Suppose that n≥2 and a1,a2,...,an are natural numbers that (a1,a2,...,an)=1. Find all strictly increasing function f:Z→R that: ∀x1,x2,...,xn∈Z:f(n∑i=1xiai)=n∑i=1f(xiai) Proposed by Navid Safaei and Ali Mirzaei
Problem
Source: Iran TST 2023 ; Exam 1 Problem 5
Tags: function, algebra
21.03.2023 14:59
Is this the only solution? Write (a2,...,an)=d,a1=d′ where (d,d′)=1. Take positive reals 0=c0<c1<c2<...<cdd′. f(x)=⌊xdd′⌋cdd′+c{xdd′}dd′ The hard part is to prove for n=2. It is obvious that f(0)=0, thus putting x1=0 gives us f(n∑i=2aixi)=n∑i=2f(aixi)⇒f(n∑i=1aixi)=n∑i=1f(aixi)=f(a1x1)+f(n∑i=2aixi) By Bezout, we have f(a1x1+dx)=f(a1x1)+f(dx) where (a2,...,an)=d and ∀x∈Z. It is also obvious that (a1,d)=1, otherwise we would have a prime dividing all ais, contradiction to the given condition. Take a1=d′ and we have f(dx+d′x′)=f(dx)+f(d′x′) for coprime d,d′ and all x,x′. Write x=dd′t+r and x′=dd′t′+r′; t,t′,r,r′ are yet to be determined. (We are not using Euclidean Division here, be careful) f(d2d′t+dr+dd′2+d′r′)=f(d2d′t+dr)+f(dd′2t′+d′r′)=f(d2d′t)+f(dr)+f(dd′2t′)+f(d′r′)=f(dd′(dt+d′t′))+f(dr+d′r′) Take y=dd′m+s where 0≤s<d′, and choose t,t′,r,r′ such that dt+d′t′=m and dr+d′r′=s by Bezout as d,d′ are coprime. We clearly have f(dd′m+s)=f(dd′m)+f(s). It is obvious that if we construct 0=c0=f(0),c1=f(1),...,cdd′=f(dd′) as we want, in increasing order, we can use the last result to finish for every integer that \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ x \pmod{dd'}}
23.03.2023 14:28
electrovector wrote: Is this the only solution? Write (a_2, ..., a_n)=d, a_1=d' where (d, d')=1. Take positive reals 0=c_0<c_1<c_2<...<c_{dd'}. f(x)= \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ \left\{ \frac{x}{dd'} \right\} dd'} I Agree with the rest but does this categorise all the functions? There are some conditions on the c_i and the f(i), you define the c_i's in terms of f(i)'s but we don't know what the f(i)'s really are. For example if the problem were to ask us which numbers satisfied x^{p-1} \equiv 1 \pmod{p} this would be like 'check all numbers between 1 and p and the numbers satisfying are whose remainders by p lie in that set'.
26.10.2023 12:37
Bump ! Can someone post a complete solution ? Not all of the solutions mentioned by electrovector verify because to verify the condition f(dx+d'x')=f(dx)+f(d'x') the c_i must verify that 1) c_{db+d'b'}=c_{db}+c_{d'b'} whenever db+d'b'<dd' 2)c_{dd'}+c_{db+d'b-dd'}=c_{db}+c_{d'b'} whenever db+d'b'\ge dd' and this are the answers to the case n=2 but i don't if you can reduce it even more to a more accurate answer and i also don't know how to solve the general case to verify (for example when d=2,d'=3, f(0)=0;f(1)=a;f(2)=b;f(3)=c;f(4)=d;f(5)=c+b;f(6)=c+d-a for every 0<a<b<c and d \in (max \left\{c,a+b \right\},b+c) so there are "weird" solutions.
30.10.2023 09:59
No one ? It is a good problem.
15.12.2024 22:45
Official Solution?