Prove that for each integer a, if a is an odd integer that is not multiple of 3,

Question

Prove that for each integer a, if a is an odd integer that is not multiple of 3, then a^2 is congruent to 1 (mod 6).

Explanation / Answer

Dear Student Thank you for using Chegg ! Let the given integer be I Now it is given that - I is an odd integer => General Term 2k+1, 1,3,5,7,9,……. - I is not a multiple of 3 => General Term 3k+1 or 3k+2, 1,2,4,5,7,8…… Combining both the conditions above we get terms like 1, 5, 7, 11,13, 17,19 Hence Generl term is (6k + 1) where for each value of k we have two integers which are not multiple of 3 and odd intergers Therefore I = 6k -1 or I = 6k+1 Squaring both sides I^2 = 36k^2 -12k + 1 or I^2 = 36k^2 + 12k + 1 Now using congruency when I^2 is divided by 6 I^2mod(6) = 1 or I^2mod(6) = 1 Hence remainder is 1 in both the cases which is equal to 1 mod(6) i.e. RHS Hence proved that for each integer a, if a is an odd integer that is not multiple of 3, then a^2 is congruent to 1 (mod 6).

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