What shall we do to find the polynom P knowing that #(X^3+1)# divides #P-1# and #(X^3-1)# divides #P+1# ?

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Answer 1 · 2017-11-10T21:29:59

By inspection, #P(X) = -X^3# is a solution.

More generally, for any polynomial #Q(X)#, a solution is:

#P(X) = Q(X)(X^6-1)-X^3#

To say that #(X^3+1)# divides #P(x)-1# is to say that #P(x)-1# is a multiple of #(X^3+1)#
To say that #(X^3-1)# divides #P(x)+1# is to say that #P(x)+1# is a multiple of #(X^3-1)#

Simplest solution

Note that:

#(X^3+1)# divides #color(blue)(-X^3)-1 = -(X^3+1)#

#(X^3-1)# divides #color(blue)(-X^3)+1 = -(X^3-1)#

So:

#P(X) = color(blue)(-X^3)#

is a solution.

Another solution

Another solution is #P(X) = color(blue)(-X^9)# since:

#(X^3+1)(-X^6+X^3-1) = color(blue)(-X^9)-1 = P(X)-1#

#(X^3-1)(-X^6-X^3-1) = color(blue)(-X^9)+1 = P(X)+1#

Similarly #P(X) = -X^(3(2k+1))# is a solution for any non-negative integer #k#.

A generalised solution

More generally still, let #Q(X)# be any polynomial and:

#P(X) = Q(X)(X^6-1)-X^3 = Q(X)(X^3-1)(X^3+1)-X^3#

We find:

#P(X)-1 = Q(X)(X^6-1)-X^3-1#

#color(white)(P(X)-1) = (X^3+1)(Q(X)(X^3-1)-1)#

So #P(X)-1# is divisible by #(X^3+1)#

Also, we find:

#P(X)+1 = Q(X)(X^6-1)-X^3+1#

#color(white)(P(X)+1) = (X^3-1)(Q(X)(X^3+1)-1)#

So #P(X)+1# is divisible by #(X^3-1)#