Good solutions to this problem were received from Tyrone of
Cyfarthfa High School in Merthyr Tydfil, and Koopa of Boston
College in the USA.

Tyrone solved the problem by relating both polynomials to
$(x+1)^2$ :

$$ \eqalign { p(x)=x^2 + 2x \Rightarrow &p
&=&(x+1)^2 - 1 \\ &p+1 &=& (x+1)^2 \\ }
$$

(1)

$$ \eqalign { q(x)=x^2 + x + 1 \Rightarrow &q
&=&(x+1)^2 - x \\ &q+x &=& (x+1)^2 \\ }
$$

(2)

$ \Rightarrow p+1=q+x $ (combining
eqns (1) and (2)).

But $x=(p+1)^{1/2}-1$ (from eqn (1)). So

$$ \eqalign { \Rightarrow p+1&=&q+ ((p+1)^{1/2}-1) \\

&=&q+ (p+1)^{1/2}-1 }$$

$$ \eqalign { p-q+2&=&(p+1)^{1/2} \\

(p-q+2)^2&=&p+1}$$.

Squaring the bracket,

$$ \eqalign { &p^2-pq+2p-pq+q^2-2q+2p-2q+4=p+1 \\ &p^2 -2pq+q^2+4p-4q+4=p+1 \\ &p^2-2pq+q^2+3p-4q+3=0 } $$.