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Let S be the set of all pairs (m,n) of relatively prime positive integers m,n with n even and m < n. For s = (m,n) \in S write n = 2^k \cdot n_o where k, n_0 are positive integers with n_0 odd and define f(s) = (n_0, m + n - n_0). Prove that f is a function from S to S and that for each s = (m,n) \in S, there exists a positive integer t \leq \frac{m+n+1}{4} such that f^t(s) = s, where f^t(s) = \underbrace{ (f \circ f \circ \cdots \circ f) }_{t \text{ times}}(s).

If m+n is a prime number which does not divide 2^k - 1 for k = 1,2, \ldots, m+n-2, prove that the smallest value t which satisfies the above conditions is \left [\frac{m+n+1}{4} \right ] where \left[ x \right] denotes the greatest integer \leq x.

Slični zadaci

Prove that the set of positive integers cannot be partitioned into three nonempty subsets such that, for any two integers x,y taken from two different subsets, the number x^2-xy+y^2 belongs to the third subset.
The function F is defined on the set of nonnegative integers and takes nonnegative integer values satisfying the following conditions: for every n \geq 0,

(i) F(4n) = F(2n) + F(n),
(ii) F(4n + 2) = F(4n) + 1,
(iii) F(2n + 1) = F(2n) + 1.

Prove that for each positive integer m, the number of integers n with 0 \leq n < 2^m and F(4n) = F(3n) is F(2^{m + 1}).
Let P be a cubic polynomial given by P(x)=ax^3+bx^2+cx+d, where a,b,c,d are integers and a\ne0. Suppose that xP(x)=yP(y) for infinitely many pairs x,y of integers with x\ne y. Prove that the equation P(x)=0 has an integer root.
Find all functions f: \mathbb{N^{*}}\to \mathbb{N^{*}} satisfying
\left(f^{2}\left(m\right)+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}
for any two positive integers m and n.

Remark. The abbreviation \mathbb{N^{*}} stands for the set of all positive integers:
\mathbb{N^{*}}=\left\{1,2,3,...\right\}.
By f^{2}\left(m\right), we mean \left(f\left(m\right)\right)^{2} (and not f\left(f\left(m\right)\right)).
For an integer m, denote by t(m) the unique number in \{1, 2, 3\} such that m + t(m) is a multiple of 3. A function f: \mathbb{Z}\to\mathbb{Z} satisfies f( - 1) = 0, f(0) = 1, f(1) = - 1 and f\left(2^{n} + m\right) = f\left(2^n - t(m)\right) - f(m) for all integers m, n\ge 0 with 2^n > m. Prove that f(3p)\ge 0 holds for all integers p\ge 0.

Proposed by Gerhard Woeginger, Austria
Let f be a non-constant function from the set of positive integers into the set of positive integer, such that a-b divides f\!\left(a\right)-f\!\left(b\right) for all distinct positive integers a, b. Prove that there exist infinitely many primes p such that p divides f\!\left(c\right) for some positive integer c.

Proposed by Juhan Aru, Estonia