# Find the number of irrational terms in the binomial expansion of $(3^{1/5}+7^{1/3})^{100}$

After expanding the above term binomially, I can well guess that the majority terms are irrational, but i'm unable to find any proper method of solving this sum, after repeated trials.

Thank You

• Cute disguise for a Diophantine equation. – Brian Tung May 7 '15 at 21:17
• Each of the $101$ terms in the expansion has the form ${100\choose i}{\sqrt[5]{3}}^i{\sqrt[3]{7}}^{100-i}$. Such a term is irrational unless $i$ is a multiple of $5$ and $100-i$ is a multiple of $3$. Can you go from there? – Steve Kass May 7 '15 at 21:19
• @BrianTung: I'd say a cute disguise for a Chinese Remainder. – Pedro M. May 7 '15 at 21:21
• @Pedro: to-MAY-to, to-MAH-to? :-) – Brian Tung May 7 '15 at 21:22
• @BrianTung: Fair enough :-). Although a "Diophantine Equation" sounds scarier, somehow. – Pedro M. May 7 '15 at 21:25

Hint: The binomial sum is $$\sum_{k = 0}^{100} {100 \choose k} 3^{k/5}7^{(100-k)/3},$$ and the rational terms are exactly the integer terms, i.e., when both $k/5$ and $(100-k)/3$ are integers.
• @InfinitelyUndefined Have you learned abound congruences? You need to solve the system $k \equiv 0 \pmod 5$ and $k \equiv 100 \pmod 3$ for $k$, by using the Chinese Remainder Theorem. – Pedro M. May 7 '15 at 22:13
• @InfinitelyUndefined: I assumed this was a problem in a class involving congruences and Chinese Remainder. However, this particular case is simple enough to allow for simpler solutions. For instance, one can write $k = 5m$ and ask when $100 - k = 5(20 - m)$ is a multiple of $3$. – Pedro M. May 7 '15 at 23:00
• @InfinitelyUndefined: You're right, well spotted! You can argue that $100 - k$ must be one of the seven multiples of $15$, hence there are $7$ rational terms. Note, however, that this works because $100$ is a multiple of $5$ (if you want to play around with the problem, you can try to see what happens if the exponent is $101$ instead). – Pedro M. May 8 '15 at 16:36