Question 577369
There is probably a very simple and elegant solution, but I will only see it after I post this messy one,
Consider the function {{{f(x)=p(x)-q(x)}}}
{{{f(1)=p(1)-q(1)=0}}}
{{{f(2)=p(2)-q(2)=0}}}
{{{f(3)=p(3)-q(3)=0}}}
There are two many options, but I would try the simplest polynomial function with those three zeros:
{{{f(x)=(x-1)(x-2)(x-3)=(x^2-3x+2)(x-3)=x^3-3x^2+2x-2x^2+9x-6=x^3-6x^2+11x-6}}}
I can split {{{f(x)}}} into {{{p(x)}}} and {{{-q(x)}}} many ways, but I need both polynomials to be of degree 3.
I'll try something. There are probably better, simpler ways.
{{{p(x)=2x^3-2x^2+2x-2=2(x^3-x^2+x-1)}}} and
{{{-q(x)=-x^3-4x^2+9x-4}}} made up so that {{{p(x)+(-q(x))=p(x)-q(x)=x^3-6x^2+11x-6}}}
{{{q(x)=x^3+4x^2-9x+4}}}
Polynomials p and q are of degree 3
{{{p(1)=2(1^3-1^2+1-1)=0}}} and {{{q(1)=1^3+4*1^2-9*1+4=1+4-9+4=0}}}, so  {{{p(1)=q(1)}}}
{{{p(2)=2(2^3-2^2+2-1)=2(8-4+2-1)=2*5-10}}} and {{{q(2)=2^3+4*2^2-9*2+4=8+4*4-18+4=8+16-18+4=10}}}, so  {{{p(2)=q(2)}}}
{{{p(3)=2(3^3-3^2+3-1)=2(27-9+3-1)=2*20=40}}} and {{{q(3)=3^3+4*3^2-9*3+4=27+4*9-27+4=27+36-27+4=40}}}, so  {{{p(3)=q(3)}}}
I  should not have to prove that, brcause there could only be 3 intersection points for two polynomials of degree 3, but ...
{{{p(4)=2(4^3-4^2+4-1)=2(64-16+4-1)=2*51=102}}} and {{{q(4)=4^3+4*4^2-9*4+4=64+64-36+4=128-36+4=96}}}, so  {{{p(4)}}} and {{{q(4)}}} are different.