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# In the land of polynomials with 500 dimensions

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

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

1 year, 7 months ago

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@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting. · 1 year, 7 months ago

does the summation mean $\dfrac{1}{2+x_1}+\dfrac{1}{3+x_2}+....+\dfrac{1}{p_{100}+x_{100}}$ · 1 year, 7 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 above · 1 year, 7 months ago

@Chinmay Sangawadekar, toitent here is $$\phi(x_i$$? · 1 year, 7 months ago

no toitent of I · 1 year, 7 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 ? · 1 year, 7 months ago

what about p then? There are 100 values of p not any constant · 1 year, 7 months ago

no no take the whole summation as constant · 1 year, 7 months ago

But I think I can do the totient one · 1 year, 7 months ago

yes · 1 year, 7 months ago

I think yes. · 1 year, 7 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. · 1 year, 7 months ago

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 $$x_1$$? · 1 year, 6 months ago

@Calvin Lin I have updated it accordingly. · 1 year, 7 months ago

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? Staff · 1 year, 7 months ago

sorry sir its pi ..denotes the ith prime no...I will update it · 1 year, 7 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 $$x_ 1$$? Staff · 1 year, 6 months ago

I have one more ,

Just replace $$p$$ with totient(_i) · 1 year, 7 months ago

@Chinmay Sangawadekar, have you solved this, or just randomly picked some functions? · 1 year, 7 months ago

So I posted it on Brilliant · 1 year, 7 months ago

I jist had an idea abt it...then I tried , but no outcome · 1 year, 7 months ago

Can we find this for small degree polynomials · 1 year, 7 months ago

No I haven't solved this yet but can this be solved? · 1 year, 7 months ago

i think not possible to find or may be i am wrong?? · 1 year, 7 months ago

u are correct, speaking analytically... · 1 year, 7 months ago

do you think totient one is possible to find? · 1 year, 7 months ago

hmm... i have not quite understand what toitent means here. eulers toitent function? · 1 year, 7 months ago