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Yes, thanks for that suggestion! For square-free numbers the count turns out to be pretty easy... the general case is a mess, though! Nobody has solved it yet... I'm counting on you!

@Otto Bretscher , do we need to find a polynomial whose roots are , $2017 , 2017^2 ,.... 2017^2016$ ?
As the inner summation is 2017 @Aareyan Manzoor help

-1 is obviously not a root. we multiply by (x+1) to find
$f(x)=x^{2017}+1=0$
one root of this expression is -1, and if we put it in the sum,i.e. $(1+x_i)=(1-1)=0$. so we can go on withput worries. we are just searching for:
$\sum_{k=1}^{2016}\sum_{i=1}^{2017} (x_i+1)^k=\sum_{i=1}^{2017} (\dfrac{(x_i+1)^{2017}-1}{x_i+1-1}-1)=\sum_{i=1} (x_i^{2016}+\binom{2017}{1}x_i^{2015}+....+\binom{2017}{2016}-1)$
gp used here. view this.yes, the sum of all 2016th to 1st power is zero. the $\binom{2017}{2016}-1=2016$is left. added 2017 times results in
$2016*2017=\boxed{4066272}$

There may be a small error in the count (or rather two errors that cancel out). Your sum actually involves 2017 $i$'s (you have added -1) but you have to subtract 1 from the summand.

@Aareyan Manzoor
–
But if you use only 2016 terms then it is no longer true that the sum of the 2016th to first powers is zero. Think about it... you will see what I mean.

@Aareyan Manzoor
–
No, there is another small error that compensates for this one... you are losing a term $-1$ along the way... the constant term in your sum should be 2017-1=2016, so it all works out... a Christmas miracle ;)

My solution is quite short and straightforward.... why should I look for another solution? ;) (I guess I could use Viète to prove that the sum of the $j$th powers of the roots of unity is 0)

I thought about it in a slightly different but equivalent way.

We observe that the $y_i=-x_i$, together with $y_0=1$, are the 2017th roots of unity. Now $\sum_{i=0}^{2016}(1-y_i)^k$$=\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=2017$ since the sum of the $j$th powers of the 2017th roots of unity is 0.

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## Comments

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TopNewest@Calvin Lin , what is your opinion for this ? What will be your approach ?

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@Otto Bretscher ,

Cool problem .... the new problem which is based on our yesterday's conversation ...

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Yes, thanks for that suggestion! For square-free numbers the count turns out to be pretty easy... the general case is a mess, though! Nobody has solved it yet... I'm counting on you!

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Bonus problem is complicated though @Otto Bretscher

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@Otto Bretscher , do we need to find a polynomial whose roots are , $2017 , 2017^2 ,.... 2017^2016$ ? As the inner summation is 2017 @Aareyan Manzoor help

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@Otto Bretscher , @Aareyan Manzoor , I have slight doubt ,

the inner summation will be definitely 2017.

So we need a polynomial whose roots are , $2017^{1} , 2017^{2} , 2017^{3} , .... , 2017^{2016}$ , isnt it ?

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nope.

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So am I right or wrong ? @Aareyan Manzoor

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@Aareyan Manzoor , but why ?

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@Aareyan Manzoor

sorry we need a polynomial , so that we can find the summation using vieta ,Log in to reply

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@Aareyan Manzoor

But can this be solved by vieta ? according to me it could notLog in to reply

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$x_i$ are zero?

Aareyan, are you not using Viete to make sure that the symmetric sums of theLog in to reply

@Aareyan Manzoor , @Otto Bretscher

Lets try again write a solution on the work that you have done , on vieta approach , then I will post mineLog in to reply

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@Otto Bretscher

I am just asking whether my approach is right ? (by vieta)Log in to reply

@Otto Bretscher , but please also see the vieta approach ,,, is it right ?

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$x^{2017}+1$... that works.

Yes, Aareyan is applying Viete to the polynomialLog in to reply

-1 is obviously not a root. we multiply by (x+1) to find $f(x)=x^{2017}+1=0$ one root of this expression is -1, and if we put it in the sum,i.e. $(1+x_i)=(1-1)=0$. so we can go on withput worries. we are just searching for: $\sum_{k=1}^{2016}\sum_{i=1}^{2017} (x_i+1)^k=\sum_{i=1}^{2017} (\dfrac{(x_i+1)^{2017}-1}{x_i+1-1}-1)=\sum_{i=1} (x_i^{2016}+\binom{2017}{1}x_i^{2015}+....+\binom{2017}{2016}-1)$ gp used here. view this.yes, the sum of all 2016th to 1st power is zero. the $\binom{2017}{2016}-1=2016$is left. added 2017 times results in $2016*2017=\boxed{4066272}$

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There may be a small error in the count (or rather two errors that cancel out). Your sum actually involves 2017 $i$'s (you have added -1) but you have to subtract 1 from the summand.

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i have said that "one root of this expression is -1...". at the first.

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$\sum_{i}$ you have 2017 terms.

So when you add upLog in to reply

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$2017^2$?

true... the answer should beLog in to reply

$-1$ along the way... the constant term in your sum should be 2017-1=2016, so it all works out... a Christmas miracle ;)

No, there is another small error that compensates for this one... you are losing a termLog in to reply

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$i$ that goes to 2017, not $k$.

One more small correction: In your double sum, it isLog in to reply

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Taking a quick look at this during our x-mas festivities, I would suspect that the inner sum is always 2017, so that the whole sum would be 2016*2017.

[added later] We observe that the $y_i=-x_i$, together with $y_0=1$, are the 2017th roots of unity. Now

$\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-y_i)^k=\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=\sum_{k=1}^{2016}2017=2016*2017$

since the sum of the $j$th powers of the 2017th roots of unity is 0 for $j\leq 2016$.

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@Otto Bretscher , please try to solve using Vieta.

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My solution is quite short and straightforward.... why should I look for another solution? ;) (I guess I could use Viète to prove that the sum of the $j$th powers of the roots of unity is 0)

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yes,you are right @Otto Bretscher, please view my comment when you have time.

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I thought about it in a slightly different but equivalent way.

We observe that the $y_i=-x_i$, together with $y_0=1$, are the 2017th roots of unity. Now $\sum_{i=0}^{2016}(1-y_i)^k$$=\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=2017$ since the sum of the $j$th powers of the 2017th roots of unity is 0.

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