Question 1150730
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If x = 1 mod 6 then  (x^2 + x + 2) mod 6 = (1 + 1 + 2)  mod 6 =  4 mod 6;

If x = 2 mod 6 then  (2^2 + 2 + 2) mod 6 = (4 + 2 + 2)  mod 6 =  8 mod 6 = 2 mod 6;

If x = 3 mod 6 then  (3^2 + 3 + 2) mod 6 = (9 + 3 + 2)  mod 6 = 14 mod 6 = 2 mod 6;

If x = 4 mod 6 then  (4^2 + 2 + 2) mod 6 = (16 + 4 + 2) mod 6 = 22 mod 6 = 4 mod 6;

If x = 5 mod 6 then  (5^2 + 2 + 2) mod 6 = (25 + 5 + 2) mod 6 = 32 mod 6 = 2 mod 6;

If x = 6 mod 6 then  (6^2 + 2 + 2) mod 6 = (36 + 6 + 2) mod 6 = 44 mod 6 = 2 mod 6.


Thus,  x^2 + x + 2  NEVER gives the remainder of 1 when divided by 6 for ALL integer x.    <U>ANSWER</U>
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Solved, answered, explained and completed.