Show-that-for-all-n-N-0-7-2n-1-1-is-an-integer-multiple-of-8-

Question Number 25173 by NECx last updated on 05/Dec/17

Show that for all nεN−{0} 7^(2n+1) +1 is an integer multiple of 8.

$${Show}\:{that}\:{for}\:{all}\:{n}\epsilon{N}−\left\{\mathrm{0}\right\}\: \\ $$$$\mathrm{7}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\:{is}\:{an}\:{integer}\:\:{multiple}\:{of} \\ $$$$\mathrm{8}. \\ $$

Answered by ajfour last updated on 05/Dec/17

7^(2n+1) +1=(8−1)^(2n+1) +1 =8^(2n+1) −^(2n+1) C_1 (8)^(2n) +...+^(2n+1) C_(2n) (8) =8m .

$$\mathrm{7}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}=\left(\mathrm{8}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1} \\ $$$$=\mathrm{8}^{\mathrm{2}{n}+\mathrm{1}} −^{\mathrm{2}{n}+\mathrm{1}} {C}_{\mathrm{1}} \left(\mathrm{8}\right)^{\mathrm{2}{n}} +…+^{\mathrm{2}{n}+\mathrm{1}} {C}_{\mathrm{2}{n}} \left(\mathrm{8}\right) \\ $$$$=\mathrm{8}{m}\:\:. \\ $$

Commented by NECx last updated on 06/Dec/17

$${please}\:{can}\:{it}\:{be}\:{proven}\:{by}\:{P}.{M}.{I} \\ $$$${i}\:{would}\:{like}\:{to}\:{see}\:{that}\:{used}. \\ $$$$ \\ $$$${Thanks}. \\ $$

Commented by jota+ last updated on 06/Dec/17

7^(2n+1) +1=8× 7^3 +1=344=8× 7^(2k+1) +1=8× hipotesis 7^2 (7^(2k+1) +1)=7^2 (8×) 7^(2(k+1)+1) +48+1=8× 7^(2(k+1)+1) +1=8×−48=8×

$$\mathrm{7}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}=\mathrm{8}× \\ $$$$\mathrm{7}^{\mathrm{3}} +\mathrm{1}=\mathrm{344}=\mathrm{8}× \\ $$$$\mathrm{7}^{\mathrm{2}{k}+\mathrm{1}} +\mathrm{1}=\mathrm{8}×\:\:\:\:\:{hipotesis} \\ $$$$\mathrm{7}^{\mathrm{2}} \left(\mathrm{7}^{\mathrm{2}{k}+\mathrm{1}} +\mathrm{1}\right)=\mathrm{7}^{\mathrm{2}} \left(\mathrm{8}×\right)\:\:\: \\ $$$$\mathrm{7}^{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} +\mathrm{48}+\mathrm{1}=\mathrm{8}× \\ $$$$\mathrm{7}^{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} +\mathrm{1}=\mathrm{8}×−\mathrm{48}=\mathrm{8}× \\ $$$$ \\ $$$$ \\ $$

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