Solve-for-positive-integers-abcd-abc-a-1-b-1-c-1-

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 :

Extra \left or missing \right

abc (d + 1) = (a + 1) (b + 1) (c + 1)

\Rightarrow abc ∣ (a + 1) (b + 1) (c + 1) \dots \dots \dots . A

\Rightarrow {\begin{cases} a ∣ (a + 1) \Rightarrow a = 1 \dots \dots \dots (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) \dots \dots \dots . 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 \dots

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} \dots \dots \dots . . 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, \dots

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 \dots