I think a nice idea is to consider the 1990-th roots of unity...and after that consider the vectors with sizes $1^2,2^2,\ldots,1990^2$ and multiples of the roots of unity. Drawing it would be easier to show the idea...=]

Ilthigore wrote: Is there an official solution that doesn't use complex numbers? Were they 'IMO syllabus' back in 1990? :S Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly.

Erken wrote: Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly. How about writing it down here please?

Use Complex numbers. Let $\{a_1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}$ and $\theta=\frac{\pi}{995}$. Then the problem is equivalent to that there exists a way to assign $a_1,a_2,\cdots,a_{1990}$ such that $\sum_{i=1}^{1990}a_i^2e^{i\theta}=0$. Note that $e^{i\theta}+e^{i\theta+\pi}=0$, then if we can find a sequence $\{b_n\}$ such that $b_i=a_p^2-a_q^2, (p\not=q)$ and $\sum_{i=1}^{995}b_ie^{i\theta}=0$, the problem will be solved. Note that $2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979$, then $3,7,\cdots,3979$ can be written as a sum from two elements in sets $A=\{3,3+4\times199,3+4\times398,3+4\times597,3+4\times796\}$ and $B=\{0,4\times1,4\times2,\cdots,4\times 198\}$. If we assign the elements in $A$ in the way that $l_{a1}=3,l_{a2}=3+4\times199,l_{a3}=3+4\times398,l_{a4}=3+4\times597,l_{a5}=3+4\times796$, then clearly $\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}=0, (j\in\{1,2,3,4,5\})$ Similarly, we could assign elements in $B$ in that way ($l_{b1}=0,l_{b2}=4,\cdots$) to $e^{i\frac{2\pi}{5}}$. Then we make $b_i$ according to the previous steps. Let $i\equiv j$ (mod 5), $i\equiv k$ (mod 199), and $b_i=l_{aj}+l_{bk}$, then each $b_i$ will be some $a_ p^2-a_q^2$ and $\sum_{i=1}^{995}b_ie^{i\theta}=\sum_{j=1}^5\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}+\sum_{k=1}^{199}\sum_{i=1}^{5}l_{bk}e^{i\frac{2\pi}{5}}=0 $ And we are done.

Set f(x)=x*(x-1)*(x^10-1)/(x^2-ε(5))*(x^1990-1)/(x^10-ε(199)). Here, ε(n) denotes e^(2πi/n). We can claim that: f(x) has 1990 terms with degree 1 through 1990. The coefficients are a permutation of all 1990-th roots of unity. f(x) is divisible by (x-1)^3. Therefore, f(1)=f''(1)=0, summing them yields the answer. Generalization: if n has at least s+1 distinct prime factors, then such a polygon with sides 1^s, 2^s, ..., n^s exists.

Solution from Twitch Solves ISL: Throughout this solution, $\omega$ denotes a primitive $995$th root of unity. We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \]being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$. Define $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$. Then notice that \begin{align*} z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdotswithin{=} \end{align*}and so summing yields the desired conclusion, as the left-hand side becomes \[ (1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0 \]and the right-hand side is the desired expression.

I like the first part of this answer: v_Enhance wrote: Solution from Twitch Solves ISL: Throughout this solution, $\omega$ denotes a primitive $995$th root of unity. We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \]being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$. But I can't follow the 2nd part, especially how we can get $\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$ from $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$ So this is my version for 2nd part: We have to prove that $\sum_{k=0}^{994} (4d_k-1) \omega^k = 0$ where $d_k \to \{1,2,3,4...,995\}$ and $\omega=\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995}$ (Define $\omega$ as a primitive 995th root of unity is also good, but I avoid it because there are many primitive 995th roots of unity) Because $\omega^k=(\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995})^k=\cos \frac{2k\pi}{995} + i \sin \frac{2k\pi}{995}$ We can show that $\omega^{995}=1$ and $\sum_{k=0}^{994}\omega^k = 0$ Because $995=5 \cdot 199$ we can also show that $\sum_{k=0}^{4}\omega^{199k} = 0$ and $\sum_{k=0}^{198}\omega^{5k} = 0$ Let's make our equation simpler: $\sum_{k=0}^{994} (4d_k-1) \omega^k = 4 \sum_{k=0}^{994}d_k \omega^k - \sum_{k=0}^{994} \omega^k=0 \iff \sum_{k=0}^{994}d_k \omega^k=0$ Now we split the sum into 5 smaller sums: $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+1} \omega^{5k+1} + \sum_{k=0}^{198}d_{5k+2} \omega^{5k+2} + \sum_{k=0}^{198}d_{5k+3} \omega^{5k+3} + \sum_{k=0}^{198}d_{5k+4} \omega^{5k+4}$ Here is the trick: we define that $d_{k+995j}=d_{k}$ (just remember that $d_k$s are the sides of 995-got) and because $\gcd(5,199)=1$, we can rewrite our group of sums to this: $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+199} \omega^{5k+199} + \sum_{k=0}^{198}d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + \sum_{k=0}^{198}d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} + \sum_{k=0}^{198}d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199}$ So: $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k} \omega^{5k} + d_{5k+199} \omega^{5k+199} + d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199})$ $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ d_{5k+199} \omega^{199} + d_{5k+2\cdot 199} \omega^{2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{4 \cdot 199}) \omega^{5k} $ Now if we choose : $d_{5k} \to \{1,6,11,...,198\cdot 5+1\}$ $d_{5k+199}=d_{5k}+1$ $d_{5k+2\cdot 199}=d_{5k}+2$ $d_{5k+3\cdot 199}=d_{5k}+3$ $d_{5k+4\cdot 199}=d_{5k}+4$ So: $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ (d_{5k}+1) \omega^{199} + (d_{5k}+2) \omega^{2 \cdot 199} + (d_{5k}+3) \omega^{3 \cdot 199} +(d_{5k}+4) \omega^{4 \cdot 199}) \omega^{5k} $ $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}(1+\omega^{199} + \omega^{2 \cdot 199} + \omega^{3 \cdot 199} + \omega^{4 \cdot 199})+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k} $ $\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}\cdot 0+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k} $ $\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\sum_{k=0}^{198} \omega^{5k} $ $\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\cdot 0 $ $\sum_{k=0}^{994}d_k \omega^k= 0 $ Q.E.D

Solved with ARCH hints and Evan's stream We use complex vectors to represent the sides of the 1990-gon. Rather than utilizing the 1990th roots of unity, since 1990 is 2 mod 4, we can add up the vectors of opposite sides in jumps of 2 and use the 995th roots of unity. Let $w$ be a 995th root of unity. Pairing consecutive perfect squares $(2k-1)^2$ and $(2k)^2$ as opposite sides, we require \[\sum_{k=1}^{995} a_n \cdot w^k = 0\] where the $a_n$ are of the form $(2m)^2 - (2m-1)^2 = 4m-1$. We finish using a clever arrangement of $a_n$. $\blacksquare$ andib2n wrote: I like the first part of this answer: v_Enhance wrote: Solution from Twitch Solves ISL: Throughout this solution, $\omega$ denotes a primitive $995$th root of unity. We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to \[ 0 = \sum_{n=0}^{994} c_n \omega^n \]being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$. But I can't follow the 2nd part, especially how we can get $\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$ from $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$ The expression for $\omega^5 z$ follows from $20\omega^{5} + 20\omega^{204} + 20\omega^{403} + 20\omega^{602} + 20\omega^{801}$ being $20\omega^5$ times the sum of the fifth roots of unity, or 0. The remaining expressions follow analogously.

In order to construct the wanted polygon, we'll deal with complex vectors for each of the sides of the polygon, where each pair of opposite sides is of the form $(4i^2,(2i-1)^2)$. Let $z$ be a $995-$th root of unity and let $d_i = 4i^2 = 4i^2 - (2i-1)^2 = 4i-1 $. We want to find a permutation of the $s_i$'s for which the following holds : $$\sum_{i=0}^{994} s_i \cdot z^i = 0$$ Since $995 = 5 \times 199$, then $z^{199}$ is a $5-$th root of unity, thus $20 z^{5i} (z^0 + z^{199}+z^{398}+z^{597}+z^{796} )=0 $ for all $0\geq i \geq 198 $ $...(**)$. Now by recursively combining property $(**)$ with the fact that $t = 3z^0 + 7z^{199} + 11z^{398} + 15z^{597} + 19z^{796}$, we can compute all of the $z^{5i}t$ to prove that the desired sum is no more than $$t\times \left(\sum_{i=0}^{198}z^{5i}\right) = 0$$ $$\mathbb{Q.E.D.}$$

Suppose that $\omega$ is a $995$th root of unity, and place the sides $(2k-1)^2$ and $(2k)^2$ opposite of each other, so that we only need to find a permutation $(c_0, c_1, \dots,c_{994})$ of $(3,7,\dots, 3975)$ such that \[\sum_{i=0}^{994} c_i \omega^i = 0.\] However, the following construction suffices. $\square$ \begin{align*} &\phantom{+} 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ &+ 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ &+ 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdots \end{align*}