Vom demonstra prin inductie.
Vom nota orice multiplu de 17 cu M17. Astfel:
M17 + M17 = M17
M17 * M17 = M17
a * M17 = M17, pentru orice a ∈ Z
Fie:
[tex]P(n):3\cdot5^{2n+1}+2^{3n+1}=\mathcal{M}_{17}\ \ , \ n\in N[/tex]
[tex]P(0)=3\cdot5^1+2^1=17=\mathcal{M}_{17}\ ,\ \text{ceea ce e adevarat}[/tex]
[tex]\text{Presupunem ca } P(k) \text{ este o propozitie adevarata:}\\
P(k):3\cdot5^{2k+1}+2^{3k+1}=\mathcal{M}_{17}\ \ \text{(adevarat)}[/tex]
Acum trebuie sa demonstram ca P(k + 1) este adevarata, tinand cont ca P(k) este adevarata:
[tex]P(k+1):3\cdot5^{2k+3}+2^{3k+4}=\mathcal{M}_{17}\ \ \text{(de demonstrat)}\\\\
3\cdot5^{2k+1}+2^{3k+1}=\mathcal{M}_{17}\rightarrow2^{3k+1}=\mathcal{M}_{17}-3\cdot5^{2k+1}\\\\
P(k+1):3\cdot5^2\cdot5^{2k+1}+2^3\cdot2^{3k+1}\\
P(k+1):75\cdot5^{2k+1}+8(\mathcal{M}_{17}-3\cdot5^{2k+1})\\
P(k+1):5^{2k+1}\cdot75-5^{2k+1}\cdot24+8\mathcal{M}_{17}\\
P(k+1):5^{2k+1}(75-24)+\mathcal{M}_{17}\\
P(k+1):2^{2k+1}\cdot51+\mathcal{M}_{17}=17(2^{2k+1}\cdot3)+\mathcal{M}_{17}=\mathcal{M}_{17}\ \ \text{q.e.d.}[/tex]
Asta inseamna ca P(n) este adevarata pentru orice n natural ≥ 0
