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Question Number 163812 by mathlove last updated on 11/Jan/22

	Answered by cortano1 last updated on 11/Jan/22

	$p (x) = x^{3} + 1$ $p (4) = 65$
	Commented by mathlove last updated on 11/Jan/22

	$h o w ?$
	Commented by Rasheed.Sindhi last updated on 11/Jan/22

	$T o w h o m$ $w h o l i k e d t h e a b o v e p o s t$ $W o u l d y o u l i k e t o t e l l t h e r e a s o n o f$ $l i k i n g p l e a s e ?$
	Commented by mr W last updated on 11/Jan/22

	$m a y b e h e j u s t a l s o w a n t e d t o k n o w$ $‘ ‘ {h o w}^{″} . i n t h i s f o r u m m a n y p e o p l e$ $u s e ‘ ‘ {l i k e}^{″} n o t t o s h o w t h a t t h e y r e a l l y$ $l i k e a p o s t . i t h i n k m a n y p e o p l e g i v e$ $t h e m s e l v e s a l i k e s u c h t h a t t h e i r$ $p o s t c a n n o t b e r e d f l a g g e d o r f o r w h a t$ $w h o k n o w s .$
	Commented by Rasheed.Sindhi last updated on 12/Jan/22

	$M a y b e s / h e h a s g i v e n a l i k e t o i n c r e a s e$ $t h e i m p o r t a n c e o f t h e p o s t b u t u l t i m a t e l y$ ${i t}^{'} l l b e t h e c a u s e t o d e c r e a s e i m p o r t a n c e$ $o f^{'} {l i k e s}^{'} i t s e l f!$
	Answered by Ar Brandon last updated on 11/Jan/22

	$p (3) = 28 therefore p (x) must be a non - zero polynomial .$ $Let p (x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{n} x^{n}, a_{n} \neq 0$ $Suppose that p (x) + p (\frac{1}{x}) = p (x) p (\frac{1}{x}) for all x \neq 0$ $Then \underset{r = 0}{\sum^{n}} a_{r} x^{r} + \underset{r = 1}{\sum^{n}} \frac{a_{r}}{x^{r}} = (\underset{r = 0}{\sum^{n}} a_{r} x^{r}) (\underset{r = 1}{\sum^{n}} \frac{a_{r}}{x^{r}})$ $Multiplying through by x^{n}, we get that$ $\underset{r = 0}{\sum^{n}} a_{r} x^{n + r} + \underset{r = 0}{\sum^{n}} a_{r} x^{n - r} = (\underset{r = 0}{\sum^{n}} a_{r} x^{r}) (\underset{r = 0}{\sum^{n}} a_{r} x^{n - r})$ $That is,$ $(a_{0} x^{n} + a_{1} x^{n + 1} + \cdot \cdot \cdot a_{n} x^{2 n}) + (a_{0} x^{n} + a_{1} x^{n - 1} + \cdot \cdot \cdot + a_{n - 1} x + a_{n})$ $= (a_{0} + a_{1} x + \cdot \cdot \cdot a_{n} x^{n}) (a_{0} x^{n} + a_{1} x^{n - 1} + \cdot \cdot \cdot + a_{n - 1} x + a_{n})$ $Equating the corresponding coefficients of powers of x,$ $we have$ $a_{n} \overset{x^{0}}{=} a_{0} a_{n}, a_{n - 1} \overset{x^{1}}{=} a_{0} a_{n - 1} + a_{1} a_{n}$ $a_{n - 2} \overset{x^{2}}{=} a_{2} a_{n} + a_{1} a_{n - 1} + a_{n - 2} a_{0}$ $2 a_{0} \overset{x^{n}}{=} a_{0}^{2} + a_{n}^{2}$ $a_{n} \overset{x^{2 n}}{=} a_{0} a_{n} \Rightarrow a_{0} = 1 (since a_{n} \neq 0)$ $a_{n - 1} = a_{0} a_{n - 1} + a_{1} a_{n} \Rightarrow a_{1} a_{n} = 0 \Rightarrow a_{1} = 0$ $a_{n - 2} = a_{2} a_{n} + a_{1} a_{n - 1} + a_{n - 2} a_{0} \Rightarrow a_{n - 2} = a_{2} a_{n} + a_{n - 2} \Rightarrow a_{2} = 0$ $Continuing this process, we get that a_{n - 1} = 0 and 2 = 1 + a_{n}^{2} .$ $Hence a_{n} = \pm 1 . Therefore$ $p (x) = 1 \pm x^{n}$ $Since we are given that p (3) = 28 we have$ $28 = 1 \pm 3^{n}$ $Therefore p (x) cannot be 1 - x^{n} . Thus, p (x) = 1 + x^{n} and$ $28 = 1 + 3^{n} and hence n = 3 . So p (x) = 1 + x^{3} and p (4) = 65 .$
	Commented by mr W last updated on 11/Jan/22

	$g o o d s o l u t i o n!$
	Commented by Ar Brandon last updated on 11/Jan/22
	��
	Commented by Tawa11 last updated on 11/Jan/22

	$Great sir$
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