AoPS Academy
, the remainder upon dividing 
by 
is a power of 
. Either prove the assertion or find (with proof) a counterexample.
![[asy]
import geometry;
unitsize(3cm);
draw(circle((0,0),1),linewidth(1.5));
for (int i = 0; i < 7; ++i) {
for (int j = 0; j < i; ++j) {
draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));
}
}
for(int i = 0; i < 7; ++i) {
dot(dir(i * 360/7),5+black);
}
[/asy]](http://latex.artofproblemsolving.com/9/2/e/92ecdecb16bc7556a4fcda2df737b19e1d80013e.png)
, where 
. Then, the desired answer is![\[\sum_{j=1}^7\sum_{k=j+1}^7\left|\omega^j-\omega^k\right|^4=\frac 12\sum_{k=1}^7\sum_{k=1}^7\left|\omega^{j-k}-1\right|^4\left|\omega^k\right|^4=\frac 72\sum_{j=1}^6\left|\omega^j-1\right|^4\]](http://latex.artofproblemsolving.com/7/d/1/7d1b739cb5a665dae8996f9db79cde32440a03e3.png)
In particular, we know![\[|\omega^j-1|^2=(\omega^j-1)\overline{(\omega^j-1)}=(\omega^j-1)\left(\frac 1{\omega^j}-1\right)\]](http://latex.artofproblemsolving.com/f/5/3/f5325e7543c3b6f6c2a5543c993964ce7c106db3.png)
Thus, the sum rewrites itself as![\[\frac72\sum_{j=1}^6(\omega^j-1)^2\left(\frac 1{\omega^j}-1\right)^2\]](http://latex.artofproblemsolving.com/a/c/a/acae28d7b313f9ff2337d3befbffda78dc246088.png)
Define the rational function![\[P(x)=\sum_{j=1}^6(x^j-1)^2\left(\frac 1{x^j}-1\right)^2\]](http://latex.artofproblemsolving.com/8/5/5/855db128186f3642b829d9ea4c8be0f507823189.png)
Note that 
, and 
for 
(as it simply re-shuffles the order of the same summands), we have![\[P(\omega)=\frac 16\sum_{j=1}^7P(\omega^j)\]](http://latex.artofproblemsolving.com/4/b/d/4bd59cb3c87959e4bcaac21134f29ec711607135.png)
Now, as the polynomial with roots is 
, we know![\[\sum_{k=1}^7\omega^{jk}=\begin{cases}7&7\mid j\\0&\mathrm{otherwise}\end{cases}\]](http://latex.artofproblemsolving.com/c/f/d/cfdff33cd6b03da5a04614260572651605517bc8.png)
In particular, this implies we only care about the terms in the expansion of 
with exponents which are multiples of powers of 
, which happens to be the constant terms. However, we note that the constant term is simply 
, so thus![\[\frac 16\sum_{j=1}^7P(\omega^j)=\frac 16\cdot 6\cdot 6\cdot 7=42\]](http://latex.artofproblemsolving.com/2/a/0/2a0410eed6edde419ecda71be90a98187ffce9dc.png)
so the answer is 
, or 
.
. Answer is half of 
. Since terms such as 
sum as 
and all annihilate. Thus the answer is 
.
. Then the answer is 
, since there are edges with length 
, with length 
, and with length 
.
, we obtain
and 
. Additionally, we know that 
. Adding, we get
so the answer is 
, which is 
.
so I'd answer C in a hurry.
, where 
. The key idea is that 
Similarly, the fourth powers of the other two lengths are 
and 
. Adding these up and reducing using 
gives 
. Since 
, the resulting sum is equal to 
. Since there are seven segments of each length, the answer is 
.
-gons inscribed in a unit circle. As a warm-up, here are the answers for 
, 
, and 
.
, the summation is just:
, the summation is just:
, the summation is just:
, 
, and 
, so by the principle of Engineer's Induction we find that the answer for general -gons is 
. Applying this for the case of 
gives our final answer of 
over all 
with 
, where 
. This is simply![\[ S: = \frac{1}{2}\sum_{m = 0}^6\sum_{n = 0}^6 |z^m - z^n|^4.\]](http://latex.artofproblemsolving.com/1/b/f/1bfac9fd79ec367f5f477d393c21fd17ce36a3e8.png)
Note that![\[ |z^m - z^n|^2 = (z^m - z^n)(\overline{z}^m - \overline{z}^n) = 2 - z^{m - n} - z^{n- m}.\]](http://latex.artofproblemsolving.com/0/5/f/05fef2ab23d54affdcd516a893905740d6748fa5.png)
Square both sides to get![\[ |z^m - z^n|^4 = 6 - 4(z^{m-n} + z^{n- m}) + z^{2m-2n}+ z^{2n -2m}.\]](http://latex.artofproblemsolving.com/5/2/f/52f369dadf53f6ed5475e74a9f3e2fdbfbd1b961.png)
Sum both sides over the set previously mentioned to get![\[ S = \frac{1}{2}\cdot 6\cdot 7^2 - 4\sum_{m = 0}^6\sum_{n = 0}^6 z^{n-m} + \sum_{m = 0}^6\sum_{n = 0}^6 (z^2)^{n-m}.\]](http://latex.artofproblemsolving.com/e/4/b/e4b3f9bbcc6acc19c0966c159db39348f70d85fe.png)
For obvious reasons 
, so![\[ S = 3\cdot 7^2 = \boxed{147}.\]](http://latex.artofproblemsolving.com/d/8/9/d89e682178cf1f30877bb193f862f07fc63c6fc7.png)
. Then the answer is , since there are edges with length , with length , and with length ., we obtain and . Additionally, we know that . Adding, we getso the answer is , which is .
, where 
.
for all 
. However, since 
for all 
, we have
.
respectively.
Denote to be a complex solution to 
. WLOG let 
.![$2\sin{\frac{2\pi}{7}}=2[\frac{1}{2}(\omega+\omega^{-1})], 2\sin{\frac{4\pi}{7}}=2[\frac{1}{2}(\omega^2+\omega^{-2})], 2\sin{\frac{6\pi}{7}}=2[\frac{1}{2}(\omega^3+\omega^{-3})]$](http://latex.artofproblemsolving.com/6/1/3/613013052904a1f00113a6868d78b8811ef83b9d.png)
.![\begin{align*}
&7[(\omega+\omega^{-1})^4+(\omega^2+\omega^{-2})^4+(\omega^3+\omega^{-3})^4] \\
&=7[(\omega^4-4\omega^2+6-4\omega^{-2}+\omega^{-4})+(\omega^8-4\omega^4+6-4\omega^{-4}+\omega^{-8})+(\omega^{12}-4\omega^6+6-4\omega^{-6}+\omega^{-12})] \\
&=7[-3(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega)+18]=7[-3(-1)+18]=\boxed{147} * \\
\end{align*}](http://latex.artofproblemsolving.com/a/e/9/ae97ab49635b30030848d9dd099cc0b35d9f0c0b.png)
so 
.
