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# Polynomial Relations

##### Stage: 5 Challenge Level:

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 }.