MathDB

3

Part of 2001 Turkey Team Selection Test

Problems(2)

S^{n+k}(a,b,c) = S^{n}(a,b,c) (mod abc)

Source: Turkey TST 2001 - P3

4/5/2013
For all integers x,y,zx,y,z, let S(x,y,z)=(xy−xz,yz−yx,zx−zy).S(x,y,z) = (xy - xz, yz-yx, zx - zy). Prove that for all integers aa, bb and cc with abc>1abc>1, and for every integer n≥n0n\geq n_0, there exists integers n0n_0 and kk with 0<k≤abc0<k\leq abc such that Sn+k(a,b,c)≡Sn(a,b,c)(modabc).S^{n+k}(a,b,c) \equiv S^n(a,b,c) \pmod {abc}. (S1=SS^1 = S and for every integer m≥1m\geq 1, Sm+1=S∘Sm.S^{m+1} = S \circ S^m. (u1,u2,u3)≡(v1,v2,v3)(modM)⟺ui≡vi(modM)(i=1,2,3).(u_1, u_2, u_3) \equiv (v_1, v_2, v_3) \pmod M \Longleftrightarrow u_i \equiv v_i \pmod M (i=1,2,3).)
modular arithmeticparameterizationnumber theory proposednumber theory
Continuous f(x-f(x))=x/2 does not exist

Source: Turkey TST 2001 - P6

4/4/2013
Show that there is no continuous function f:R→Rf:\mathbb{R}\rightarrow \mathbb{R} such that for every real number xx f(x−f(x))=x2.f(x-f(x)) = \dfrac x2.
functionalgebra proposedalgebra