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

S h o w t h a t f o r a l l n ϵ N - {0}

7^{2 n + 1} + 1 i s a n i n t e g e r m u l t i p l e o f

8 .

Answered by ajfour last updated on 05/Dec/17

7^{2 n + 1} + 1 = {(8 - 1)}^{2 n + 1} + 1

= 8^{2 n + 1} -^{2 n + 1} C_{1} {(8)}^{2 n} + \dots +^{2 n + 1} C_{2 n} (8)

= 8 m .

Commented by NECx last updated on 06/Dec/17

p l e a s e c a n i t b e p r o v e n b y P . M . I

i w o u l d l i k e t o s e e t h a t u s e d .

T h a n k s .

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

7^{2 n + 1} + 1 = 8 \times

7^{3} + 1 = 344 = 8 \times

7^{2 k + 1} + 1 = 8 \times h i p o t e s i s

7^{2} (7^{2 k + 1} + 1) = 7^{2} (8 \times)

7^{2 (k + 1) + 1} + 48 + 1 = 8 \times

7^{2 (k + 1) + 1} + 1 = 8 \times - 48 = 8 \times

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