Determine-all-triplets-a-b-c-of-positive-integers-which-satisfy-1-a-1-b-1-c-1-2-

Question Number 152753 by mathdanisur last updated on 01/Sep/21

Determine all triplets (a; b; c) of positive

integers which satisfy :

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}

Commented by Rasheed.Sindhi last updated on 01/Sep/21

Determine all triplets (a; b; c) of positive

integers which satisfy :

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}

∙ C 1 : a = b = c

a = b = c = k (say)

\frac{1}{k} + \frac{1}{k} + \frac{1}{k} = \frac{1}{2}

\frac{3}{k} = \frac{1}{2} \Rightarrow k = 6

a = b = c = 6

∙ C 2 : b = c

b = c = k (say)

\frac{1}{a} + \frac{1}{k} + \frac{1}{k} = \frac{1}{2}

\frac{k + 2 a}{ak} = \frac{1}{2}

2 k + 4 a = ak

2 (k + 2 a) = ak

2 ∣ a \lor 2 ∣ k \lor (2 ∣ a \land 2 ∣ k)

★ \underset{―}{2 ∣ a \lor 2 ∣ k \lor (2 ∣ a \land 2 ∣ k) :}

L e t a = 2 m

2 (k + 2 a) = ak \Rightarrow 2 (k + 2 (2 m)) = (2 m) k

2 (k + 4 m) = 2 mk

k + 4 m = mk

m = \frac{k}{k - 4}

possible values of k for which m \in Z^{+}

k = 5, 6, 8 \Rightarrow m = 5, 3, 2

b = c = k = 5 \Rightarrow a = 2 m = 10

b = c = k = 6 \Rightarrow a = 2 m = 6

b = c = k = 8 \Rightarrow a = 2 m = 4

o ★ \underset{―}{2 ∣ a \lor 2 ∣ k \lor (2 ∣ a \land 2 ∣ k) :}

k = 2 m

2 (k + 2 a) = ak \Rightarrow 2 (2 m + 2 a) = 2 ma

\underset{(m or a \in E^{+})}{2 m + 2 a = ma}

a = \frac{2 m}{m - 2}

possible positive integral values of m

for which a is also positive integer .

m = 3, 4, 6 \Rightarrow a = 6, 4, 3

b = c = k = 2 m = 6, 8, 12

(6, 6, 6), (4, 8, 8), (3, 12, 12)

{a, b, c}

= {3, 12, 12}, {4, 8, 8}, {10, 5, 5}, {6, 6, 6}

★ \underset{―}{2 ∣ a \lor 2 ∣ k \lor (2 ∣ a \land 2 ∣ k) :}

a = 2 m, k = 2 n

2 (k + 2 a) = ak \Rightarrow 2 (2 n + 2 . 2 m) = 2 m . 2 n

4 (n + 2 m) = 4 mn

n + 2 m = mn

mn - 2 m = n

m = \frac{n}{n - 2}

possible values of n

n = 3, 4 \Rightarrow m = 3, 2

a = 2 m = 6, 4; b = c = k = 2 n = 6, 8

(a, b, c) = (6, 6, 6), (4, 8, 8)

{a, b, c}

= {3, 12, 12}, {4, 8, 8}, {10, 5, 5}, {6, 6, 6}

∙ C 3 : a \neq b \neq c \neq a

a = k, b = pk, c = qk; p, q \in Z^{+}

\frac{1}{k} + \frac{1}{pk} + \frac{1}{qk} = \frac{1}{2}

\frac{p + q + pq}{pqk} = \frac{1}{2}

2 (p + q + pq) = pqk

2 ∣ p : p = 2 m

2 (2 m + q + pq) = (2 m) qk

2 m + q + 2 mq = qmk

2 m (1 + q) = qmk - q

2 m (1 + q) = q (mk - 1)

\frac{q}{q + 1} = \frac{2 m}{mk - 1}

\dots . .

\dots

(6, 12, 4) \dots

Continue

Commented by mathdanisur last updated on 02/Sep/21

Very nice S er thaankyou

Commented by Rasheed.Sindhi last updated on 02/Sep/21

P l w r i t e a l l a n s w e r s, i f y o u h a v e .

Commented by mathdanisur last updated on 02/Sep/21

(3, 7, 42); (3, 8, 24); (3, 9, 18); (3, 10, 15)

(3, 42, 12); (4, 5, 20); (4, 6, 12); (4, 8, 8)

(5, 5, 10); (6, 6, 6) and permutations

Commented by Rasheed.Sindhi last updated on 02/Sep/21

T han k Y o ⋓ S r!

Commented by Rasheed.Sindhi last updated on 02/Sep/21

m a t h d a n i s u r s e r, a r e y o u s t u d y i n g ?

Commented by mathdanisur last updated on 02/Sep/21

Yes Ser, I will finish the highest level

Commented by Rasheed.Sindhi last updated on 03/Sep/21

S e r, W h y {d o n}^{'} t y o u t r y t o s o l v e

a n y q u e s t i o n i n t h i s f o r u m ? (A l t h o u g h

y o u h a v e g i v e n a n s w e r s t o s o m e

q u e s t i o n s b u t n o t c o m p l e t e s o l u t i o n s ?)

Commented by mathdanisur last updated on 03/Sep/21

I^{'} ve solved, well, I know the solution

to most of the questions I sak, I just

want different solutions, I^{'} m learning,

{it}^{'} s a beautiful place, there are educated

people (like you) and I like your solutions,

thank you very much for your solutions \dots

Commented by Rasheed.Sindhi last updated on 03/Sep/21

^{∙} Th \propto nk You very much!

^{∙} Your one solution seems incorrect :

(3, 42, 12) . Pl correct it .

^{∙} I wish you share your solution for

this question because my solutions

are not fine enough .

^{∙} Others also want variety in solutions

so if you know how to solve some

questions pl share their

solutions . (Particularly solution for

this question .)

Commented by mathdanisur last updated on 03/Sep/21

Good S er, thank you, sorry (3, 12, 12)

Commented by peter frank last updated on 03/Sep/21

your selfish if you have soln then

you dont what to share it this

not good

Answered by Rasheed.Sindhi last updated on 01/Sep/21

\underset{(i)}{\underset{⏟}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}}}; a, b, c \in Z^{+}; (a, b, c) = ?

\frac{1}{a} + \frac{1}{b} = \frac{1}{2} - \frac{1}{c}

\frac{1}{a} + \frac{1}{b} > 0 \Rightarrow \frac{1}{2} - \frac{1}{c} > 0 \Rightarrow c \neq 1, 2

∵ (i) i s s y m m e t r i c i n a, b, c

∴ a, b, c \neq 1, 2 i - e a, b, c > 2

N o w,

(i) \Rightarrow 2 (ab + bc + ca) = abc

\Rightarrow A t l e a s t o n e o f a, b, c i s even

L e t a = 2 k

2 (2 bk + bc + 2 ck) = 2 kbc

2 bk + bc + 2 ck = kbc

2 k (b + c) = bc (k - 1)

\Rightarrow A t l e a s t o n e o f b, c, (k - 1) i s even .

If (1) (k - 1) \in E^{+} \Rightarrow k \in O^{+} \Rightarrow a i s a

singly even n u m b e r i n

2 (ab + bc + ca) = abc

\dots \dots .

If (2) b i s even . S a y b = 2 m

2 k (b + c) = bc (k - 1)

2 k (2 m + c) = 2 mc (k - 1)

k (2 m + c) = mc (k - 1)

k ∣ k - 1 \Rightarrow k = 2 \Rightarrow a = 2 k = 4

\Rightarrow

\dots \dots . .

\dots \dots . .

Commented by mathdanisur last updated on 02/Sep/21

Nice S er thank you

Answered by Rasheed.Sindhi last updated on 03/Sep/21

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \dots \dots \dots \dots . (★)

\frac{1}{a} + \frac{1}{b} = \frac{1}{2} - \frac{1}{c}

∵ \frac{1}{a} + \frac{1}{b} > 0 \Rightarrow \frac{1}{2} - \frac{1}{c} > 0 \Rightarrow c > 2

∵ T h e e q u a t i o n i s s y m m e t r i c i n

a, b, c \Rightarrow a, b, c > 2

(★) \Rightarrow 2 (ab + bc + ca) = abc

\Rightarrow A t l e a s t o n e o f a, b, c i s even .

C 1 : o n l y o n e o f a, b, c i s even :

\frac{1}{2 l} + \frac{1}{2 m + 1} + \frac{1}{2 n + 1} = \frac{1}{2}

▸ \overset{\times}{2} {(2 l) (2 m + 1) + (2 m + 1) (2 n + 1) + (2 n + 1) (2 l)}

= \overset{\times}{2} l (2 m + 1) (2 n + 1)

▸ (2 l) (2 m + 1) + (2 m + 1) (2 n + 1) + (2 n + 1) (2 l)

= l (2 m + 1) (2 n + 1)

▸ 4 l m + 2 l + 4 m n + 2 m + 2 n + 1 + 4 n l + 2 l

= 4 l m n + 2 l m + 2 l n + l

▸ 4 l m + 4 m n + 4 l n + 4 l + 2 m + 2 n + 1

- 4 l m n - 2 l m - 2 l n - l = 0

▸ 2 l m + 4 m n + 2 l n + 2 m + 2 n + 3 l + 1 = 4 l m n

▸ 2 (l m + 2 m n + l n + m + n) + (3 l + 1) = 4 l m n

\Rightarrow 3 l + 1 i s even \Rightarrow l i s odd

L e t l = 2 k + 1

▸ 2 {(2 k + 1) m + 2 m n + (2 k + 1) n + m + n} + (3 (2 k + 1) + 1) = 4 (2 k + 1) m n

▸ 2 {2 k m + 2 m n + 2 k n + 2 n + 2 m} + 6 k + 4 = 4 (2 k + 1) m n

▸ \overset{\times}{4} ({2 k m + 2 m n + 2 k n + 2 n + 2 m} + 3 k + 2) = \overset{\times}{4} (2 k + 1) m n

▸ 2 k m + m n + 2 k n + 2 n + 2 m + 2 + 3 k = 2 k m n

▸ 2 (k m + k n + n + m + 1) + 3 k + m n = 2 k m n

\Rightarrow 3 k + m n i s even .

\Rightarrow {\begin{cases} k, m & n a r e odd \\ k i s even & (m o r n i s even) \end{cases}

C o n t i n u e

C 1 : o n l y t w o o f a, b, c i s even :

\frac{1}{2 l} + \frac{1}{2 m} + \frac{1}{2 n + 1} = \frac{1}{2}

\frac{m (2 n + 1) + l (2 n + 1) + l m}{2 l m (2 n + 1)} = \frac{1}{2}

\frac{m (2 n + 1) + l (2 n + 1) + l m}{l m (2 n + 1)} = 1

\frac{1}{l} + \frac{1}{m} + \frac{1}{2 n + 1} = 1

C 1 : A l l o f a, b, c a r e even :

\frac{1}{2 l} + \frac{1}{2 m} + \frac{1}{2 n} = \frac{1}{2}

\frac{l m + m n + n l}{2 l m n} = \frac{1}{2}

l m + m n + n l = l m n

l ∣ m n \land m ∣ n l \land n ∣ l m

\frac{1}{l} + \frac{1}{m} + \frac{1}{n} = 1