Let be the product of all prime divisors of .
Find the sum of all positive integer values so that .
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If N > 1 then g ( N ) > 1 divides N .
If n > 1 is such that g ( n 2 − n + 1 ) divides n + 1 , then g ( n 2 − n + 1 ) must divide both n 2 − n + 1 and n + 1 . Since n 2 − n + 1 = ( n − 2 ) ( n + 1 ) + 3 , this means that g ( n 2 − n + 1 ) divides 3 , and hence must be equal to 3 . Thus we deduce that n 2 − n + 1 = 3 K for some K ≥ 1 . If n ≡ 0 ( m o d 3 ) or n ≡ 1 ( m o d 3 ) it is easy to see that n 2 − n + 1 ≡ 1 ( m o d 3 ) . Thus we must have n ≡ 2 ( m o d 3 ) . But if we write n = 3 m − 1 we see that n 2 − n + 1 = 9 m ( m − 1 ) + 3 is divisible by 3 but not by 9 . Thus we deduce that K = 1 and hence that n 2 − n + 1 = 3 . Thus ( n − 2 ) ( n + 1 ) = 0 and so (since n > 1 ), n = 2 . It is easy to see that n = 2 is indeed a solution. Thus the only possible answer, and hence the sum of all possible answers, is 2 .