n 2 0 1 5 + n 1 0 + 1 Find the sum of all positive integers n such that the above expression is prime. If there are infinitely many such n , enter 666.
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Yes, exactly! For the benefit of other members, can you please explain how you know that n 2 + n + 1 divides n 2 0 1 5 + n 1 0 + 1
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Let x be a root of n 2 + n + 1 then x 3 = 1 from this the we can conclude x is a zero of the given expression.
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Yes, exactly! The key observation is that the third root of unity ω is a root of x 2 0 1 5 + x 1 0 + 1 since ω 2 0 1 5 + ω 1 0 + 1 = ω 2 + ω + 1 = 0
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@Otto Bretscher – what do we do after we get to know that n^2+n+1 divides the number ?
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@Avn Bha – That shows that the number isn't a prime since n 2 + n + 1 is a proper divisor if n > 1
I observe that 2015 has a remainder of 2, 10 has a remainder of 1, and 0 has a remainder of 0, when all divided by 3.
Therefore, if ω is the non-one cube root of unity, then when n = ω , the expression would be zero, making ( n 3 − 1 ) / ( n − 1 ) = n 2 + n + 1 a factor.
The only way for the expression to be prime is when n = 1 .
ingenuously 1 2 0 1 5 + 1 1 0 +1=3 and 3 is a prime
Why is that the only way for the expression to be prime?
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^^Frankly it's the first and the only i could think of.
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n 2 + n + 1 divides the number.