If $|G|=105$ then there is only one Sylow $5$-subgroup or only one Sylow $7$-subgroup I'm having some trouble with the following question:

Let $G$ be a group with $|G|=105$. Prove that there is only one Sylow $5$-subgroup or only one Sylow $7$-subgroup.

If $s_5$ and $s_7$ are the number of Sylow $5$-subgroup and Sylow $7$-subgroup, because fo Sylow's 3rd theorem we have that: $$s_5 = 1 \text{ or } 21$$ $$s_7 = 1 \text{ or } 15$$
The thing that I'm not understanding is: How does the $s_5$ affect the number $s_7$? What can we take from this info?
 A: Given the $|G|=105=3×5×7$
According to Sylow third theorem $s_5=5k+1|21$
Only possible if $s_5=1 \text{ and }21$
Also $s_7=7k+1|15$ only possible if $s_7=1,15$
Now suppose the case if $s_7=15$ and $s_5=21$
Then number of elements of order $5$ are $84$
and number of elements of order $7$ are $90$
But the order of group is $105$
It means either one of them $s_7$ and $s_5$ is equal to $1$
Since a p-sylow subgroup is Normal iff it is unique
We are done
A: As mentioned in the comment - if G had 21 5-sylow subgroups. Each 5-sylow subgroup is of order 5. Every 5-sylow subgroup has 4 elements of order 5 and any two 5-sylow subgroups have a trivial intersection (why?). Hence totally, we would have $4 \times 21 = 84$ elements of order 5. Along with this, an identity elements. So, totally we covered 85. We are left with only 20 elements. If G had fifteen 7-sylow subgroups, then, G would have to have $15\times 6 = 90$ elements of order 7(why?), which is not possible.
A similar argument shows, if G has fifteen 7-sylow subgroups, then it cannot have 21 5-sylow subgroups.
