Every integer can be written in the form 5k, 5k1, 5k2

 

We are given that pa³+qa²+ra+s is divisible by 5 for some integer a. It is given that s is not divisible by 5 (otherwise, the problem would be trivial). Hence a  5k.

a = 5kt, where t = 1 or 2

Case I:Now assume that a is of the form 5k+1

Then p+q+r+s is divisible by 5. In that case if we choose b = 5m+1

Then sb³+rb²+qb+p = multiple of 5 + (p+q+r+s) which is given to be a multiple of 5. So, for this choice of b, sb³+rb²+qb+p is divisible by 5.

 

Case II: a is of the form 5k-1. Hence -p+q-r+s is divisible by 5.

            It is easy to see that again b = 5m-1 is the choice in this case

Case III: a is of the form 5k+2

             Hence 8p+4q+2r+s is divisible by 5

            Now choose b of the form 5k-2. The residue is -8s+4r-2q+p and our task is to prove that this expression is divisible by 5.

 

Consider 2(-8s+4r-2q+p ) + (8p+4q+2r+s) = -15s+10r+10p = 5n where n is an integer.

 

Also 8p+4q+2r+s = 5l

 

Hence  2(-8s+4r-2q+p ) = 5(n-l)

Since 5 does not divide 2, it divides (-8s+4r-2q+p )

 

Hence if a = 5k+2, choose b = 5m-2

 

Case IV: a = 5k-2

From the above working we can see that b must be of the form 5m+2

 

Hence for every choice of a, such that pa³+qa²+ra+s is divisible by 5, we can find infinite integers such that sb³+rb²+qb+p is divisible by 5.