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

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

## Comments

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TopNewest@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting. – Chinmay Sangawadekar · 1 year, 9 months ago

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– Aareyan Manzoor · 1 year, 9 months ago

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

– Chinmay Sangawadekar · 1 year, 9 months ago

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

@Chinmay Sangawadekar, toitent here is \(\phi(x_i\)? – Aareyan Manzoor · 1 year, 9 months ago

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– Chinmay Sangawadekar · 1 year, 9 months ago

no toitent of ILog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

I think this could be possible , take the summation as any constant and then compare it with 1/provided ... then apply vieta ? isnt it ?Log in to reply

– Dev Sharma · 1 year, 9 months ago

what about p then? There are 100 values of p not any constantLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

no no take the whole summation as constantLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

But I think I can do the totient oneLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

yesLog in to reply

– Akshat Sharda · 1 year, 9 months ago

I think yes.Log in to reply

– Aareyan Manzoor · 1 year, 9 months ago

this can't be analytically solved i think.... it could have been if the polynomial at the negative of the primes attained the same value.Log in to reply

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 \(x_1\)? – Anupam Nayak · 1 year, 8 months ago

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@Calvin Lin I have updated it accordingly. – Chinmay Sangawadekar · 1 year, 9 months ago

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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? – Calvin Lin Staff · 1 year, 9 months ago

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– Chinmay Sangawadekar · 1 year, 9 months ago

sorry sir its pi ..denotes the ith prime no...I will update itLog in to reply

– Calvin Lin Staff · 1 year, 9 months ago

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 \( x_ 1 \)?Log in to reply

I have one more ,

Just replace \(p\) with totient(_i) – Chinmay Sangawadekar · 1 year, 9 months ago

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@Chinmay Sangawadekar, have you solved this, or just randomly picked some functions? – Aareyan Manzoor · 1 year, 9 months ago

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– Chinmay Sangawadekar · 1 year, 9 months ago

So I posted it on BrilliantLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

I jist had an idea abt it...then I tried , but no outcomeLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

Can we find this for small degree polynomialsLog in to reply

– Chinmay Sangawadekar · 1 year, 9 months ago

No I haven't solved this yet but can this be solved?Log in to reply

i think not possible to find or may be i am wrong?? – Dev Sharma · 1 year, 9 months ago

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– Aareyan Manzoor · 1 year, 9 months ago

u are correct, speaking analytically...Log in to reply

– Dev Sharma · 1 year, 9 months ago

do you think totient one is possible to find?Log in to reply

– Aareyan Manzoor · 1 year, 9 months ago

hmm... i have not quite understand what toitent means here. eulers toitent function?Log in to reply

– Dev Sharma · 1 year, 9 months ago

yeah........Log in to reply