A Game of Polynomials

We can modify the equation ${ x }^{ 2 }+x+2014$ with this: ${ x }^{ 2 }+Cx+2014+P$ , where $C$ is the total sum of Calvin and $P$ the sum of Peter. According to the equation, the roots are:

$x=\frac { -C\pm \sqrt { { C }^{ 2 }-8056-4P } }{ 2 } =-\frac { C }{ 2 } \pm \frac { \sqrt { { C }^{ 2 }-8056-4P } }{ 2 }$

Note that $C$ must be a multiple of 2 ( $2n$ ) and ${ C }^{ 2 }-8056-4P$ must be a perfect square ( $(a{ n })^{ 2 }$ ), where $a$ is a parameter (both $a$ and $n$ are integers too); with this, the roots are modified into this

$x=-\frac { 2n }{ 2 } \pm \frac { \sqrt { { (2n) }^{ 2 }-8056-4P } }{ 2 } =-n\pm \sqrt { { n }^{ 2 }-2014-P }$

Given that ${ C }^{ 2 }-8056-4P$ or ${ n }^{ 2 }-2014-P$ must be a perfect square ( $(a{ n })^{ 2 }$ ), then

${ n }^{ 2 }-2014-P={ (an) }^{ 2 }\longrightarrow P=(1-{ a }^{ 2 }){ n }^{ 2 }-2014$ .

If we substitute this in the roots

$-n\pm an\longrightarrow -n(1+a)\quad or\quad n(a-1)$ , given that $C=2n$ , then the roots are $-\frac{C}{2}$ $(1+a)$ or $\frac{C}{2}$ $(a-1)$

With this (and the note below), Calvin could win in a finite number of steps.

Note: Look that $P+2014=(1-{ a }^{ 2 }){ n }^{ 2 }$ , you can substitute this in the modified quadratic equation and you´ll get this

${ x }^{ 2 }+Cx+2014+P={ x }^{ 2 }+2nx+{ (1-a }^{ 2 }){ n }^{ 2 }$

Solving the quadratic formula you can get the roots written above. You´ll get two integer solutions if $a\neq 0$ .