Find the sum of all integers n such that 3 n + 5 n 2 + 1 3 n + 2 is an integer.
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I didn't understand the 2nd step in which you omitted 9 from denominator
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I multiplied ( 3 n + 5 ) ( 3 1 n + 9 3 4 ) − 9 1 5 2 by 9 .
Y did u omit 9 from the denominator ?
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I multiplied ( 3 n + 5 ) ( 3 1 n + 9 3 4 ) − 9 1 5 2 by 9 .
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3 n + 5 ∣ n 2 + 1 3 n + 2 ⟺ 3 n + 5 ∣ ( 3 n + 5 ) ( 3 1 n + 9 3 4 ) − 9 1 5 2 ⟺ 3 n + 5 ∣ ( 3 n + 5 ) ( 3 n + 3 4 ) − 1 5 2 ⟺ 3 n + 5 ∣ 1 5 2
Note that one of the steps I used required g cd ( 3 n + 5 , 9 ) = 1 , which is true for all n ∈ Z .
Since 3 n + 5 ∣ 1 5 2 , we have the posibilities − 2 7 , − 8 , − 3 , − 2 , − 1 , 1 , 1 1 , 4 9 .
Add those up and the answer is: 2 0 .