What are the number of possible integral values of a polymonial such that $f (x) = a x^{3} + b x^{2} + c x + d, ((b, c, d \in Z) e q u a l s t o 1 a t x = 19 a n d e q u a l t o 2 a t x = 62$ .

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Solution

$a (19)^{3} + b (19)^{2} + c (19) + d = 1$
$a (62)^{3} + b (62)^{2} + c (62) + d = 2$
$\Rightarrow a (62^{3} - 19^{3}) + b (62^{2} - 19^{2}) + c (62 - 19) = 1$
$\Rightarrow 43 (i n t e g e r) = 1$
Not possible

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