Matematické Fórum

stuart clark · 19. 10. 2011 21:37

Determine all triplets $(a, b, c)$ in $a + b + c + (ab + bc + ca) = abc +1$ .

Where $a, b, c$ are positive integer

vanok · 21. 10. 2011 00:08 — Editoval vanok (23. 10. 2011 13:33)

Hi ↑ stuart clark:,

For a, b, c positive integer $a + b + c + (ab + bc + ca) = abc +1$ $(E_1)$ is equivalent in
$(1 + \frac1a)(1 + \frac1b)(1 + \frac1c)= \frac2{abc} +2$

Without loss of generality let $a\le b\le c$ then if we have $a\ge5$
$(1 + \frac1a)(1 + \frac1b)(1 + \frac1c)- \frac2{abc} \le (1 + \frac1a)(1 + \frac1b)(1 + \frac1c) \le (\frac65)^3=\frac{216}{125}< 2$

So it remains to study that 4 cases for $a\in \{1; 2; 3; 4\}$

For $a=1$ $(E_1)$ becomes
$1 + b + c + b + bc + c = bc +1$ or
$b + c = 0$
which has no solution in positive integers.

For $a=2$ $(E_1)$ becomes
$2 + b + c + 2b + bc + 2c = 2bc +1$ or
$3b + 3c = bc-1$ or
$\frac3b +\frac3c + \frac1{ab}=1$
Without loss of generality let $a=2; b\le c$ gives for $b\ge 7$
$\frac3b +\frac3c + \frac1{ab} < 1$
So it remains to study that 5 cases for $a=2$ and $b\in \{ 2; 3; 4; 5; 6\}$
....
This leads us easily towards 2 solutions (2; 4; 13)and (2; 5; 8}

For a=3 (E_1) becomes
$3 + b + c + 3b + bc + 3c = 3bc +1$ or
$2 + 4b + 4c = 2bc$ or
$1 + 2b + 2c = bc$ or still
$\frac1{bc} + \frac2b + \frac2c =1$
Without loss of generality let $a=3; b\le c$ gives for $b\ge5$
$\frac3b +\frac3c + \frac1{ab} < 1$
So it remains to study that 2 cases for $a=3$ and $b\in \{ 3; 4\}$
......
This leads us easily towards one solution (3; 3; 7)

Finally for a=4 (E_1) becomes
$4 + b + c + 4b + bc + 4c= 4bc +1$ or
$3 + 5b + 5c =3bc$ or
$\frac3{bc} + \frac5b + \frac5c =3$
Without loss of generality let $a=4; b\le c$ gives for $b\ge4$
$\frac3{bc} + \frac5b + \frac5c <3$
....
This shows that we have no solutions in that case

Conlusion: the set of solutions is
$S=$
$\{ (2; 4; 13); (2; 13; 4); (4; 2; 13); (4, 13; 2); (13; 2; 4);(13; 4; 2); (2; 5; 8) \}$ $U$ $\{ (2; 5; 8); (5; 2; 8); (5; 8; 2); (8;2;5); (8; 5; 2); (3; 3; 7); (3; 7; 3); 7; 3; 3)\}$

Sincerely Vanok

Pavel · 21. 10. 2011 07:21

↑ vanok:

It is a realy nice solution with one exception - $(2,3,7)$ does not solve the problem.

vanok · 21. 10. 2011 07:38 — Editoval vanok (23. 10. 2011 13:33)

↑ Pavel:
Thank you Pavel, me corrected my night-works
Sincerely Vanok

stuart clark · 21. 10. 2011 09:59

Thanks for very nice solution vanok.

check_drummer · 21. 10. 2011 17:03

Autor skryl část textu. Zobrazit/skrýt!

vanok · 23. 10. 2011 11:37 — Editoval vanok (23. 10. 2011 13:33)

Some remarks:
He can be interesting to study this equation in $Z_n$
For example:
It is clear that modulo 2 we have 4 solution triplets (1; 1; 1); (1; 0; 0); (0; 1; 0) et (0; 0; 1)

On the other hand, which are solutions "prohibited" by the statement

Sincerely Vanok

Matematické Fórum

#1 19. 10. 2011 21:37

all triplets

#2 21. 10. 2011 00:08 — Editoval vanok (23. 10. 2011 13:33)

Re: all triplets

#3 21. 10. 2011 07:21

Re: all triplets

#4 21. 10. 2011 07:38 — Editoval vanok (23. 10. 2011 13:33)

Re: all triplets

#5 21. 10. 2011 09:59

Re: all triplets

#6 21. 10. 2011 17:03

Re: all triplets

#7 23. 10. 2011 11:37 — Editoval vanok (23. 10. 2011 13:33)

Re: all triplets

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