Number of positive integers x for which f(x)=x^3 −8x^2 +20x−13 is a prime number are ?


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Question Number 31003 by rahul 19 last updated on 01/Mar/18

$N u m b e r o f p o s i t i v e i n t e g e r s x f o r$ $w h i c h f (x) = x^{3} - 8 x^{2} + 20 x - 13 i s a$ $p r i m e n u m b e r a r e ?$
	Answered by MJS last updated on 07/Mar/18

	$f (x) = (x - z_{1}) (x - z_{2}) (x - z_{3})$ $this can only be prime if only one$ $of the brackets contain a prime$ $number, and the two others are$ $worth 1 or - 1$ $so we need the zeros of f (x)$ $z_{1} z_{2} z_{3} = 13, so {let}^{'} s try$ $x = \pm 1 or x = \pm 13$ $f (1) = 0$ $f (x) = (x - 1) (x^{2} - 7 x + 13)$ $[x^{2} - 7 x + 13 = 0$ $\frac{7}{2} \pm \sqrt{\frac{49}{4} - 13} = \frac{7}{2} \pm \sqrt{- \frac{3}{4}}$ $no solution \in R]$ $so (x - 1) = \pm 1 \land (x^{2} - 7 x + 13) is prime$ $or (x - 1) is prime \land (x^{2} - 7 x + 13) = \pm 1$ $x - 1 = 1 \Rightarrow x = 2$ $2^{2} - 7 \cdot 2 + 13 = 3 \Rightarrow f (2) is prime$ $x - 1 = - 1 \Rightarrow x = 0$ $0^{2} - 7 \cdot 0 + 13 = 13 \Rightarrow f (0) is prime$ $but x must be a positive integer$ $x^{2} - 7 x + 13 = 1$ $\frac{7}{2} \pm \sqrt{\frac{49}{4} - 12} = \frac{7}{2} \pm \frac{1}{2}$ $x_{1} = 3; (3 - 1) = 2 \Rightarrow f (3) is prime$ $x_{2} = 4; (4 - 1) = 3 \Rightarrow f (4) is prime$ $x^{2} - 7 x + 13 = - 1$ $has no solution \in R$ $number of positive integers is 3$
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