What is the sum of the integer values of for which is also an integer?
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We first rewrite n − 1 3 n 2 − 1 4 n + 2 3 = n − 1 3 ( n − 1 ) ( n − 1 3 ) + 1 0 = n − 1 + n − 1 3 1 0 .
Now ( n − 1 3 ) is a factor of 1 0 , i.e. n − 1 3 = ± { 1 , 2 , 5 , 1 0 } .
Since the sum of all 8 possible values of ( n − 1 3 ) is zero, we deduce that the sum of all n = 1 3 × 8 = 1 0 4 .