Question 1207386
<br>
Let {{{P(x)=ax^2+bx+c}}}<br>
Then<br>
{{{P(P(x))-P(x)^2=(a(ax^2+bx+c)^2+b(ax^2+bx+c)+c)-(ax^2+bx+c)^2}}}<br>
{{{P(P(x))-P(x)^2=(a-1)(ax^2+bx+c)^2+b(ax^2+bx+c)+c}}}<br>
We need to have<br>
{{{P(P(x))-P(x)^2=x^2+x+2016}}}<br>
{{{(a-1)(ax^2+bx+c)^2+b(ax^2+bx+c)+c=x^2+x+2016}}}<br>
On the left hand side of the equation, {{{(ax^2+bx+c)^2}}} is going to give {{{x^4}}} and {{{x^3}}} terms; but there are no terms of that degree on the right hand side.  That means we must have (a-1)=0, or a=1.<br>
Then<br>
{{{b(ax^2+bx+c)+c=x^2+x+2016}}}<br>
Equating the coefficients of x^2 on the two sides of the equation, we need to have<br>
{{{ab=1}}}<br>
and since a=1, b=1 also.  And now, with a=1 and b=1, we have<br>
{{{x^2+x+c+c=x^2+x+2016}}}
{{{2c=2016}}}
{{{c=1008}}}<br>
And we have the polynomial we need.<br>
ANSWER: {{{P(x)=x^2+x+1008}}}<br>