solutions (in the notation of the problem) if the LCM doesn't divide $n$, and $Q+1$ solutions if the LCM divides $n$ (in which case $q$ divides $n$ as well, so the solutions start at $i=0$). When the LCM doesn't divide $n$, you need to find the first solution to decide if the result will be $Q$ or $Q+1$. This is the cause of "exceptions" like $\left(2 + 3^{1/4}\right)^6$.

The count, moreover, depends on there not being any other rational terms. I think this is guaranteed only if $a$ and $b$ are squarefree and coprime.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.