MathDB

3

Part of 2014 Iran Team Selection Test

Problems(3)

question 3

Source: iran tst 2014 second exam

5/22/2014
prove for all k>1k> 1 equation (x+1)(x+2)...(x+k)=y2(x+1)(x+2)...(x+k)=y^{2} has finite solutions.
algebrapolynomialnumber theory proposednumber theory
question 3

Source: iran tst 2014 first exam

4/13/2014
we named a n∗nn*n table selfishselfish if we number the row and column with 0,1,2,3,...,n−10,1,2,3,...,n-1.(from left to right an from up to down) for every {i,j∈0,1,2,...,n−1 i,j\in{0,1,2,...,n-1}} the number of cell (i,j)(i,j) is equal to the number of number ii in the row jj. for example we have such table for n=5n=5 1 0 3 3 4 1 3 2 1 1 0 1 0 1 0 2 1 0 0 0 1 0 0 0 0 prove that for n>5n>5 there is no selfishselfish table
LaTeXcombinatorics unsolvedcombinatorics
hard question 3

Source: iran tst 2014 third exam

5/22/2014
let m,n∈Nm,n\in \mathbb{N} and p(x),q(x),h(x)p(x),q(x),h(x) are polynomials with real Coefficients such that p(x)p(x) is Descending. and for all x∈Rx\in \mathbb{R} p(q(nx+m)+h(x))=n(q(p(x))+h(x))+mp(q(nx+m)+h(x))=n(q(p(x))+h(x))+m . prove that dont exist function f:R→Rf:\mathbb{R}\rightarrow \mathbb{R} such that for all x∈Rx\in \mathbb{R} f(q(p(x))+h(x))=f(x)2+1f(q(p(x))+h(x))=f(x)^{2}+1
algebrapolynomialfunctionalgebra unsolved