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If $a$, $b$, $c$ some integers and consider $f(x)$ = $($$x^2$ + $bx$+ $c$) $\div$ $(x + a)$ in the domain (-$\infty$, a) $\cup$ (-a, $\infty$) then prove the following:

(A) For $a^2$ - $ab$ + $c$ = $1$, then the graph $f(x)$ contains exactly four integral points. Namely, $(-a + 1, b – 2a + 2)$, $(-a + 1, b – 2a-2)$, $(-a - 1, b – 2a + 2)$ and $(-a -1, b – 2a – 2)$.

(B) For $a^2$ - $ab$ + $c$ = $-1$, then the graph $f(x)$ contains exactly two integral points. Namely, $(-a + 1, b – 2a)$, $(-a - 1, b – 2a)$.

Thanks to all members of this site.

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Do you mean -a instead of a in the domain? – Ishan Banerjee Jan 30 '13 at 10:22

Note that $$(x^2+bx+c)=(x+a)(x+b-a)+(a^2-ab+s). $$

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! I need still more explanation. Because, if we divide both sides by (x+a) for your equation, we get (x+b-a) + (a^2 -ab + s)/(x+a). How this is related to those four and two integral points of (A) and (B) respectively. – rr282828 Jan 31 '13 at 5:01

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