\[\displaystyle \sum_{ i=1}^{500} \frac{1}{p_i+x_i}\]

Where \(p_i\) denotes the \(i^{th}\) prime number , for a polynomial mentioned below.

\(x^{500} + x^{300} - x^{3} + x - 1729\)

has roots \(x_{1} , x_{2} , ..... , x_{500}\)

Then evaluate the above summation.

See this

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## Comments

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TopNewest@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting.

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does the summation mean \[\dfrac{1}{2+x_1}+\dfrac{1}{3+x_2}+....+\dfrac{1}{p_{100}+x_{100}}\]

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I have been working on this problem as you said it cant be solved analytically yeah see for the totient I have mentioned it above

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@Chinmay Sangawadekar, toitent here is \(\phi(x_i\)?

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But I think I can do the totient one

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yes

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I think yes.

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I think that the question is not complete in it's statement. I mean, how do we define the pairing of a root with a prime number? What is \(x_1\)?

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@Calvin Lin I have updated it accordingly.

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I am assuming that you want "where \(p_i \) denotes the ith prime number"?

If \(p\) is allowed to vary , then this problem is not well phrased because there isn't a (canonical) order to the roots of a polynomial. E.g. Assume that we only have 2 roots which are \(1+i \) and \( 1 - i \). What is the sum that we're interested in?

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sorry sir its pi ..denotes the ith prime no...I will update it

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I understand. But, what is the order of the roots? Since we (most likely) have complex roots, there is no (canonical) order to the roots that we can set. In particular, which of the roots is \( x_ 1 \)?

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I have one more ,

Just replace \(p\) with totient(_i)

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@Chinmay Sangawadekar, have you solved this, or just randomly picked some functions?

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So I posted it on Brilliant

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I jist had an idea abt it...then I tried , but no outcome

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Can we find this for small degree polynomials

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No I haven't solved this yet but can this be solved?

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i think not possible to find or may be i am wrong??

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u are correct, speaking analytically...

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do you think totient one is possible to find?

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