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Q-167612-reposted-Determine-all-the-possible-triples-a-b-c-of-positive-integers-for-which-ab-c-bc-a-and-ca-b-are-powers-of-2-

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Question Number 167725 by Rasheed.Sindhi last updated on 23/Mar/22
Q#167612 reposted. Determine all the possible triples (a,b,c) of positive integers for which ab−c,bc−a and ca−b are powers of 2.
$${Q}#\mathrm{167612}\:{reposted}. \\ $$$$\mathcal{D}{etermine}\:{all}\:{the}\:{possible}\:{triples} \\ $$$$\left({a},{b},{c}\right)\:{of}\:{positive}\:{integers}\:{for}\:{which} \\ $$$${ab}−{c},{bc}−{a}\:{and}\:{ca}−{b}\:{are}\:{powers}\:{of} \\ $$$$\mathrm{2}. \\ $$
Commented by Tinku Tara last updated on 24/Mar/22
a=b=c=2
$${a}={b}={c}=\mathrm{2} \\ $$
Commented by Tinku Tara last updated on 24/Mar/22
if a^2 −1 is odd then 2^x +2^y a need not be odd. It could be 2(a^2 −1)k
$$\mathrm{if}\:{a}^{\mathrm{2}} −\mathrm{1}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{then}\:\mathrm{2}^{{x}} +\mathrm{2}^{{y}} {a}\:\mathrm{need}\:\mathrm{not}\:\mathrm{be} \\ $$$$\mathrm{odd}.\:\mathrm{It}\:\mathrm{could}\:\mathrm{be}\:\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right){k} \\ $$
Commented by Tinku Tara last updated on 24/Mar/22
subtract (1) from (2) ab−c=2^α ac−b=2^β a(b−c)−(b−c)=2^a (2^b −1) (b−c)(a−1)=2^a (2^b −1)
$$\mathrm{subtract}\:\left(\mathrm{1}\right)\:\mathrm{from}\:\left(\mathrm{2}\right) \\ $$$${ab}−{c}=\mathrm{2}^{\alpha} \\ $$$${ac}−{b}=\mathrm{2}^{\beta} \\ $$$${a}\left({b}−{c}\right)−\left({b}−{c}\right)=\mathrm{2}^{{a}} \left(\mathrm{2}^{{b}} −\mathrm{1}\right) \\ $$$$\left({b}−{c}\right)\left({a}−\mathrm{1}\right)=\mathrm{2}^{{a}} \left(\mathrm{2}^{{b}} −\mathrm{1}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 24/Mar/22
There′s no c on the right side. That means it has no effect on the result?
$$\mathcal{T}{here}'{s}\:{no}\:{c}\:{on}\:{the}\:{right}\:{side}. \\ $$$$\mathcal{T}{hat}\:{means}\:{it}\:{has}\:{no}\:{effect}\:{on}\:{the} \\ $$$${result}? \\ $$
Answered by Rasheed.Sindhi last updated on 24/Mar/22
A Try... (e represents even and o represents odd here) Case0:No one of a,b,c is odd. Let a,b,c are denoted as e_a ,e_b ,e_c respectively. ab−c=e_a e_b −e_c =2^l e−e=e e=e bc−a=e_b e_c −e_a =2^m e−e=e e=e ca−b=e_c e_a −e_b =2^n e−e=e e=e No contradiction! Case1:Only one of a,b,c is odd. Let a,b,c are denoted as o_a ,e_b ,e_c respectively. ab−c=o_a e_b −e_c =2^l bc−a=e_b e_c −o_a =2^m e−o=e o=e Contrdiction! ca−b=e_c o_a −e_b =2^n Case2:Only two of a,b,c are odd. Let a,b,c are denoted as o_a ,o_b ,e_c respectively. ab−c=o_a o_b −e_c =2^l o−e=e o=e Contrdiction! Case3:All of a,b,c are odd. Let a,b,c are denoted as o_a ,o_b ,o_c respectively. ab−c=o_a o_b −o_c =2^l o−o=e e=e bc−a=o_b o_c −o_a =2^m o−o=e e=e ca−b=o_c o_a −o_b =2^l o−o=e e=e No contradiction! ∴ Either all the three are odd or even. determinant (( determinant (((a,b,c ∈ E)))_ determinant (((or)))_( determinant (((a,b,c ∈_ ^ O)))) ))
$$\mathcal{A}\:\mathcal{T}{ry}… \\ $$$$\left({e}\:{represents}\:{even}\:{and}\:{o}\:{represents}\:{odd}\:{here}\right) \\ $$$${Case}\mathrm{0}:{No}\:{one}\:{of}\:{a},{b},{c}\:{is}\:{odd}. \\ $$$${Let}\:{a},{b},{c}\:{are}\:{denoted}\:{as}\:{e}_{{a}} ,{e}_{{b}} ,{e}_{{c}} \\ $$$${respectively}. \\ $$$${ab}−{c}={e}_{{a}} {e}_{{b}} −{e}_{{c}} =\mathrm{2}^{{l}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}−{e}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$${bc}−{a}={e}_{{b}} {e}_{{c}} −{e}_{{a}} =\mathrm{2}^{{m}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}−{e}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$${ca}−{b}={e}_{{c}} {e}_{{a}} −{e}_{{b}} =\mathrm{2}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}−{e}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$$\mathcal{N}{o}\:{contradiction}! \\ $$$${Case}\mathrm{1}:{Only}\:{one}\:{of}\:{a},{b},{c}\:{is}\:{odd}. \\ $$$${Let}\:{a},{b},{c}\:{are}\:{denoted}\:{as}\:{o}_{{a}} ,{e}_{{b}} ,{e}_{{c}} \\ $$$${respectively}. \\ $$$${ab}−{c}={o}_{{a}} {e}_{{b}} −{e}_{{c}} =\mathrm{2}^{{l}} \\ $$$${bc}−{a}={e}_{{b}} {e}_{{c}} −{o}_{{a}} =\mathrm{2}^{{m}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{e}−{o}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{o}={e} \\ $$$${Contrdiction}! \\ $$$${ca}−{b}={e}_{{c}} {o}_{{a}} −{e}_{{b}} =\mathrm{2}^{{n}} \\ $$$${Case}\mathrm{2}:{Only}\:{two}\:{of}\:{a},{b},{c}\:{are}\:{odd}. \\ $$$${Let}\:{a},{b},{c}\:{are}\:{denoted}\:{as}\:{o}_{{a}} ,{o}_{{b}} ,{e}_{{c}} \\ $$$${respectively}. \\ $$$${ab}−{c}={o}_{{a}} {o}_{{b}} −{e}_{{c}} =\mathrm{2}^{{l}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{o}−{e}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{o}={e} \\ $$$${Contrdiction}! \\ $$$${Case}\mathrm{3}:{All}\:{of}\:{a},{b},{c}\:{are}\:{odd}. \\ $$$${Let}\:{a},{b},{c}\:{are}\:{denoted}\:{as}\:{o}_{{a}} ,{o}_{{b}} ,{o}_{{c}} \\ $$$${respectively}. \\ $$$${ab}−{c}={o}_{{a}} {o}_{{b}} −{o}_{{c}} =\mathrm{2}^{{l}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{o}−{o}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$${bc}−{a}={o}_{{b}} {o}_{{c}} −{o}_{{a}} =\mathrm{2}^{{m}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{o}−{o}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$${ca}−{b}={o}_{{c}} {o}_{{a}} −{o}_{{b}} =\mathrm{2}^{{l}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{o}−{o}={e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}={e} \\ $$$$\mathcal{N}{o}\:{contradiction}! \\ $$$$ \\ $$$$\therefore\:\mathcal{E}{ither}\:{all}\:{the}\:{three}\:{are}\:{odd}\:{or} \\ $$$${even}. \\ $$$$\begin{array}{|c|}{\begin{array}{|c|}{{a},{b},{c}\:\in\:\mathbb{E}}\\\hline\end{array}_{\begin{array}{|c|}{\mathrm{or}}\\\hline\end{array}_{\:\:\begin{array}{|c|}{{a},{b},{c}\:\in_{} ^{} \:\mathbb{O}}\\\hline\end{array}} } }\\\hline\end{array} \\ $$

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