[tex] 2^{a+b} [/tex] ≤ 40 => a+b <6
a+b=0 => [tex] 2^{b+c} =39[/tex], imposibil
a+b=1 => [tex] 2^{b+c} =38[/tex], imposibil
a+b=2 => [tex] 2^{b+c} =36[/tex], imposibil
a+b=3 => [tex] 2^{b+c} =32 => b+c=5[/tex]
a+b=4 => [tex] 2^{b+c} =24[/tex], imposibil
a+b=5 => [tex] 2^{b+c} =8 => b+c=3[/tex]
Deci avem 2 cazuri:
I. a+b=3 si b+c=5, cu solutiile (a,b,c)∈{(0,3,2);(1,2,3);(2,1,4);(3,0,5)}
II. a+b=5 si b+c=3, cu solutiile (a,b,c)∈{(2,3,0);(3,2,1);(4,1,2);(5,0,3)}
