MathDB

4

Part of 2017 Iran Team Selection Test

Problems(3)

Number Theory from Iranian TST 2017

Source: Iranian TST 2017, first exam day 2, problem 4

4/6/2017
We arranged all the prime numbers in the ascending order: p1=2<p2<p3<⋯p_1=2<p_2<p_3<\cdots. Also assume that n1<n2<⋯n_1<n_2<\cdots is a sequence of positive integers that for all i=1,2,3,⋯i=1,2,3,\cdots the equation xni≡2(modpi)x^{n_i} \equiv 2 \pmod {p_i} has a solution for xx. Is there always a number xx that satisfies all the equations?
Proposed by Mahyar Sefidgaran , Yahya Motevasel
number theoryIranIranian TSTprime numbersmodular arithmetic
2017 Iran TST2 day2 p4

Source: 2017 Iran TST second exam day2 p4

4/24/2017
A n+1n+1-tuple (h1,h2,⋯ ,hn+1)\left(h_1,h_2, \cdots, h_{n+1}\right) where hi(x1,x2,⋯ ,xn)h_i\left(x_1,x_2, \cdots , x_n\right) are nn variable polynomials with real coefficients is called good if the following condition holds: For any nn functions f1,f2,⋯ ,fn:R→Rf_1,f_2, \cdots ,f_n : \mathbb R \to \mathbb R if for all 1≤i≤n+11 \le i \le n+1, Pi(x)=hi(f1(x),f2(x),⋯ ,fn(x))P_i(x)=h_i \left(f_1(x),f_2(x), \cdots, f_n(x) \right) is a polynomial with variable xx, then f1(x),f2(x),⋯ ,fn(x)f_1(x),f_2(x), \cdots, f_n(x) are polynomials.
a)a) Prove that for all positive integers nn, there exists a good n+1n+1-tuple (h1,h2,⋯ ,hn+1)\left(h_1,h_2, \cdots, h_{n+1}\right) such that the degree of all hih_i is more than 11.
b)b) Prove that there doesn't exist any integer n>1n>1 that for which there is a good n+1n+1-tuple (h1,h2,⋯ ,hn+1)\left(h_1,h_2, \cdots, h_{n+1}\right) such that all hih_i are symmetric polynomials.
Proposed by Alireza Shavali
algebrapolynomialIranIranian TSTfunction
Iran TST 2017 third exam

Source: 2017 Iran TST third exam day2 p4

4/27/2017
There are 66 points on the plane such that no three of them are collinear. It's known that between every 44 points of them, there exists a point that it's power with respect to the circle passing through the other three points is a constant value kk.(Power of a point in the interior of a circle has a negative value.) Prove that k=0k=0 and all 66 points lie on a circle.
Proposed by Morteza Saghafian[/I]
IranIranian TSTgeometrycombinatorics