Z {\displaystyle \mathbb {Z} } - מספר שלם.
6 ∗ 5 2 n + 1 + 2 3 n + 2 34 = Z {\displaystyle {\frac {6*5^{2n+1}+2^{3n+2}}{34}}=\mathbb {Z} }
6 ∗ 5 2 n + 1 + 2 3 n + 2 34 = Z 6 ∗ 5 2 + 1 + 2 3 + 2 34 = 23 1 = z √ {\displaystyle {\begin{aligned}&{\frac {6*5^{2n+1}+2^{3n+2}}{34}}=\mathbb {Z} \\&{\frac {6*5^{2+1}}{+}}2^{3+2}{34}=23\\&1=\mathbb {z} \surd \\\end{aligned}}}
6 ∗ 5 2 k + 1 + 2 3 k + 2 34 = Z {\displaystyle {\frac {6*5^{2k+1}+2^{3k+2}}{34}}=\mathbb {Z} }
6 ∗ 5 2 k + 3 + 2 3 k + 5 34 = Z 6 ∗ 5 2 k + 1 ∗ 5 2 + 2 3 k + 2 ∗ 2 3 34 = Z 6 ∗ 5 2 k + 1 ∗ ( 17 + 8 ) + 2 3 k + 2 ∗ 8 34 = Z 6 ∗ 5 2 k + 1 ∗ 8 + 2 3 k + 2 ∗ 8 34 + 6 ∗ 5 2 k + 1 ∗ 17 34 = Z 8 ( 6 ∗ 5 2 k + 1 + 2 3 k + 2 ) 34 + 6 ∗ 5 2 k + 1 2 = Z 8 ∗ Z + 3 ∗ 5 2 k + 1 = Z Z + Z = Z {\displaystyle {\begin{aligned}&{\frac {6*5^{2k+3}+2^{3k+5}}{34}}=\mathbb {Z} \\&{\frac {6*5^{2k+1}*5^{2}+2^{3k+2*2^{3}}}{34}}=\mathbb {Z} \\&{\frac {6*5^{2k+1}*(17+8)+2^{3k+2}*8}{34}}=\mathbb {Z} \\&{\frac {6*5^{2k+1}*8+2^{3k+2}*8}{34}}+{\frac {6*5^{2k+1}*17}{34}}=\mathbb {Z} \\&{\frac {8(6*5^{2k+1}+2^{3k+2})}{34}}+{\frac {6*5^{2k+1}}{2}}=\mathbb {Z} \\&8*\mathbb {Z} +3*5^{2k+1}=\mathbb {Z} \\&\mathbb {Z} +\mathbb {Z} =\mathbb {Z} \\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
