m(p+1)=a ((q(p+1))/(r+1))=b & ((p(p+1))/(q+1))=c and ((r(p+1))/(m+1))=d find either of p,q,r,m in terms of a,b,c,d.


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Question Number 67337 by ajfour last updated on 25/Aug/19

$m (p + 1) = a$ $\frac{q (p + 1)}{r + 1} = b & \frac{p (p + 1)}{q + 1} = c$ $a n d \frac{r (p + 1)}{m + 1} = d$ $f i n d e i t h e r o f p, q, r, m i n t e r m s o f$ $a, b, c, d .$
	Answered by mr W last updated on 25/Aug/19
	$m=(a/(p+1)) q=((p(p+1))/c)−1 r=((q(p+1))/b)−1=[((p(p+1))/c)−1]((p+1)/b)−1 r(p+1)=d(m+1) {[((p(p+1))/c)−1]((p+1)/b)−1}(p+1)=((a/(p+1))+1)d let p+1=t {[(t−1)t−c]t−bc}t^2 =(a+t)bcd ⇒t^5 −t^4 −ct^3 −bct^2 −bcdt−abcd=0 ......???$
	$m = \frac{a}{p + 1}$ $q = \frac{p (p + 1)}{c} - 1$ $r = \frac{q (p + 1)}{b} - 1 = [\frac{p (p + 1)}{c} - 1] \frac{p + 1}{b} - 1$ $r (p + 1) = d (m + 1)$ ${[\frac{p (p + 1)}{c} - 1] \frac{p + 1}{b} - 1} (p + 1) = (\frac{a}{p + 1} + 1) d$ $l e t p + 1 = t$ ${[(t - 1) t - c] t - b c} t^{2} = (a + t) b c d$ $\Rightarrow t^{5} - t^{4} - {c t}^{3} - {b c t}^{2} - b c d t - a b c d = 0$ $. . . . . . ? ? ?$
	Commented by ajfour last updated on 26/Aug/19

	$t r y i n g a g a i n t o s o l v e t h e g e n e r a l$ $q u i n t i c i h a v e s t u c k u p o n t h i s S i r!$
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