Solve for positive integers: abcd + abc = (a+1)(b+1)(c+1)


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Question Number 155204 by mathdanisur last updated on 26/Sep/21

$Solve for positive integers :$ $abcd + abc = (a + 1) (b + 1) (c + 1)$
	Answered by MJS_new last updated on 27/Sep/21

	$a ⩽ b ⩽ c$ $all possible solutions for a b c d are$ $1 1 1 7$ $1 1 2 5$ $1 1 4 4$ $1 2 3 3$ $1 3 8 2$ $1 4 5 2$ $2 2 3 2$ $2 4 15 1$ $2 5 9 1$ $2 6 7 1$ $3 3 8 1$ $3 4 5 1$
	Commented by ajfour last updated on 27/Sep/21

	$s i r, i h a d e m a i l e d u c o u p l e o f$ $w e e k s b a c k; u d i n t r e p l y!$
	Commented by MJS_new last updated on 27/Sep/21
	Sorry I received no email from you, but this happened before, might be an issue of my provider.
	Commented by ajfour last updated on 30/Sep/21

	$d i n t u c h a s e o u r p r i z e f u r t h e r$ $s i r ? (m o r e o r l e s s i w a s a s k i n g$ $t h i s) a t l e a s t u c a n r e p l y h e r e$ $S i r M j S .$
	Commented by mathdanisur last updated on 27/Sep/21

	$Yes S er, thank you, solution if possible$
	Answered by Rasheed.Sindhi last updated on 28/Sep/21
	$Solve for positive integers: abcd + abc = (a+1)(b+1)(c+1) [_(−) abc(d+1)=(a+1)(b+1)(c+1) ⇒abc ∣ (a+1)(b+1)(c+1)..........A ⇒ { ((a∣(a+1)⇒a=1.........(i))),((a∣(b+1))),((a∣(c+1))),((a∣(a+1)(b+1))),((a∣(b+1)(c+1))),((a∣(c+1)(a+1))),((a∣(a+1)(b+1)(c+1))) :} a=1:⇒bc ∣ 2(b+1)(c+1)..........B ⇒ { ((b∣2⇒b=1,2)),((b∣(b+1)⇒b=1)),((b∣(c+1))),((b∣2(b+1))),((b∣2(c+1))),((b∣(b+1)(c+1))),((b∣2(b+1)(c+1))) :} a=1,b=1: B⇒c ∣ 4(c+1) ⇒ { ((c∣4⇒c=1,2,4)),((c∣(c+1)⇒c=1)),((c∣4(c+1))) :} a=1,b=1,c=1: d+1=(((a+1)(b+1)(c+1))/(abc)) ⇒d+1=((2.2.2)/(1.1.1))⇒d=7 a=1,b=2,c=1: d+1=((2.3.2)/(1.2.1))=6⇒d=5 a=1,b=1,c=2: ⇒d+1=((2.2.3)/(1.1.2))=6⇒d=5 Continue...$
	$Solve for positive integers :$ $Extra \left or missing \right$ $abc (d + 1) = (a + 1) (b + 1) (c + 1)$ $\Rightarrow abc ∣ (a + 1) (b + 1) (c + 1) . . . . . . . . . . A$ $\Rightarrow {\begin{cases} a ∣ (a + 1) \Rightarrow a = 1 . . . . . . . . . (i) \\ a ∣ (b + 1) \\ a ∣ (c + 1) \\ a ∣ (a + 1) (b + 1) \\ a ∣ (b + 1) (c + 1) \\ a ∣ (c + 1) (a + 1) \\ a ∣ (a + 1) (b + 1) (c + 1) \end{cases}$ $a = 1 :\Rightarrow bc ∣ 2 (b + 1) (c + 1) . . . . . . . . . . B$ $\Rightarrow {\begin{cases} b ∣ 2 \Rightarrow b = 1, 2 \\ b ∣ (b + 1) \Rightarrow b = 1 \\ b ∣ (c + 1) \\ b ∣ 2 (b + 1) \\ b ∣ 2 (c + 1) \\ b ∣ (b + 1) (c + 1) \\ b ∣ 2 (b + 1) (c + 1) \end{cases}$ $a = 1, b = 1 : B \Rightarrow c ∣ 4 (c + 1)$ $\Rightarrow {\begin{cases} c ∣ 4 \Rightarrow c = 1, 2, 4 \\ c ∣ (c + 1) \Rightarrow c = 1 \\ c ∣ 4 (c + 1) \end{cases}$ $a = 1, b = 1, c = 1 :$ $d + 1 = \frac{(a + 1) (b + 1) (c + 1)}{abc}$ $\Rightarrow d + 1 = \frac{2.2.2}{1.1.1} \Rightarrow d = 7$ $a = 1, b = 2, c = 1 :$ $d + 1 = \frac{2.3.2}{1.2.1} = 6 \Rightarrow d = 5$ $a = 1, b = 1, c = 2 :$ $\Rightarrow d + 1 = \frac{2.2.3}{1.1.2} = 6 \Rightarrow d = 5$ $Continue . . .$
	Answered by Rasheed.Sindhi last updated on 29/Sep/21

	$Solve for positive integers :$ $abcd + abc = (a + 1) (b + 1) (c + 1)$ $d + 1 = \frac{(a + 1) (b + 1) (c + 1)}{abc} . . . . . . . . . . . A$ $since d \in Z^{+}$ $∴ \frac{(a + 1) (b + 1) (c + 1)}{abc} \in Z^{+}$ $One of many possibilities is a ∣ (a + 1)$ $and we start from it$ $a ∣ (a + 1) \Rightarrow a = 1$ $Let a ⩽ b ⩽ c$ $a = 1 : A \Rightarrow d + 1 = \frac{(1 + 1) (b + 1) (c + 1)}{1 . bc}$ $= \frac{2 (b + 1) (c + 1)}{bc} \Rightarrow b = 1, 2, . . .$ $b = 1 : d + 1 = \frac{2 (1 + 1) (c + 1)}{1 . c}$ $= \frac{4 (c + 1)}{c} \Rightarrow c = 1, 2, 4$ $d (a, b, c) = \frac{(a + 1) (b + 1) (c + 1)}{abc} - 1$ $d (1, 1, 1) = \frac{(1 + 1) (1 + 1) (1 + 1)}{1.1.1} - 1 = 7$ $d (1, 1, 2) = \frac{(1 + 1) (1 + 1) (2 + 1)}{1.1.2} - 1 = 5$ $d (1, 1, 4) = \frac{(1 + 1) (1 + 1) (4 + 1)}{1.1.4} - 1 = 4$ $a = 1 : d + 1 = \frac{(1 + 1) (b + 1) (c + 1)}{1 . b . c}$ $= \frac{2 (b + 1) (c + 1)}{b . c} \Rightarrow b = 1, 2$ $b = 2 :$ $d + 1 = \frac{2 (2 + 1) (c + 1)}{1.2 . c} = \frac{3 (c + 1)}{c} \Rightarrow c = 1, 3$ $d (1, 2, 3) = \frac{(1 + 1) (2 + 1) (3 + 1)}{1.2.3} - 1 = 3$ $Continue . . .$
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