1. General term of the binomial expansion

For $\left(3^{\frac12}+5^{\frac14}\right)^{680},$ the general term is $T_{r+1}=\binom{680}{r}\left(3^{\frac12}\right)^{680-r}\left(5^{\frac14}\right)^r, \qquad r=0,1,2,\dots,680.$

So, $T_{r+1}=\binom{680}{r}3^{\frac{680-r}{2}}5^{\frac r4}.$

2. Condition for an integral term

Since $\binom{680}{r}$ is always an integer, the term will be an integer only when the powers of both $3$ and $5$ are integers.

So we need: $\frac{680-r}{2}\in \mathbb Z$ and $\frac r4\in \mathbb Z.$

3. Simplify the conditions

• $\dfrac r4\in \mathbb Z \Rightarrow r$ must be divisible by $4$.
• If $r$ is divisible by $4$, then it is automatically even, hence $680-r$ is even, so $\frac{680-r}{2}\in \mathbb Z$ is also satisfied.

Thus the only effective condition is: $r\equiv 0 \pmod 4.$

4. Count such values of $r$

We need multiples of $4$ from $0$ to $680$ inclusive: $r=0,4,8,\dots,680.$

This is an arithmetic progression with number of terms $\frac{680-0}{4}+1=170+1=171.$

5. Final answer

Hence, the number of integral terms is $\boxed{171}.$