Aratati ca, pentru orice numar natural n:
3×5^(2n+1)+2^(3n+1) se divide cu 17
Rezolvare:
I.Verificare:
p(0):17/ 3×5^(2×0+1)+2^(3×0+1) (A)

ll.Demonstratie:
presupunem P(n)(A) si dem ca P(n+1)(A)
P(n): 17/3×5^(2n+1)+2^(3n+1)
–––––––––/––––––––-
P(n+1):17/3×5^(2n+3)+2^(3n+4)
–––––––––//––––––––-

17/3×5^(2n+3)+2^(3n+4)<=>
<=> 3×5^(2n+1)x5^2+2^(3n+1)x2^3<=>
<=> 3×5^(2n+1)x25+2^(3n+1)x8<=>
<=> 3×5^(2n+1)x(24+1)+2^(3n+1)x(7+1)<=>
<=>24x3x5^(2n+1)+7×2^(3n+1)+2^(3n+1)<=>
{3×5^(2n+1)+2^(3n+1)}+{(24x3x5^(2n+1)+7×2^(3n+1)}

3×5^(2n+1)+2^(3n+1)– partea aceasta se divide cu 17 ( ipoteza); 17k
(24x3x5^(2n+1)+7×2^(3n+1)– partea aceasta trebuie sa o demonstrez prin alta inductie ca este divizibila cu 17?