Find-all-ordered-pairs-a-b-so-that-ab-a-b-is-an-integer-a-and-b-are-integers-

Question Number 10944 by 314159 last updated on 03/Mar/17

Find all ordered pairs (a, b) so that \frac{ab}{a + b} is an integer .

(a and b are integers) .

Commented by FilupS last updated on 03/Mar/17

n = \frac{a b}{a + b}, n, a, b \in Z

n = \frac{a b}{a + b} \Rightarrow a b ⩾ a + b

∵ \forall (a b < a + b) \Rightarrow 0 < n < 1 \Rightarrow n \notin Z

a b ⩾ a + b

a (b - 1) = b

a = \frac{b}{b - 1}

∴ \frac{b}{b - 1} \in Z

if b = 2 \Rightarrow a = \frac{2}{2 - 1} = 2

∴ (2, 2) is a solution .

I am unsure if this is the only solution

because this is a bad attempt

Commented by mrW1 last updated on 04/Mar/17

i t h i n k t h e r e s h o u l d b e i n f i n i t e n u m b e r p a i r s :

2 / 2

3 / 6

4 / 4 4 / 12

5 / 20

6 / 6 6 / 12 6 / 30

7 / 42

8 / 8 8 / 24 8 / 56

9 / 18 9 / 72

10 / 10 10 / 15 10 / 40

11 / 110

12 / 12 12 / 24 12 / 36 12 / 60 12 / 132

13 / 156

14 / 14 14 / 84 14 / 182

\dots \dots

Answered by mrW1 last updated on 04/Mar/17

l e t \frac{a b}{a + b} = n

l e t b ⩾ a a n d b = k a (k ⩾ 1)

\Rightarrow \frac{a k a}{(1 + k) a} = n

a = \frac{1 + k}{k} \times n = n + \frac{n}{k}

w i t h \frac{n}{k} = i o r k = \frac{n}{i}

\Rightarrow a = n + i

b = k a = \frac{n}{i} (n + i) = n (1 + \frac{n}{i})

w i t h \frac{n}{i} = j o r n = j i

\Rightarrow a = n + i = j i + i = (1 + j) i

\Rightarrow b = n (1 + j) = j (1 + j) i

i . e . t h e s o l u t i o n i s

{\begin{cases} a = (1 + j) i \\ b = j (1 + j) i \end{cases}

i, j \in N

e . g .

i = 1, j = 1, 2, 3, 4 \dots

a / b = 2 / 2, 3 / 6, 4 / 12, 5 / 20, 6 / 30 \dots \dots

i = 2, j = 1, 2, 3, 4 \dots

a / b = 4 / 4, 6 / 12, 8 / 24, 10 / 40, 12 / 60 \dots \dots

i = 3, j = 1, 2, 3, 4 \dots

a / b = 6 / 6, 9 / 18, 12 / 36, 15 / 60, 18 / 90 \dots \dots

\dots . .

W e s e e i n t h e s o l u t i o n t h a t a i s t h e

p r o d u c t o f 2 n u m b e r s a n d b i s t h e

p r o d u c t o f 3 n u m b e r s a n d b = a \times j .

F o r e v e r y a ⩾ 2,

i f a i s a p r i m e n u m b e r, t h e r e i s o n e

c o r r e s p o n d i n g v a l u e f o r b, a n d

i f a i s n o p r i m e n u m b e r, t h e r e a r e t w o

o r m o r e c o r r e s p o n d i n g v a l u e s f o r b .

E . g . a = 7 (p r i m e n u m b e r) w h i c h c a n

o n l y b e e x p r e s s e d a s

7 \times 1 (j = 6)

\Rightarrow t h e r e i s o n l y o n e v a l u e f o r b :

b = 7 \times 6 = 42

E . g . a = 12 w h i c h c a n b e e x p r e s s e d a s

12 \times 1 (j = 11)

3 \times 4 (j = 2)

4 \times 3 (j = 3)

2 \times 6 (j = 1)

6 \times 2 (j = 5)

\Rightarrow t h e r e a r e 5 v a l u e s f o r b :

b = 12 \times 1 = 12

b = 12 \times 2 = 24

b = 12 \times 3 = 36

b = 12 \times 5 = 60

b = 12 \times 11 = 132

t h e p a i r s o f a a n d b f o r a f r o m 2 t i l l 14

s e e c o m m e n t a b o v e .

Commented by FilupS last updated on 04/Mar/17

I Tried a slimilar subtitution, but couldnt

work with it . Awesome job

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