Prime expression
Find the sum of all integers n for which n 4 − 3 n 2 + 9 is a prime.
The answer is 0.
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3 solutions
Is the set finite?
If the set is not finite, it's not reasonable to talk about the sum of its elements. Even if − x is in the set for every element x .
Let me expand: an infinite sum is taken to be the limit of the partial sums. And a series that doesn't converge absolutely can approach all sorts of limits depending on how its terms are ordered. Since the set in question is a subset of the integers, either the set itself is finite, or its terms are arbitrarily large. If the terms are arbitrarily large, it will not converge absolutely, and there isn't a unique way to assign a value to the sum "of the set".
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Well put. Let me put a disclaimer here by saying that I assumed that the number of solutions is finite.
I did this the same way as @Pi Han Goh , but as an alternative, note that n 4 − 3 n 2 + 9 = ( n 2 + 3 n + 3 ) ( n 2 − 3 n + 3 )
so the quantity can only be prime if one of these is equal to ± 1 ; this is only the case when n ∈ { − 2 , − 1 , 1 , 2 } (and all of these cases do indeed give primes).
Since f ( n ) = n 4 − 3 n 2 + 9 is aven, if f ( m ) is a prime, f ( − m ) is also the same prime. And f ( 0 ) = 9 is not a prime. Therefore, the sum of all n such that f ( n ) is a prime is 0 .
Disclaimer: This solution is incomplete. We do not know whether the number of solutions of n is finite or not. If it is not, then my solution fails dramatically.
Let S denote the set of all integers n that satisfy the given condition. If n = t is a solution, then so is n = − t . The sum of elements this set S must be 0 .