More fun in 2015, Part 19

n 2015 + n 10 + 1 n^{2015}+n^{10}+1 Find the sum of all positive integers n n such that the above expression is prime. If there are infinitely many such n n , enter 666.


The answer is 1.

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3 solutions

Mietantei Conan
Oct 5, 2015

n 2 + n + 1 n^2+n+1 divides the number.

Yes, exactly! For the benefit of other members, can you please explain how you know that n 2 + n + 1 n^2+n+1 divides n 2015 + n 10 + 1 n^{2015}+n^{10}+1

Otto Bretscher - 5 years, 8 months ago

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Let x x be a root of n 2 + n + 1 n^2+n+1 then x 3 = 1 x^3=1 from this the we can conclude x x is a zero of the given expression.

mietantei conan - 5 years, 8 months ago

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Yes, exactly! The key observation is that the third root of unity ω \omega is a root of x 2015 + x 10 + 1 x^{2015}+x^{10}+1 since ω 2015 + ω 10 + 1 = ω 2 + ω + 1 = 0 \omega^{2015}+\omega^{10}+1=\omega^{2}+\omega+1=0

Otto Bretscher - 5 years, 8 months ago

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@Otto Bretscher what do we do after we get to know that n^2+n+1 divides the number ?

avn bha - 5 years, 8 months ago

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@Avn Bha That shows that the number isn't a prime since n 2 + n + 1 n^2+n+1 is a proper divisor if n > 1 n>1

Otto Bretscher - 5 years, 8 months ago
Kenny Lau
Oct 15, 2015

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 \omega is the non-one cube root of unity, then when n = ω n=\omega , the expression would be zero, making ( n 3 1 ) / ( n 1 ) = n 2 + n + 1 (n^3-1)/(n-1)=n^2+n+1 a factor.

The only way for the expression to be prime is when n = 1 n=1 .

ingenuously 1 2015 1^{2015} + 1 10 1^{10} +1=3 and 3 is a prime

Why is that the only way for the expression to be prime?

Calvin Lin Staff - 5 years, 8 months ago

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^^Frankly it's the first and the only i could think of.

Anh Đức Trần - 5 years, 8 months ago

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