Win first, take 4 pills for the first time, and then ensure that the total number of pills taken by each cycle and opponents is 5 according to the number taken by opponents. Subtract the remaining 1 tablet from the total of 80, and the remaining 79 tablets. After the first person takes it, the rest is a multiple of 4+ 1=5 to ensure victory, that is, four pieces are taken first, and 75 pieces are left (a multiple of 5), and there are actually 76 pieces left. The number taken at the back is 5 MINUS the number taken by the opponent, which can guarantee that 655 pieces will be left to the opponent at last.

The square difference of any two odd numbers is a multiple of 8.

Prove: Let any odd number 2n+ 1, 2m+ 1, (m, n∈N).

(2m+ 1)2-(2n+ 1)2

=(2m+ 1+2n+ 1)*(2m-2n)

=4(m+n+ 1)(m-n)

When both m and n are odd or even, m-n is even and divisible by 2.

When m and n are odd and even, m+n+ 1 is even and divisible by 2.

So (m+n+ 1)(m-n) is a multiple of 2.

Then 4(m+n+ 1)(m-n) must be a multiple of 8.

(Note: 0 can be divisible by 2, so 0 is an even number, and 0 can also be divisible by 8, so 0 is a multiple of 8. )