Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Well, first of all i´m sorry, but I don't know how to use Latex, anyway the key of the solution is the function f(x)=(1+x+x^2)(1+x^2+x^4) ... (1+x^(p-1)+x^2(p-1)) Let c_n be the factor multiplying x^n in the polynomic form of the function and it's clearly the number of ways to sum n with: x_1 + 2(x_2) + ... + (p-1) x_(p-1) = n where x_i is either 0, 1 or 2 so E(0,1,2)=c_p + c_2p + c_3p + ... Let r be a nontrivial pth-root of the unity, then 1 + z^n + z^(2n) + ... + z^((p-1)n) equals p if p|n and 0 otherwise. Hence we have that: f(1)+ f(z)+f(z^2)+...+f(z^(p-1))=p(c_p+c_2p + ... )= p ( E{0,1,2}) We also have that f(1)=3^(p-1) and if we expres 1+y+y^2 as (y^3-1)/(y-1) we find thet f(z^k)=1 for 0<k<p, because all factors cancell out, so we have that: E(0,1,2)=(3^(p-1)+p-1)/p Now let us consider g(x)=(1+x+x^3)(1+x^2+x^6) ... (1+x^(p-1)+x^3(p-1)) Everything work the same and we get that E(0,1,3)=(3^(p-1) + (p-1)g(z))/p and the given inequality transforms into g(z)>=1. But as E(0,1,3) is clearly an integer, g(z) mod p = 1 and if we take: (1+z^j+z^3j) and (1+z^k+z^3k) where j+k=p the they are complex conjugates so their product is nonnegative, so g(z)>=0, and that completes the proof that g(z)>=1 However I wasn't able to proove that g(z)=1 iff p=5, I mean when p=5 is easy to see that g(z)=1 because things cancell out but I don't know how to proove that if g(z)=1 => p=5, if anyone knows please tell me. Bye, Julian PS: I'm sure it's not the best way to do it, maybe there exists a nice bijection but anyway it nice to find the real numbers.

assume that: $f(x)=1+x+x^2$ and $F(x)=f(x)f(x^2)...f(x^{p-1})$ thus $F(x)=\sum\limits_{n=0}^{p(p-1)}a_nx^n$ now notice that $a_n$ is equal to the number of $(x_1,...,x_{p-1})$ wich $x_j\in {0,1,2}$ and we know that $x_1+2x_2+...+(p-1)x_{p-1}=n$ there for $|E({0,1,2})|$ is equal to the sum of $a_n$ which $n$ is divisible by $p$ assume that $\omega=\cos(\frac{2\pi}{p})+i\sin(\frac{2\pi}{p})$ it yields easyly that: $1+\omega^j+\omega^2j+...+\omega^{(p-1)j}=\{\begin{array}{cc}p&\mbox{ if } p\mid j\\0&\mbox{ if } p\not\mid j\end{array}$ there for if we add $F(x)$ for $x=1,\omega,...,\omega^{p-1}$: $F(1)+F(\omega)+...+F(\omega^{p-1})=p|E({0,1,2})|$ notice that $F(1)=3^{p-1}$ and also notice that if $j$ is not divisible by $p$ then: $(1,\omega^j,\omega^{2j},...,\omega^{(p-1)j})$ is a permutation of numbers $1,\omega,\omega^2,...,\omega^{p-1}$ now with using this point for $j=3$ then we have: $F(\omega)=F(\omega^2)=...=F(\omega^{p-1})=(1+\omega+\omega^2)...(1+\omega^{p-1}+\omega^{2(p-1)})=(\frac{1-\omega^2}{1-\omega})(\frac{1-\omega^6}{1-\omega^2})...(\frac{1-\omega^{2(p-1)}}{1-\omega^{p-1}})=1$ therefor $|E({0,1,2})|=\frac{3^{p-1}+p-1}{p}$ now assume that $g(x)=1+x^2+x^3$ and $G(x)=g(x)g(x^2)...g(x^{p-1})$ now we can use the soloution before to get that: $|E({0,1,3})|=\frac{G(1)+(p-1)G(\omega)}{p}=\frac{3^{p-1}+(p-1)G(\omega)}{p}$ there for we most show that $1\leq G(\omega)$ and it equals only for $p=5$ now if $h(x)=x^3+x+1$ then we can say that $h(x)=(x-\lambda)(x-\mu)(x-\nu)$ wich $\mu,\lambda,\nu$ are complex numbers and $0<x$,$0<h(x)$ and also $\lambda+\mu+\nu=0$ there for $h(x)$ has a negative real root like $\lambda$ $\mu , \nu=\bar{\mu}$ are two complex roots of $h$ with positive real parts notice that: $G(\omega)=\prod\limits_{j=1}^{p-1}(1+\omega^j+\omega^{3j})=\prod\limits_{j=1}^{p-1}(\omega^j-\lambda)(\omega^j-\mu)(\omega^j-\nu)=(\frac{\lambda^p-1}{\lambda-1})(\frac{\mu^p-1}{\mu-1})(\frac{\nu^p-1}{\nu-1})$ (for the last equality we used $\prod\limits_{j=1}^{p-1}(x-\omega^j)=\frac{x^p-1}{x-1}$ for $x=\lambda,\mu,\nu$) and also $(\lambda-1)(\mu-1)(\nu-1)=-h(1)=-3$ in the other hand we have $\lambda^3+\lambda^2+1=0$ therefor for all natural numbers $k$,$\lambda^{k+3}+\lambda^{k+1}+\lambda^k=0$ and teh equality is for $\mu ,\nu$ also now if we add these equalities and use induction we`ll have that for every natural number $r$,$\lambda^r+\mu^r+\nu^r \in \mathbb {Z}$ now assume that $G(\omega)=1$, therefor $(\lambda^p-1)(\mu^p-1)(\nu^p-1)=-3$ and $\lambda^p,\mu^p,\nu^p$ are the roots of $m(x)=x^3-qx^2+(1+q)x+1=0$ wich $q=\lambda^p+\mu^p+\nu^p \in \mathbb{Z}$ and also $\lambda$ is the real root of $x^3+x+1=0$ and $-1<\lambda<-\frac{1}{2}$ thus if $q<0$ for $x=\lambda^p$,$0<x^3-qx^2$ and $0\leq (1+q)x$ therefor $m(\lambda^p)\not= 0$ also if $q=0$,$\lambda^p=\lambda$ and therefor $p=1$ and then again we`ll reach a contradiction there for $1\leq q$ for all $-1\leq x\leq 0$,$q(x^2-x)$ is positive and $x^3+x+1$ is only increasing and also $\lambda^3+\lambda+1=0$ $\lambda^p$,the $x$component of point $P$,is the point that the two functions $y=x^3+x+1$ and $y=q(x^2-x)$ gathere to gether and it bvelongs to $]-1,0[$ therefor when $p$ gets greater ($\lambda^p$ nearly reaches $0$) $q$ grows greater for $p=5,g(\omega)=1+\omega+\omega^2=-\omega^2(1+\omega^2)$ and therefor: $G(\omega)=\prod\limits_{j=1}^{p-1}(1+\omega^{2j})=\prod\limits_{j=1}^{p-1}(1+\omega^j)=\frac{1-(-1)^p}{1-(-1)}=1$ notice that $\lambda^5=-\lambda^2(\lambda+1)=-\lambda^2+\lambda+1$ $q=-(\lambda^2+\mu^2+\nu^2)+(\lambda+\mu+\nu)+3=5$ because: $\lambda^2+\mu^2+\nu^2=(\lambda+\mu+\nu)^2-2(\lambda\mu+\mu\nu+\nu\lambda)=-2(\lambda\mu+\mu\nu+\nu\lambda)=-2$ therefor $5\leq q$ and $q=5$ happens only for $p=5$ we know that $6\leq q$ and $m(x)=x^3+x+1-q(x^2-x)$ $m(-1)<0$,$m(0)>0$,$m(2)<0$ and for a great enogh $x$,$m(x)>0$ therefor $m(x)=0$ has three different real roots but these roots are $\lambda^p,\mu^p,\nu^p=\bar{\mu^p}$ and if $\mu^p \in \mathbb{R}$,then $\nu^p=\bar{\mu^p}=\mu^p$ it means that $m(x)$has a repetitive root and it would be a contradiction there for $6>q$ therefor $q=5,p=5$ is the only way that $G(\omega)=1$ from $|E({0,1,3})|=\frac{3^{p-1}+(p-1)G(\omega)}{p}$ and that $0\leq G(\omega)$ (this inequality comes from $\bar{g(\omega^j)}=g(\omega^{p-j}$) there for $1\leq G(\omega)$ there for: $|E({0,1,3})|\geq |E({0,1,2})|$ and the equality happens only for $p=5$

I think I found another proof that \[\prod ^{p-1}_{t=1}(1+\zeta ^t +\zeta^{3t})\ge 1\] has equality if and only if $p=5$ where $\beta=e^{2\pi i/p}$. Since the 'if' part is a simple computation, it is equivalent to proving that \[\prod ^{p-1}_{t=1}(1+\zeta ^t +\zeta^{3t})> 1\] if $p>5$. Notice that this is equivalent to proving that \[\prod ^{p-1}_{t=1}|1+\zeta ^t +\zeta^{3t}|=\prod ^{p-1}_{t=1}\sqrt{(1+\cos t\theta+\cos 3t\theta)^2+(\sin t\theta +\sin 3t\theta)^2}> 1\] Where $\theta=2\pi/p$. Or more simply: \[\prod ^{p-1}_{t=1}(3+2\cos t\theta+2\cos 2t\theta +2\cos 3t\theta)=\prod ^{p-1}_{t=1}(8x_t^3+(2x_t-1)^2)>1\] Where $x_t=\cos t\theta$. The roots of the equation \[p(x_t)=8x_t^3+(2x_t-1)^2=1\] Are $-1,0,1/2$ and thus we know that $p(x_t)>1$ if $x_t>1/2$ ($t\theta \in (-\pi/3,\pi/3)$) or $x\in (-1,0)$ ($t\theta \in (\pi/2,3\pi/2)$). Using calculus, one can verify that $p(x_t)>1/3$ if $x_t\in (0,1/2)$ ($t\theta \in (\pi/3,\pi/2)\cup (-\pi/2,-\pi/3)$) And due to the fact that $p$ is cubic and that $p(\sqrt{2}/2)=3$, $p(x_t)\ge 3$ if $t\theta \in (-\pi/4,\pi/4)$. Thus, \[\prod ^{p-1}_{t=1}p(x_t)>\frac{1}{3^{2\lfloor p/4\rfloor-2\lfloor p/6\rfloor}}\cdot 3^{2\lfloor p/8\rfloor}\] Thus, it is sufficient to prove that \[\lfloor p/8\rfloor \ge \lfloor p/4\rfloor-\lfloor p/6\rfloor\] For $p>5$. For primes $p\in [7,23]$ it is easy to verify. If $p>24$, then $p=p'+24$ with $p'>0$ and \[3+\lfloor p'/8\rfloor \ge 2+\lfloor p'/4\rfloor -\lfloor p'/6\rfloor\] Which is true because \[\lfloor p'/8\rfloor +1\ge \lfloor p'/12\rfloor +1 \ge \lfloor p'/4\rfloor -\lfloor p'/6\rfloor\] as desired.

My solution was same up to proving that $$P = \prod_{i=1}^{p-1} (1 + \omega^i + \omega^{3i}) \ge 1.$$ Lemma. $P > 0$. Proof. Write $$P = \prod_{i=1}^\frac{p-1}{2} (1 + \omega^i + \omega^{3i})(1 + \omega^{p-i} + \omega^{3(p-i)}) = \prod_{i=1}^\frac{p-1}{2} |1 + \omega^i + \omega^{3i}|^2.$$ Lemma 2. Let $a, b, c$ be the roots of $x^3 + 3x + 1$. Then $$P = \frac{-1}{3} (a^p - 1)(b^p - 1)(c^p - 1).$$ Proof. $$3P = \prod_{i=0}^{p-1} (1 + \omega^i + \omega^{3i}) = \prod_{i=1}^{p-1} (\omega^i - a)(\omega^i - b)(\omega^i - c)$$$$= (-1)^{3p} \prod_{j=0}^{p-1} (a - \omega^j)(b - \omega^j)(c - \omega^j) = -(a^p - 1)(b^p - 1)(c^p - 1).$$ Now, notice that because $g(x) = x^3 + 3x + 1$ is monotonic increasing (because two of the terms are and one term is constant), it has exactly one real root $a$ and two roots which are conjugates, $b, c$. Because $g(-\dfrac{1}{\sqrt{2}}) < 0 < g(0)$, we deduce that $|a| < \dfrac{1}{\sqrt{2}}$. Because the magnitude of the other two roots are equal and because the product of the roots equals $-1$ by Vieta's formula, $|b|$ and $|c|$ both are greater than $\sqrt[4]{2}.$ Thus, for $p \ge 7$, we have $|a|^7 > \sqrt[4]{128} > 3$ and so $$P = \frac{-1}{3} |a^p - 1| |b^p - 1| |c^p - 1| \ge \frac{1}{3} (1 + |a|^p)(|b|^p - 1)(|c|^p - 1) > \frac{1}{3} (3 - 1)(3 - 1) > 1,$$using the fact that $a$ is negative and the triangle inequality. Thus the strict inequality is proved for $p \ge 7$.

Let \[f_T(x)=\prod\limits_{i=1}^{p-1}\left(\sum\limits_{j\in T}\left(x^T\right)\right)=\sum_{i=1}^{N}\left(a_i x^i\right)\]where $a_i$ is the number of $(p-1)$-tuples $(x_1, \ldots ,x_{p-1} )$, where each $x_i \in T$ and $x_1+2x_2+ \ldots + (p-1)x_{p-1}=i$ and $N$ is large enough. $~$ Let $\omega$ be a primitive $p$th root of unity. Note that \[1+\omega^{i}+\omega^{2i}+\dots+\omega^{(p-1)i}\]is equal to $0$ if $i$ is not divisible by $p$, and equal to $p$ otherwise. Therefore, \[f_T(1)+f_T(\omega)+\dots+f_T(\omega^{p-1}) = \sum_{i=1}^{N}\left(a_i \left(1+\omega^i+\omega^{2i}+\dots+\omega^{(p-1)i}\right)\right)\]Since for all $i$ not divisible by $p$, the summand is zero, and for all $i$ divisble by $p$, it is $pa_i$, we have that the sum of functions is equal to $p|E(T)|$ $~$ Now we search for a second representation of the sum $f_T(1)+f_T(\omega)+\dots+f_T(\omega^{p-1})$ through direct computation. Note that $f_T(1)=|T|^{p-1}$ and that $f_T(\omega^i)=f_T(\omega)$ for all $i$ not divisible by $p$. This is because multiplying $(1,2,3,4,\dots,p-1)$ by $i$ maintains the same set of residues. Therefore, \[|E(T)|=\frac{|T|^{p-1}+(p-1)f_T(\omega)}{p}\]This implies that $f_T(\omega)$ is always real. It suffices to show that $f_{\{0,1,3\}}(\omega)\ge f_{\{0,1,2\}}(\omega)$ with equality if and only if $p=5$. We have \[f_{\{0,1,2\}}(\omega)=\prod_{i=1}^{p-1}\left(\frac{\omega^3-1}{\omega-1}\right)\]but the numerators are just a permutation of the denominators, so it is equal to one. $~$ Denote $T$ by $\{0,1,3\}$ for the remainder of this problem. We can pair up factors in $f_T(\omega)$ to get that \begin{align*} f_T(\omega) &= \prod_{i=1}^{p-1}(1+\omega^i+\omega^{3i}) \\ &= \prod_{i=1}^{\frac{p-1}{2}}(1+\omega^i+\omega^{3i})(1+\omega^{p-i}+\omega^{3(p-i)}) \\ &= \prod_{i=1}^{\frac{p-1}{2}}(1+\omega^i+\omega^{3i})\overline{(1+\omega^{i}+\omega^{3i})} \\ &= \prod_{i=1}^{\frac{p-1}{2}}\left|1+\omega^i+\omega^{3i}\right|^2 \\ &> 0 \end{align*}Observe that $(p-1)f_T(\omega)=p|E(T)|-3^{p-1}$, which is an integer and so it is at least $p-1$, so $f_T(\omega)\ge 1$. Now we tackle equality. If $p=5$ then \begin{align*} f_T(\omega)&=(1+\omega+\omega^3)(1+\omega^2+\omega^6)(1+\omega^3+\omega^9)(1+\omega^4+\omega^{12})\\ &= (1+\omega+\omega^3)(1+\omega+\omega^2)(1+\omega^3+\omega^4)(1+\omega^4+\omega^2) \\ &= 1 \end{align*}by expanding and bashing. Now let $f_T(\omega)=1$. Observe that $(x-\omega)(x-\omega^2)\dots(x-\omega^{p-1}) = 1+x+x^2+\dots+x^{p-1} = \tfrac{1-x^p}{1-x}$. Let $x^3+x+1 = (x-r)(x-s)(x-t)$ then \[f_T(\omega)=\frac{(1-r^p)(1-s^p)(1-t^p)}{(1-r)(1-s)(1-t)}=\frac{(1-r^p)(1-s^p)(1-t^p)}{3}\]so $(1-r^p)(1-s^p)(1-t^p)=3$. $~$ We also have $r^ps^pt^p=-1$. Let $r^p+s^p+t^p=k$ then $r^ps^p+s^pt^p+t^pr^p=k+1$ so $r^p,s^p,t^p$ are roots of the polynomial $x^3-kx^2+(k+1)x+1=0$. Note that $k$ is an integer. Of $r$, $s$, and $t$, we know that exactly one is real, and that it is between $-1$ and $0$ exclusive by the intermediate value theorem. Therefore, there is also a negative root to $x^3-kx^2+(k+1)x+1=0$. It is clear that if $k<0$ then there are no negative roots, so that's not possible. If $k=0$ then $r^p, s^p, t^p$ are also roots to $x^3+x+1$ but that implies $p=1$. Therefore, $k\ge 1$. Two of $r^p,s^p,t^p$ are nonreal, and so $x^3-kx^2+(k+1)x+1$ cannot have three real roots. $~$ We use disriminant of a cubic, $a^2b^2 + 18abc - 4b^3 - 4a^3c - 27c^2$. We have \[k^2(k+1)^2-18k(k+1) - 4(k+1)^3 + 4k^3 - 27<0\]It is easy to see that $k\le 5$ is necessary for this to satisfy, and we previously got that $k\ge 1$. $~$ Now we use Newton's Sums on $x^3+x+1$. Let $P_n=r^n+s^n+t^n$ then $P_0=3,P_1=0,P_2=-2,P_3=-3$, and $P_i=-P_{i-2}-P_{i-3}$ so $P_4=2,P_5=5,P_6=1,P_7=-7,P_8=-6,P_9=6,P_{10}=13,P_{11}=0,P_{12}=-19,P_{13}=-13$. From here it is easy to inductively prove a contradiction for all $p\ge 7$.