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Question Number 152753 by mathdanisur last updated on 01/Sep/21
Determine all triplets (a;b;c) of positive integers which satisfy: (1/a) + (1/b) + (1/c) = (1/2)
$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{triplets}\:\left(\mathrm{a};\mathrm{b};\mathrm{c}\right)\:\mathrm{of}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{which}\:\mathrm{satisfy}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/21
Determine all triplets (a;b;c) of positive integers which satisfy: (1/a) + (1/b) + (1/c) = (1/2) •C1:a=b=c a=b=c=k(say) (1/k)+(1/k)+(1/k)=(1/2) (3/k)=(1/2)⇒k=6 a=b=c=6 •C2:b=c b=c=k(say) (1/a)+(1/k)+(1/k)=(1/2) ((k+2a)/(ak))=(1/2) 2k+4a=ak 2(k+2a)=ak 2∣a ∨ 2∣k ∨ (2∣a ∧ 2∣k) ★ 2∣a ∨ 2∣k ∨ (2∣a ∧ 2∣k): Let a=2m 2(k+2a)=ak⇒2(k+2(2m))=(2m)k 2(k+4m)=2mk k+4m=mk m=(k/(k−4)) possible values of k for which m∈Z^+ k=5,6,8⇒m=5,3,2 b=c=k=5⇒a=2m=10 b=c=k=6⇒a=2m=6 b=c=k=8⇒a=2m=4 o ★ 2∣a ∨ 2∣k ∨ (2∣a ∧ 2∣k): k=2m 2(k+2a)=ak⇒2(2m+2a)=2ma 2m+2a=ma_((m or a ∈E^+ )) a=((2m)/(m−2)) possible positive integral values of m for which a is also positive integer. m=3,4,6⇒a=6,4,3 b=c=k=2m=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} ★ 2∣a ∨ 2∣k ∨ (2∣a ∧ 2∣k): a=2m,k=2n 2(k+2a)=ak⇒2(2n+2.2m)=2m.2n 4(n+2m)=4mn n+2m=mn mn−2m=n m=(n/(n−2)) possible values of n n=3,4⇒m=3,2 a=2m=6,4; b=c=k=2n=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} •C3: a≠b≠c≠a a=k,b=pk,c=qk ;p,q∈Z^+ (1/k)+(1/(pk))+(1/(qk))=(1/2) ((p+q+pq)/(pqk))=(1/2) 2(p+q+pq)=pqk 2∣p: p=2m 2(2m+q+pq)=(2m)qk 2m+q+2mq=qmk 2m(1+q)=qmk−q 2m(1+q)=q(mk−1) (q/(q+1))=((2m)/(mk−1)) ..... ... (6,12,4)... Continue
$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{triplets}\:\left(\mathrm{a};\mathrm{b};\mathrm{c}\right)\:\mathrm{of}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{which}\:\mathrm{satisfy}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\bullet\mathrm{C1}:\mathrm{a}=\mathrm{b}=\mathrm{c} \\ $$$$\:\:\:\:\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{k}\left(\mathrm{say}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{k}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\mathrm{k}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{k}=\mathrm{6} \\ $$$$\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{6} \\ $$$$\bullet\mathrm{C2}:\mathrm{b}=\mathrm{c} \\ $$$$\mathrm{b}=\mathrm{c}=\mathrm{k}\left(\mathrm{say}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{k}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{k}+\mathrm{2a}}{\mathrm{ak}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2k}+\mathrm{4a}=\mathrm{ak} \\ $$$$\mathrm{2}\left(\mathrm{k}+\mathrm{2a}\right)=\mathrm{ak} \\ $$$$\mathrm{2}\mid\mathrm{a}\:\vee\:\mathrm{2}\mid\mathrm{k}\:\vee\:\left(\mathrm{2}\mid\mathrm{a}\:\wedge\:\mathrm{2}\mid\mathrm{k}\right) \\ $$$$\bigstar\:\underline{\mathrm{2}\mid\mathrm{a}\:\vee\:\mathrm{2}\mid\mathrm{k}\:\vee\:\left(\mathrm{2}\mid\mathrm{a}\:\wedge\:\mathrm{2}\mid\mathrm{k}\right):} \\ $$$${Let}\:\mathrm{a}=\mathrm{2m} \\ $$$$\mathrm{2}\left(\mathrm{k}+\mathrm{2a}\right)=\mathrm{ak}\Rightarrow\mathrm{2}\left(\mathrm{k}+\mathrm{2}\left(\mathrm{2m}\right)\right)=\left(\mathrm{2m}\right)\mathrm{k} \\ $$$$\mathrm{2}\left(\mathrm{k}+\mathrm{4m}\right)=\mathrm{2mk} \\ $$$$\mathrm{k}+\mathrm{4m}=\mathrm{mk} \\ $$$$\mathrm{m}=\frac{\mathrm{k}}{\mathrm{k}−\mathrm{4}} \\ $$$$\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{k}\:\mathrm{for}\:\mathrm{which}\:\mathrm{m}\in\mathbb{Z}^{+} \\ $$$$\mathrm{k}=\mathrm{5},\mathrm{6},\mathrm{8}\Rightarrow\mathrm{m}=\mathrm{5},\mathrm{3},\mathrm{2} \\ $$$$\mathrm{b}=\mathrm{c}=\mathrm{k}=\mathrm{5}\Rightarrow\mathrm{a}=\mathrm{2m}=\mathrm{10} \\ $$$$\mathrm{b}=\mathrm{c}=\mathrm{k}=\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{2m}=\mathrm{6} \\ $$$$\mathrm{b}=\mathrm{c}=\mathrm{k}=\mathrm{8}\Rightarrow\mathrm{a}=\mathrm{2m}=\mathrm{4} \\ $$$${o}\:\bigstar\underline{\:\mathrm{2}\mid\mathrm{a}\:\vee\:\mathrm{2}\mid\mathrm{k}\:\vee\:\left(\mathrm{2}\mid\mathrm{a}\:\wedge\:\mathrm{2}\mid\mathrm{k}\right):} \\ $$$$\mathrm{k}=\mathrm{2m} \\ $$$$\mathrm{2}\left(\mathrm{k}+\mathrm{2a}\right)=\mathrm{ak}\Rightarrow\mathrm{2}\left(\mathrm{2m}+\mathrm{2a}\right)=\mathrm{2ma} \\ $$$$\:\:\:\:\:\underset{\left(\mathrm{m}\:\mathrm{or}\:\mathrm{a}\:\in\mathbb{E}^{+} \right)} {\mathrm{2m}+\mathrm{2a}=\mathrm{ma}} \\ $$$$\:\:\:\:\:\mathrm{a}=\frac{\mathrm{2m}}{\mathrm{m}−\mathrm{2}} \\ $$$$\mathrm{possible}\:\mathrm{positive}\:\mathrm{integral}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m} \\ $$$$\mathrm{for}\:\mathrm{which}\:\mathrm{a}\:\mathrm{is}\:\mathrm{also}\:\mathrm{positive}\:\mathrm{integer}. \\ $$$$\mathrm{m}=\mathrm{3},\mathrm{4},\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{6},\mathrm{4},\mathrm{3} \\ $$$$\mathrm{b}=\mathrm{c}=\mathrm{k}=\mathrm{2m}=\mathrm{6},\mathrm{8},\mathrm{12} \\ $$$$\left(\mathrm{6},\mathrm{6},\mathrm{6}\right),\left(\mathrm{4},\mathrm{8},\mathrm{8}\right),\left(\mathrm{3},\mathrm{12},\mathrm{12}\right) \\ $$$$\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \\ $$$$=\left\{\mathrm{3},\mathrm{12},\mathrm{12}\right\},\left\{\mathrm{4},\mathrm{8},\mathrm{8}\right\},\left\{\mathrm{10},\mathrm{5},\mathrm{5}\right\},\left\{\mathrm{6},\mathrm{6},\mathrm{6}\right\} \\ $$$$\:\bigstar\underline{\:\mathrm{2}\mid\mathrm{a}\:\vee\:\mathrm{2}\mid\mathrm{k}\:\vee\:\left(\mathrm{2}\mid\mathrm{a}\:\wedge\:\mathrm{2}\mid\mathrm{k}\right):} \\ $$$$\mathrm{a}=\mathrm{2m},\mathrm{k}=\mathrm{2n} \\ $$$$\mathrm{2}\left(\mathrm{k}+\mathrm{2a}\right)=\mathrm{ak}\Rightarrow\mathrm{2}\left(\mathrm{2n}+\mathrm{2}.\mathrm{2m}\right)=\mathrm{2m}.\mathrm{2n} \\ $$$$\mathrm{4}\left(\mathrm{n}+\mathrm{2m}\right)=\mathrm{4mn} \\ $$$$\mathrm{n}+\mathrm{2m}=\mathrm{mn} \\ $$$$\mathrm{mn}−\mathrm{2m}=\mathrm{n} \\ $$$$\mathrm{m}=\frac{\mathrm{n}}{\mathrm{n}−\mathrm{2}} \\ $$$$\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{n}=\mathrm{3},\mathrm{4}\Rightarrow\mathrm{m}=\mathrm{3},\mathrm{2} \\ $$$$\mathrm{a}=\mathrm{2m}=\mathrm{6},\mathrm{4};\:\mathrm{b}=\mathrm{c}=\mathrm{k}=\mathrm{2n}=\mathrm{6},\mathrm{8} \\ $$$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\left(\mathrm{6},\mathrm{6},\mathrm{6}\right),\left(\mathrm{4},\mathrm{8},\mathrm{8}\right) \\ $$$$\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \\ $$$$=\left\{\mathrm{3},\mathrm{12},\mathrm{12}\right\},\left\{\mathrm{4},\mathrm{8},\mathrm{8}\right\},\left\{\mathrm{10},\mathrm{5},\mathrm{5}\right\},\left\{\mathrm{6},\mathrm{6},\mathrm{6}\right\} \\ $$$$\bullet\mathrm{C3}:\:\mathrm{a}\neq\mathrm{b}\neq\mathrm{c}\neq\mathrm{a} \\ $$$$\mathrm{a}=\mathrm{k},\mathrm{b}=\mathrm{pk},\mathrm{c}=\mathrm{qk}\:;\mathrm{p},\mathrm{q}\in\mathbb{Z}^{+} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{pk}}+\frac{\mathrm{1}}{\mathrm{qk}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{p}+\mathrm{q}+\mathrm{pq}}{\mathrm{pqk}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{p}+\mathrm{q}+\mathrm{pq}\right)=\mathrm{pqk} \\ $$$$\mathrm{2}\mid\mathrm{p}:\:\mathrm{p}=\mathrm{2m} \\ $$$$\mathrm{2}\left(\mathrm{2m}+\mathrm{q}+\mathrm{pq}\right)=\left(\mathrm{2m}\right)\mathrm{qk} \\ $$$$\mathrm{2m}+\mathrm{q}+\mathrm{2mq}=\mathrm{qmk} \\ $$$$\mathrm{2m}\left(\mathrm{1}+\mathrm{q}\right)=\mathrm{qmk}−\mathrm{q} \\ $$$$\mathrm{2m}\left(\mathrm{1}+\mathrm{q}\right)=\mathrm{q}\left(\mathrm{mk}−\mathrm{1}\right) \\ $$$$\frac{\mathrm{q}}{\mathrm{q}+\mathrm{1}}=\frac{\mathrm{2m}}{\mathrm{mk}−\mathrm{1}} \\ $$$$….. \\ $$$$… \\ $$$$\left(\mathrm{6},\mathrm{12},\mathrm{4}\right)… \\ $$$$\mathrm{Continue} \\ $$
Commented by mathdanisur last updated on 02/Sep/21
Very nice Ser thaankyou
$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thaankyou} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Sep/21
Pl write all answers, if you have.
$${Pl}\:{write}\:{all}\:{answers},\:{if}\:{you}\:{have}. \\ $$
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
$$\left(\mathrm{3},\mathrm{7},\mathrm{42}\right);\left(\mathrm{3},\mathrm{8},\mathrm{24}\right);\left(\mathrm{3},\mathrm{9},\mathrm{18}\right);\left(\mathrm{3},\mathrm{10},\mathrm{15}\right) \\ $$$$\left(\mathrm{3},\mathrm{42},\mathrm{12}\right);\left(\mathrm{4},\mathrm{5},\mathrm{20}\right);\left(\mathrm{4},\mathrm{6},\mathrm{12}\right);\left(\mathrm{4},\mathrm{8},\mathrm{8}\right) \\ $$$$\left(\mathrm{5},\mathrm{5},\mathrm{10}\right);\left(\mathrm{6},\mathrm{6},\mathrm{6}\right)\:\mathrm{and}\:\mathrm{permutations} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Sep/21
Thank Yo⋓ S r!
$$\mathbb{T}\mathrm{han}\Bbbk\:\mathbb{Y}\mathrm{o}\Cup\:\mathbb{S} \mathrm{r}! \\ $$
Commented by Rasheed.Sindhi last updated on 02/Sep/21
mathdanisur ser, are you studying?
$${mathdanisur}\:{ser},\:{are}\:{you}\:{studying}? \\ $$
Commented by mathdanisur last updated on 02/Sep/21
Yes Ser, I will finish the highest level
$$\mathrm{Yes}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{I}\:\mathrm{will}\:\mathrm{finish}\:\mathrm{the}\:\mathrm{highest}\:\mathrm{level} \\ $$
Commented by Rasheed.Sindhi last updated on 03/Sep/21
Ser,Why don′t you try to solve any question in this forum?(Although you have given answers to some questions but not complete solutions?)
$${Ser},{Why}\:\:{don}'{t}\:{you}\:{try}\:{to}\:{solve} \\ $$$${any}\:{question}\:{in}\:{this}\:{forum}?\left({Although}\right. \\ $$$${you}\:{have}\:{given}\:{answers}\:{to}\:{some} \\ $$$$\left.{questions}\:{but}\:{not}\:{complete}\:{solutions}?\right) \\ $$
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...
$$\mathrm{I}'\mathrm{ve}\:\mathrm{solved},\:\mathrm{well},\:\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{to}\:\mathrm{most}\:\mathrm{of}\:\mathrm{the}\:\mathrm{questions}\:\mathrm{I}\:\mathrm{sak},\:\mathrm{I}\:\mathrm{just} \\ $$$$\mathrm{want}\:\mathrm{different}\:\mathrm{solutions},\:\mathrm{I}'\mathrm{m}\:\mathrm{learning}, \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{beautiful}\:\mathrm{place},\:\mathrm{there}\:\mathrm{are}\:\mathrm{educated} \\ $$$$\mathrm{people}\:\left(\mathrm{like}\:\mathrm{you}\right)\:\mathrm{and}\:\mathrm{I}\:\mathrm{like}\:\mathrm{your}\:\mathrm{solutions}, \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{for}\:\mathrm{your}\:\mathrm{solutions}… \\ $$
Commented by Rasheed.Sindhi last updated on 03/Sep/21
^• Th∝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.)
$$\:^{\bullet} \mathrm{Th}\propto\mathrm{nk}\:\mathrm{You}\:\mathrm{very}\:\mathrm{much}! \\ $$$$\:^{\bullet} \mathrm{Your}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{seems}\:\mathrm{incorrect}: \\ $$$$\:\:\:\left(\mathrm{3},\mathrm{42},\mathrm{12}\right).\mathrm{Pl}\:\mathrm{correct}\:\mathrm{it}. \\ $$$$\:^{\bullet} \mathrm{I}\:\mathrm{wish}\:\mathrm{you}\:\mathrm{share}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{for} \\ $$$$\mathrm{this}\:\mathrm{question}\:\mathrm{because}\:\mathrm{my}\:\mathrm{solutions} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{fine}\:\mathrm{enough}. \\ $$$$\:\:^{\bullet} \mathrm{Others}\:\mathrm{also}\:\mathrm{want}\:\mathrm{variety}\:\mathrm{in}\:\mathrm{solutions} \\ $$$$\:\mathrm{so}\:\mathrm{if}\:\mathrm{you}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{some} \\ $$$$\:\mathrm{questions}\:\mathrm{pl}\:\mathrm{share}\:\mathrm{their} \\ $$$$\mathrm{solutions}.\left(\mathrm{Particularly}\:\mathrm{solution}\:\mathrm{for}\right. \\ $$$$\left.\mathrm{this}\:\mathrm{question}.\right) \\ $$
Commented by mathdanisur last updated on 03/Sep/21
Good Ser, thank you, sorry (3,12,12)
$$\mathrm{Good}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you},\:\mathrm{sorry}\:\left(\mathrm{3},\mathrm{12},\mathrm{12}\right) \\ $$
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
$$\mathrm{your}\:\mathrm{selfish}\:\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{soln}\:\mathrm{then} \\ $$$$\mathrm{you}\:\mathrm{dont}\:\mathrm{what}\:\mathrm{to}\:\mathrm{share}\:\mathrm{it}\:\mathrm{this} \\ $$$$\mathrm{not}\:\mathrm{good}\: \\ $$
Answered by Rasheed.Sindhi last updated on 01/Sep/21
(1/a) + (1/b) + (1/c) = (1/2) _((i)) ; a,b,c∈Z^+ ;(a,b,c)=? (1/a) + (1/b)=(1/2)−(1/c) (1/a) + (1/b)>0⇒(1/2)−(1/c)>0⇒c≠1,2 ∵ (i) is symmetric in a,b,c ∴ a,b,c≠1,2 i-e a,b,c>2 Now, (i)⇒2(ab+bc+ca)=abc ⇒At least one of a,b,c is even Let a=2k 2(2bk+bc+2ck)=2kbc 2bk+bc+2ck=kbc 2k(b+c)=bc(k−1) ⇒At least one of b,c,(k−1) is even. If(1) (k−1) ∈ E^+ ⇒k∈O^+ ⇒ a is a singly even number in 2(ab+bc+ca)=abc ....... If(2) b is even. Say b=2m 2k(b+c)=bc(k−1) 2k(2m+c)=2mc(k−1) k(2m+c)=mc(k−1) k∣k−1⇒k=2⇒a=2k=4 ⇒ ........ ........
$$\underset{\left({i}\right)} {\underbrace{\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\:};\:{a},{b},{c}\in\mathbb{Z}^{+} ;\left({a},{b},{c}\right)=? \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{c}} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}>\mathrm{0}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{c}}>\mathrm{0}\Rightarrow{c}\neq\mathrm{1},\mathrm{2} \\ $$$$\because\:\left({i}\right)\:{is}\:{symmetric}\:{in}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}} \\ $$$$\therefore\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\neq\mathrm{1},\mathrm{2}\:\mathrm{i}-{e}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}>\mathrm{2} \\ $$$${Now}, \\ $$$$\:\left({i}\right)\Rightarrow\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)=\mathrm{abc} \\ $$$$\Rightarrow{At}\:{least}\:{one}\:{of}\:\mathrm{a},\mathrm{b},\mathrm{c}\:{is}\:\boldsymbol{\mathrm{even}} \\ $$$${Let}\:\mathrm{a}=\mathrm{2k} \\ $$$$\:\:\:\mathrm{2}\left(\mathrm{2bk}+\mathrm{bc}+\mathrm{2ck}\right)=\mathrm{2kbc} \\ $$$$\:\:\:\mathrm{2bk}+\mathrm{bc}+\mathrm{2ck}=\mathrm{kbc} \\ $$$$\:\:\:\:\mathrm{2k}\left(\mathrm{b}+\mathrm{c}\right)=\mathrm{bc}\left(\mathrm{k}−\mathrm{1}\right) \\ $$$$\Rightarrow{At}\:{least}\:{one}\:{of}\:\mathrm{b},\mathrm{c},\left(\mathrm{k}−\mathrm{1}\right)\:{is}\:\boldsymbol{\mathrm{even}}. \\ $$$$\boldsymbol{\mathrm{If}}\left(\mathrm{1}\right)\:\left(\mathrm{k}−\mathrm{1}\right)\:\in\:\mathbb{E}^{+} \Rightarrow\mathrm{k}\in\mathbb{O}^{+} \Rightarrow\:\mathrm{a}\:{is}\:{a} \\ $$$$\boldsymbol{\mathrm{singly}}\:\boldsymbol{\mathrm{even}}\:{number}\:{in} \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)=\mathrm{abc} \\ $$$$\: \\ $$$$……. \\ $$$$\boldsymbol{\mathrm{If}}\left(\mathrm{2}\right)\:\mathrm{b}\:{is}\:\mathrm{even}.\:\mathrm{S}{ay}\:\mathrm{b}=\mathrm{2m} \\ $$$$\mathrm{2k}\left(\mathrm{b}+\mathrm{c}\right)=\mathrm{bc}\left(\mathrm{k}−\mathrm{1}\right) \\ $$$$\mathrm{2k}\left(\mathrm{2m}+\mathrm{c}\right)=\mathrm{2mc}\left(\mathrm{k}−\mathrm{1}\right) \\ $$$$\mathrm{k}\left(\mathrm{2m}+\mathrm{c}\right)=\mathrm{mc}\left(\mathrm{k}−\mathrm{1}\right) \\ $$$$\mathrm{k}\mid\mathrm{k}−\mathrm{1}\Rightarrow\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{2k}=\mathrm{4} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\:\:\:\:\:…….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…….. \\ $$
Commented by mathdanisur last updated on 02/Sep/21
Nice Ser thank you
$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Rasheed.Sindhi last updated on 03/Sep/21
(1/a) + (1/b) + (1/c) = (1/2).............(★) (1/a) + (1/b)=(1/2)−(1/c) ∵(1/a) + (1/b)>0⇒(1/2)−(1/c)>0⇒c>2 ∵ The equation is symmetric in a,b,c⇒a,b,c>2 (★)⇒2(ab+bc+ca)=abc ⇒At least one of a,b,c is even. C1:only one of a,b,c is even: (1/(2l)) + (1/(2m+1)) + (1/(2n+1)) = (1/2) ▶2^× {(2l)(2m+1)+(2m+1)(2n+1)+(2n+1)(2l)} =2^(×) l(2m+1)(2n+1) ▶(2l)(2m+1)+(2m+1)(2n+1)+(2n+1)(2l) =l(2m+1)(2n+1) ▶4lm+2l+4mn+2m+2n+1+4nl+2l =4lmn+2lm+2ln+l ▶4lm+4mn+4ln+4l+2m+2n+1 −4lmn−2lm−2ln−l=0 ▶2lm+4mn+2ln+2m+2n+3l+1=4lmn ▶2(lm+2mn+ln+m+n)+(3l+1)=4lmn ⇒3l+1 is even⇒l is odd Let l=2k+1 ▶2{(2k+1)m+2mn+(2k+1)n+m+n}+(3(2k+1)+1)=4(2k+1)mn ▶2{2km+2mn+2kn+2n+2m}+6k+4=4(2k+1)mn ▶4^(×) ({2km+2mn+2kn+2n+2m}+3k+2)=4^(×) (2k+1)mn ▶2km+mn+2kn+2n+2m+2+3k=2kmn ▶2(km+kn+n+m+1)+3k+mn=2kmn ⇒3k+mn is even. ⇒ { ((k,m & n are odd)),((k is even & (m or n is even))) :} Continue C1:only two of a,b,c is even: (1/(2l))+(1/(2m))+(1/(2n+1))=(1/2) ((m(2n+1)+l(2n+1)+lm)/(2lm(2n+1)))=(1/2) ((m(2n+1)+l(2n+1)+lm)/(lm(2n+1)))=1 (1/l)+(1/m)+(1/(2n+1))=1 C1:All of a,b,c are even: (1/(2l))+(1/(2m))+(1/(2n))=(1/2) ((lm+mn+nl)/(2lmn))=(1/2) lm+mn+nl=lmn l∣mn ∧ m∣nl ∧ n∣lm (1/l)+(1/m)+(1/n)=1
$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}………….\left(\bigstar\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{c}} \\ $$$$\because\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}>\mathrm{0}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{c}}>\mathrm{0}\Rightarrow\mathrm{c}>\mathrm{2} \\ $$$$\because\:{The}\:{equation}\:{is}\:{symmetric}\:{in} \\ $$$${a},{b},{c}\Rightarrow{a},{b},{c}>\mathrm{2} \\ $$$$\left(\bigstar\right)\Rightarrow\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)=\mathrm{abc} \\ $$$$\Rightarrow\boldsymbol{{At}}\:\boldsymbol{{least}}\:\boldsymbol{{one}}\:\boldsymbol{{of}}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\boldsymbol{{is}}\:\boldsymbol{\mathrm{even}}. \\ $$$$\boldsymbol{\mathrm{C}}\mathrm{1}:\boldsymbol{{only}}\:\boldsymbol{{one}}\:{of}\:\mathrm{a},\mathrm{b},\mathrm{c}\:{is}\:\boldsymbol{\mathrm{even}}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{l}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{m}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\blacktriangleright\overset{×} {\mathrm{2}}\left\{\left(\mathrm{2}{l}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)+\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{l}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\overset{×} {\mathrm{2}}{l}\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\blacktriangleright\left(\mathrm{2}{l}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)+\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{l}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={l}\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\blacktriangleright\mathrm{4}{lm}+\mathrm{2}{l}+\mathrm{4}{mn}+\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{1}+\mathrm{4}{nl}+\mathrm{2}{l} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{lmn}+\mathrm{2}{lm}+\mathrm{2}{ln}+{l} \\ $$$$\blacktriangleright\mathrm{4}{lm}+\mathrm{4}{mn}+\mathrm{4}{ln}+\mathrm{4}{l}+\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{4}{lmn}−\mathrm{2}{lm}−\mathrm{2}{ln}−{l}=\mathrm{0} \\ $$$$\blacktriangleright\mathrm{2}{lm}+\mathrm{4}{mn}+\mathrm{2}{ln}+\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{3}{l}+\mathrm{1}=\mathrm{4}{lmn} \\ $$$$\blacktriangleright\mathrm{2}\left({lm}+\mathrm{2}{mn}+{ln}+{m}+{n}\right)+\left(\mathrm{3}{l}+\mathrm{1}\right)=\mathrm{4}{lmn} \\ $$$$\Rightarrow\mathrm{3}{l}+\mathrm{1}\:{is}\:\boldsymbol{\mathrm{even}}\Rightarrow{l}\:{is}\:\boldsymbol{\mathrm{odd}} \\ $$$${Let}\:{l}=\mathrm{2}{k}+\mathrm{1} \\ $$$$\blacktriangleright\mathrm{2}\left\{\left(\mathrm{2}{k}+\mathrm{1}\right){m}+\mathrm{2}{mn}+\left(\mathrm{2}{k}+\mathrm{1}\right){n}+{m}+{n}\right\}+\left(\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{1}\right)=\mathrm{4}\left(\mathrm{2}{k}+\mathrm{1}\right){mn} \\ $$$$\blacktriangleright\mathrm{2}\left\{\mathrm{2}{km}+\mathrm{2}{mn}+\mathrm{2}{kn}+\mathrm{2}{n}+\mathrm{2}{m}\right\}+\mathrm{6}{k}+\mathrm{4}=\mathrm{4}\left(\mathrm{2}{k}+\mathrm{1}\right){mn} \\ $$$$\blacktriangleright\overset{×} {\mathrm{4}}\left(\left\{\mathrm{2}{km}+\mathrm{2}{mn}+\mathrm{2}{kn}+\mathrm{2}{n}+\mathrm{2}{m}\right\}+\mathrm{3}{k}+\mathrm{2}\right)=\overset{×} {\mathrm{4}}\left(\mathrm{2}{k}+\mathrm{1}\right){mn} \\ $$$$\blacktriangleright\mathrm{2}{km}+{mn}+\mathrm{2}{kn}+\mathrm{2}{n}+\mathrm{2}{m}+\mathrm{2}+\mathrm{3}{k}=\mathrm{2}{kmn} \\ $$$$\blacktriangleright\mathrm{2}\left({km}+{kn}+{n}+{m}+\mathrm{1}\right)+\mathrm{3}{k}+{mn}=\mathrm{2}{kmn} \\ $$$$\Rightarrow\mathrm{3}{k}+{mn}\:{is}\:\boldsymbol{\mathrm{even}}. \\ $$$$\Rightarrow\begin{cases}{{k},{m}\:\&\:{n}\:{are}\:\boldsymbol{\mathrm{odd}}}\\{{k}\:{is}\:\boldsymbol{\mathrm{even}}\:\&\:\left({m}\:{or}\:{n}\:{is}\:\boldsymbol{\mathrm{even}}\right)}\end{cases} \\ $$$${Continue} \\ $$$$\boldsymbol{\mathrm{C}}\mathrm{1}:\boldsymbol{{only}}\:\boldsymbol{{two}}\:{of}\:\mathrm{a},\mathrm{b},\mathrm{c}\:{is}\:\boldsymbol{\mathrm{even}}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{l}}+\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{m}\left(\mathrm{2}{n}+\mathrm{1}\right)+{l}\left(\mathrm{2}{n}+\mathrm{1}\right)+{lm}}{\mathrm{2}{lm}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{m}\left(\mathrm{2}{n}+\mathrm{1}\right)+{l}\left(\mathrm{2}{n}+\mathrm{1}\right)+{lm}}{{lm}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{l}}+\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{C}}\mathrm{1}:\boldsymbol{{All}}\:{of}\:\mathrm{a},\mathrm{b},\mathrm{c}\:{are}\:\boldsymbol{\mathrm{even}}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{l}}+\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}{n}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{lm}+{mn}+{nl}}{\mathrm{2}{lmn}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lm}+{mn}+{nl}={lmn} \\ $$$${l}\mid{mn}\:\wedge\:{m}\mid{nl}\:\wedge\:{n}\mid{lm} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{l}}+\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{{n}}=\mathrm{1} \\ $$

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