Your "isPrime" function is flawed! You need to add 1 to the upper end of the range or else it returns true for all perfect squares.

The given implementation can be improved by just changing the upperbound of the range to be int(math.sqrt(n))+2

Still probably not the most efficient approach, but there's no point in checking numbers above the sqrt(n)+1.

Edit:
Actually I think it does not matter since the return statement will exit the function before those higher values are tested.

Your code is much more efiicient

What you said is correct, if the number has a large factor, it will be time consuming.Please refer to my solution.

I tried to check if n is a prime number when each time n changes. Hope this can help.

This implementation takes too long when n is big and can only be factorized like (2)(n') where n' is half of n(which is a big prime number).
for example, it takes a long long time to find the primeFactors(76587634333).

My implementation checkes if n is a prime number after each time n changes and it works well for all cases, although it seems long.

Hope the idea of 'checking if n is a prime number' would help you improve your algorithm.