happy new year {a;b;c}∈Z−{0} p(x)=ax^2 +bx+c p(a)=0 p(b)=0 p(1)=?


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Question Number 162787 by amin96 last updated on 01/Jan/22
$happy new year {a;b;c}∈Z−{0} p(x)=ax^2 +bx+c p(a)=0 p(b)=0 p(1)=?$
$happy new year$ ${a; b; c} \in Z - {0}$ $p (x) = {a x}^{2} + b x + c$ $p (a) = 0$ $p (b) = 0$ $p (1) = ?$
	Answered by alephzero last updated on 01/Jan/22
	$p(a) = aa^2 + ba + c = a^3 + ba + c = 0 p(b) = ab^2 + bb + c = ab^2 + b^2 + c = 0 p(1) = a^2 + b + c = ? { ((a^3 + ba + c = 0)),((ab^2 + b^2 + c =0)) :} ⇒ { ((a^3 + ba = −c)),((ab^2 + b^2 = −c)) :} ⇒ a^3 + ba = ab^2 + b^2 ⇒ a(a^2 + b) = b^2 (a + 1) (a_1 ; b_1 ) = (−(1/2); −(1/2)) (a_2 ; b_2 ) = (−2; 4) (a_3 ; b_3 ) = (0; 0) But {a; b; c} ∈ Z−{0} ⇒ (a; b) = (−2; 4) ⇒ { ((−2^3 + (−2) × 4 = −c)),((−2 × 4^2 + 4^2 = −c)) :} { ((−8 + (−8) = −c)),((−36 + 16 = −c)) :} ⇒ −16 = −c ⇒ c = 16 a^2 + b + c = (−2)^2 + 4 + 16 = 4 + 4 + + 16 = 8 + 16 = 24 p(1) = 24$
	$p (a) = {a a}^{2} + b a + c = a^{3} + b a + c = 0$ $p (b) = {a b}^{2} + b b + c = {a b}^{2} + b^{2} + c = 0$ $p (1) = a^{2} + b + c = ?$ ${\begin{cases} a^{3} + b a + c = 0 \\ {a b}^{2} + b^{2} + c = 0 \end{cases}$ $\Rightarrow {\begin{cases} a^{3} + b a = - c \\ {a b}^{2} + b^{2} = - c \end{cases}$ $\Rightarrow a^{3} + b a = {a b}^{2} + b^{2}$ $\Rightarrow a (a^{2} + b) = b^{2} (a + 1)$ $(a_{1}; b_{1}) = (- \frac{1}{2}; - \frac{1}{2})$ $(a_{2}; b_{2}) = (- 2; 4)$ $(a_{3}; b_{3}) = (0; 0)$ $But {a; b; c} \in Z - {0}$ $\Rightarrow (a; b) = (- 2; 4)$ $\Rightarrow {\begin{cases} - 2^{3} + (- 2) \times 4 = - c \\ - 2 \times 4^{2} + 4^{2} = - c \end{cases}$ ${\begin{cases} - 8 + (- 8) = - c \\ - 36 + 16 = - c \end{cases}$ $\Rightarrow - 16 = - c$ $\Rightarrow c = 16$ $a^{2} + b + c = {(- 2)}^{2} + 4 + 16 = 4 + 4 +$ $+ 16 = 8 + 16 = 24$ $p (1) = 24$
	Commented by mr W last updated on 01/Jan/22

	$s o l u t i o n i s n o t u n i q u e .$ $f (x) = x^{2} + x - 2 i s a l s o o k .$ $f (1) = 0 .$ $i n f a c t a n y a = b = n, c = - n^{2} (n + 1) w i t h$ $n \in Z i s o k .$
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