Find the sum of all integer values of n such that the number n 4 + 4 is a prime number.
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Look, Victor Loh i think this would have been a better solution for this problem: From Sophie Germain Identity we know that n 4 + 4 can be factored as followed:
n 4 + 4 = ( n 2 + 2 n + 2 ) ( n 2 − 2 n + 2 ) . But n 4 + 4 is a prime number and therefore one of the factors must be equal to 1 .Thus, we have two cases: First case:
n 2 + 2 n + 2 = 1 → n = 1
Second case:
n 2 − 2 n + 2 = 1 → n = − 1 . Thus, n = ± 1 and the sum is 0 .
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You made a small mistake, In the FIRST CASE, it should be n = ( − 1 ) and in the SECOND CASE, it should be n = 1 . Correct me if I'm wrong.
Right, Loh, this problem becomes trivial if "integer values" includes negative integers. Otherwise, it's a harder problem. Methinks the problem should have asked for only positive integer values.
Yes I agree. Thanks
n 4 + 4
( n 2 + 2 ) 2 − 4 n 2
( n 2 + 2 n + 2 ) ( n 2 − 2 n + 2 )
A prime number has only two factors, 1 and the number itself. So, we have two cases:
If n 2 + 2 n + 2 = 1 ⟹ n = − 1
If n 2 − 2 n + 2 = 1 ⟹ n = 1
Therefore, the sum of all integer values of n is 0
Since n is raised to the power 4, if for any positive value of n , we get n 4 + 4 as a prime no., also the negative value of n satisfies that n 4 + 4 is a prime no. When all the values of n that satisfies n 4 + 4 is a prime number are added, the corresponding positive and negative values of n cancel out each other and the resultant sum is 0
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Let x be a value of n that satisfies the equation.
Then − x will also satisfy the equation as any number a squared = − a squared.
Hence the sum of all integer values of n will cancel out to be 0 .