# Integer points on graphical functions

If $a$, $b$, $c$ some integers and consider $f(x)$ = $($$x^2 + bx+ c) \div (x + a) in the domain (-\infty, a) \cup (-a, \infty) then prove the following: (A) For a^2 - ab + c = 1, then the graph f(x) contains exactly four integral points. Namely, (-a + 1, b – 2a + 2), (-a + 1, b – 2a-2), (-a - 1, b – 2a + 2) and (-a -1, b – 2a – 2). (B) For a^2 - ab + c = -1, then the graph f(x) contains exactly two integral points. Namely, (-a + 1, b – 2a), (-a - 1, b – 2a). Thanks to all members of this site. - Do you mean -a instead of a in the domain? – Ishan Banerjee Jan 30 '13 at 10:22 ## 1 Answer Note that$$(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