I am still a little bit confused. How do you know there can only be two cases, b=3,4? 

By Simon's Favorite Factoring Trick,

If b=5: 1/c+1/d=3/10

10c+10d=3cd so 3cd-10c-10d=0, so (c-10/3)(d-10/3)=10/9. Now multiply both sides by 9 to get (3c-10)(3d-10)=10 c=4 and d=5 work, but now d=b so there are no solutions.

But of course it is not possible to do this for every integer b>5.

How do you prove it must be b≤4?

Thanks,

Tina Jin