Problem 1-(Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco) Let $ a,b,c$ positive reals such that $ ab + bc + ca = 3$, show that: $ \displaystyle a^2 + b^2 + c^2 + 3 \ge \frac {a(3 + bc)^2}{(c + b)(b^2 + 3)} + \frac {b(3 + ca)^2}{(a + c)(c^2 + 3)} + \frac {c(3 + ab)^2}{(b + a)(a^2 + 3)}$ Problem 2- (Daniel Kohen, University of Buenos Aires - Buenos Aires,Argentina) Find all non-negative integers $ x,y,z$ such that $ 5^x + 7^y = 2^z$. Problem 3-(Paolo Leonetti, Università Bocconi - Pescara,Italy) Let $ 0 < a_1 < a_2 < a_3 < ... < a_{10000} < 20000$ be integers such that $ gcd(a_i,a_j) < a_i, \forall i < j$ ; is $ 500 < a_1$ (always) true ? Good work.. edit: in the last problem i had added "always", altough i think you understand..

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Contestants who give me solutions: Honey_S (partial) Mehdi Cherif (partial) Inequality_Master (partial) fede90 (partial) dima_ukraine (partial) Rigel (partial) [...]

it answered also Akashnil and giove and DDs wife..( i cannot edit the post before..) as soon as possibile we 'll do ranks and contact all you Link to problems: Problem1 Problem1 Problem1 I have also added other two that had not be used: Fake edition#1 Fake edition#2

Sorry for not participating I just realized that I lost the final round because I had some problems and I had not opened my mailbox for 10 days I posted my solution to the first problem at the inequalities section (the link above)

We are very sorry to realize that our solution of number 2 was wrong.. we've got only some steps, but not the whole solution; we're thinking about a new problem to substitute to it, it will follow more information.. Solution Problem 1 $ \sum_{cyc}{a^2}=\sum_{cyc}{a(\frac{a+c-c}{a+b})(a+b)}=$ $ \sum_{cyc}{\frac{a(a+c)(a+b)}{a+b}-ac} \ge \sum_{cyc}{\frac{a(a+\sqrt{bc})^2}{a+b}-ac}\ge$ $ \sum_{cyc}{\frac{a(a+\frac{2}{\frac{1}{b}+\frac{1}{c}})^2}{a+b}-ac}$ $ =\sum_{cyc}{\frac{a(ab+ac+2bc)^2}{(c+b)^2(a+b)}-ac}=$ $ \sum_{cyc}{\frac{a(3+bc)^2}{(c+b)(b^2+3)}-ac}$ The first follows by CS, the second one by GM-HM Solution Problem 3 It is enough to show that if we have $ 2n>3^k$ distinct positive integers s.t. noone divides another one, so the smallest element is at least $ 2^k$. Let $ \upsilon_p(x)$ be the $ p-$adic valuation of $ x$, so naturally there exist a bijective function from $ \{\frac{a_i}{\upsilon_2(a_i)}\}$ to $ \{2i-1\}$, where $ 1 \le i \le 10^4$. Consider the numbers s.t. $ \upsilon_p(a_i)=0, \forall p>3$: if $ \upsilon_2(a_i) \le \upsilon_2(a_j)$ and $ \upsilon_3(a_i) \le \upsilon_3(a_j)$ we have contradiction: it follows that $ \upsilon_2(a_i) + \upsilon_3(a_i) \ge k$, so the only number $ a_h$ with $ \frac{a_h}{2^{\upsilon_2(a_h)}}=1$ has the property that $ \upsilon_2(a_h) \ge k$. Now it is enough to prove that $ h =1$, but it is quite easy considering the number in the form $ 2^a3^bp$ and using similar tecnique as above. (Some steps on problem 2) If $ x$ and $ y$ are $ 0$ then $ z=1$. If $ x=0$ and $ y>0$ $ 1+7^y=2^z$ If $ y$ is even then $ 2^z\equiv2\bmod{4}$ so $ z=1$ and we obtain the previous solution. If $ y$ is odd then $ 7^{y}+1 \equiv8\bmod{16}$ so $ z=3$ and $ y=1$. If $ y=0$ and $ x>0$ Looking $ mod 4$ we find no new solutions. Now let $ x$,$ y$,$ z$ be positive. We can check some small values of $ z$ and it will not be too difficult to see that $ z>4$ By looking $ mod3$ you easily obtain that $ x$ is even and $ z$ is odd. By looking $ mod4$ we get that $ y$ is odd. Then looking $ mod 16$ we find that $ x=4a+2$ for some $ a$ nonnegative integer. We could now the following: $ 5^{4a+2} + 7^{2b+1}=2^{2c+1}$, $ 5^{4a+2}=2^{2c+1}-7^{2c+1}+ 7^{2c+1}-7^{2b+1}$. The left side is multiple of $ 25$ at least. $ 2^{2c+1}-7^{2c+1}$ is a multiple of $ 5$ so $ 7^{2c+1}-7^{2b+1}=7^{2b+1}(49^{c-b}-1)$ is also a multiple of $ 5$. We find that $ c-b$ But then it is easy to see that $ 7^{2c+1}-7^{2b+1}$ is a multiple of $ 25$ so back into $ 5^{4a+2}=2^{2c+1}-7^{2c+1}+ 7^{2c+1}-7^{2b+1}$. We obtain that $ 2^{2c+1}-7^{2c+1}$ is a multiple of $ 25$ so by the lifting the exponent lemma we have that $ 2c+1=5(2d+1)$ ($ 2c+1$ is a multiple of $ 5$). The equation turns into: $ 25^{2a+1}+7^{2b+1}=32^{2d+1}$. If $ d=0$ we find the solution $ a=0$ $ b=0$ which is plugging back into the original $ (2,1,5)$. We have to prove that $ 5^{2(2a+1)}+7^{2b+1}=2^{5(2d+1)}$ has no positive solutions. Let $ j$ the number s.t. $ j^2-2=0$ and begin to work in $ \mathbb{Z}[j]$. the equation becomes $ (5^{2a+1}-j^{5(2d+1)})(5^{2a+1}+j^{5(2d+1)})=-7^{2b+1}$. We know that $ \mathbb{Z}[j]$ is a euclidean ring, so the unique factorization theorem works (unless the order of primes and adding units). Lemma. Let $ x \in k$ ring. If $ N(x)$ is a prime integer, so $ x$ is prime in $ K$, (easily by contradiction). Since $ 7=(3+j)(3+j)$ and $ N(3+j)=N(3-j)=7$, so we can factorize RHS as $ -(3+j)^{2b+1}(3-j)^{2b+1}$. Now let $ p=3+j$ prime in $ \mathbb{Z}[j]$ and $ a=5^{2a+1}-j^{5(2d+1)}$. The original equation is $ a \overline{a}= - (p \overline{p})^{2b+1}$. $ a$ cannot be a unit, indeed otherwise $ N(a\overline{a})=N(a)N(\overline{a})=1$,absurd. We can note that $ gcd(p,\overline{p})=gcd(a,\overline{a})=1$. So, unless adding units, $ a=p^{2b+1}$ or $ a=(\overline{p})^{2b+1}$. So, non radical parts needs to be equal: we obtain: $ 5^{2a+1}= \frac{(3+j)^{2b+1}+(3-j)^{2b+1}}{2}$. But it is know that $ x+y|x^{2n+1}+y^{2n+1}$ so it follows that $ 3 | 5^{2a+1}$, contradiction. (but it is wrong because in $ \mathbb{Z}[j]$ every element has infinite associate others to it! )

When do we have this next Oliforum Contest? Please tell us.

Someone please reply. Will this contest be held next year too? {That is this September}.

@Agr_94, yes it can be however, for all contestant, on 19/03 i'll post another problem for substituting that one of number theory.. i'll post at the usual hour (12:00 at rome meridian!) and you'll have 12 hours to complete and send me the solution see you for the last problem bye

Waiting for the problem to be posted.

Problem. Let $ S$ be a set of $ 23 \cdot 26$ distinct positive integers that are not greater than $ 2323$. Show that there exist $ a_1 < a_2 < a_3 < ... < a_9$ element of $ S$ such that the system: $ a_1x + a_2y + a_3z = 0$ $ a_4x + a_5y + a_6z = 0$ $ a_7x + a_8y + a_9z = 0$ has a solution in nonzero integers. please send me a solution until 24 hours from now (that is 12.00 at Rome meridian) Edit: please not that i have changed the bound from $ 23 \cdot 23$ to $ 23 \cdot 26$