Problem Let $f : \mathbb{N}\to\mathbb{N}$ be a strictly increasing function such that $f(2) = 2$ and $f(mn) = f(m) \cdot f(n)$ for every relatively prime pair of positive integers $m$ and $n$. Prove that $f(n) = n$ for every positive integer $n$. My approach I tried to prove this using induction. For the base case, we need to find $f(3)$. We have $$f(3) \cdot f(5)&lt;2f(9)&lt;4f(5)$$ $$\implies f(3)&lt;4$$ And since $f(3)&gt;f(2)=2$, we have $f(3)=3$. Now, we assume that $f(k-1)=k-1$ and $f(k)=k$. Since $k$ and $k-1$ are relatively prime, we have $f(k(k-1))=f(k)f(k-1)=k(k-1)$. But this does not prove.  So, how do I complete the proof?
6905
Let $(a_n)$ be a strictly increasing sequence of positive integers such   that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.   This is a result apparently due to Paul Erdős, and  has a proof by induction.  I tried like this, $a_{10}=a_2a_5$. After this what we can do?    [Editor's comment] Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22&lt;a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$. [/comment, JL]
37471
We know that $f(x)=x^4+1$ is a .  My question is:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ has a linear factor over $\mathbb{Q}$?   Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?   Thanks in advance!
39523
If $A=\left\{1,2,3,4,5\right\}$ Then     Find Number of functions $f : A \to A$ such that $f(f(x))=f(x)$   Case $1.$ if $f$ is injective then $f(f(x))=f(x)$ $\implies$ $f(x)=x$, hence there is only one injective function which is an identity function.  Case $2.$ When $f$ is many to one function  Let us assume that $f(1)=f(2)=f(3)=f(4)=f(5)=k$ Then clearly for any $k$ from the set $A$, $f(f(x))=f(x)$  hence there are five such functions.  Are there any other possibilities?
97084
If $f: \mathbb N \mapsto \mathbb N$ is one-to-one and $f(mn) = f(m)f(n)$, what is the smallest possible value of $f(999)$?  Easily $f(1)=1$, and I think $f(n)=n$ must be the only map, but not able to prove it. The other alternative is if $f(999) = f(3)^3f(37) \ge 24$, which looks somehow odd.
123135
Let $n \in \mathbb{N}$ and $p$ be a prime number. Let $f(x),g(x) \in \mathbb{Z}_{p^n}[x]$ such that, in $\mathbb{Z}_{p^n}[x]$, $f(x)g(x)=0$. Prove that $a_ib_j=0$ for all $i,j$, where $a_i$ and $b_j$ are the coefficients of $f(x)$ and $g(x)$.  I have the set of equations (for degree $r$ and degree $s$): $$a_rb_s=0, a_rb_{s-1}+a_{r-1}b_s=0,...,a_0b_0=0.$$ From these, how do I conclude $a_ib_j=0$ for all $i,j$? Also, I found that for $n=1$, the polynomial ring is an integral domain, so the conclusion holds. I had an idea to use induction on $n$, but do not know how to proceed.
84180
Let $f:[0,\infty)\rightarrow [0,\infty) $ be a continuous strictly increasing function and $g=f^{-1}$. Let $a,b&gt;0$ and $a\neq b$. Then how $\displaystyle\int_{0}^af(x)dx+\int_0^bg(y)dy\geq ab.$ Any hint will be appreciated.
83266
For any positive integer $n$, let $\mathrm{PF}(n)$ denote the set of all distinct prime factors of $n$ (not counted with multiplicity).  The question (using the above notation) is: Is it true that for any two positive integers $m$ and $n$, $\phi(mn)=m\phi(n)$ if and only if $\mathrm{PF}(m) \subseteq \mathrm{PF}(n)$, where $\phi$ is the totient function?  Hint for the if direction: Any prime factor of $mn$ must either be a prime factor of $m$ or a prime factor of $n$ (by Euclid's lemma). But since any prime dividing $m$ also divides $n$, $mn$ has the same distinct prime factors as $n$, and hence any integer coprime to one of them is also coprime to the other.  Note that it is possible for $\mathrm{PF}(m) \subseteq \mathrm{PF}(n)$ to be true while $m \not\mid n$. For example, $4 \not\mid 2$, but they both have $2$ as the only prime factor. In this case, $\phi(8)=4=4 \cdot 1=4\phi(2)$.
36912
Theorem. Let $f\colon I \to J$, where $I$ is an interval and $J$ is the image $f(I)$, be a function such that:   $f$ is strictly increasing on $I$. $f$ is continuous on $I$.   Then $J$ is an interval, and $f$ has an inverse function $f^{-1}\colon J\to I$ such that:   $f^{-1}$ is strictly increasing on $J$. $f^{-1}$ is continuous on $J$.   I need to prove only that $f^{-1}$ is continuous on $J$.  Proof. Let $y\in J$, and (for simplicity) assume that $y$ is not an end-point of $J$. Then $y=f(x)$, for some $x\in I$, and we want to prove that $$y_n\to y\implies f^{-1}(y_n)\to f^{-1}(y)=x.$$ Thus we want to deduce that for each $\varepsilon&gt;0$, there is some number $X$ such that $$x-\varepsilon&lt;f^{-1}(y_n)&lt;x+\varepsilon\qquad \forall n&gt;X\qquad(1).$$ Since $f$ is strictly increasing, we know that $$f(x-\varepsilon)&lt;f(x)&lt;f(x+\varepsilon).$$ also, since $y_n\to y=f(x)$, there is some number $X$ such that $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ Then, applying the strictly increasing function $f^{-1}$ to these inequalities, we obtain (1).  Edit  I don't understand why, since $y_n\to y=f(x)$, there is some number $X$ such that  $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ I would have written the last line as follows: $$f(x)-\varepsilon&lt;y_n&lt;f(x)+\varepsilon, \qquad\forall n&gt;X.$$ Applying the $f^{-1}$ to my last line, which I think it's correct, I obtain $$f^{-1}[f(x)-\varepsilon]&lt;f^{-1}[y_n]&lt;f^{-1}[f(x)+\varepsilon], \qquad\forall n&gt;X.$$ which is not (1).  What am I misunderstanding? Why the author of this proof wrote that last line and not my version?  Thank you.
111869
The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out?
97943
Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?  The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?
68776
Possible Duplicate:             For positive integers $m$ and $n$ where $d=gcd(m,n)$, show that $$\phi(mn) = \phi(m)\phi(n)\frac{d}{\phi(d)}$$   This is just the generalization of the multiplicative property of phi function.I have tried to solve this in the same way as the proof of multiplicative property but couldn't get far.Please help.
51194
Let $f$ be a function from $X=\{1,2,3,...,n\}$ to itself. We say $f$ is increasing if $a\le b$ then $f(a)\le f(b)$. How do we find the number of increasing functions?    I think if we can define $f(1)$ then we can count. But it is very difficult.
92798
So the statement is:  Prove: $$F^2_{n+1} - F^2_{n-1} = F_{2n}$$  How could this be proven with induction?  What about without induction?  For induction I got here: (already tested $n=1$, and assumed $n=k$ held true)  $$F^2_{k+2} - F^2_{k} = F_{2k + 2}$$ $$(F_{k+1} + F_{k})^2 - (F_{k-1} + F_{k-2})^2 = F_{2k+1} + F_{2k}$$ Then you get some cancelation based off the $n=k$ case $$2F_{k+1}F_k + F_k^2 - 2F_{k-1}F_{k-2} - F_{k-2}^2 = F_{2k+1}$$  This is where I am stuck for the induction proof.  I have no idea where to begin with the non-induction proof.
51245
Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)?
120311
This proof seems odd to me. I have come to the conclusion that I will use induction. I would like to see a smoother way or just some improvements on my technique.     Let $f:A\rightarrow B$ be a function between two finite sets of equal cardinality. Show that $f$ is surjective if and only if it  is injective.   To start, I will show that a surjection implies an injection using induction. I will dismiss the cases that both sets are empty or contain one element as being trivial (essentially vacuously true).   Assume $|A| \geq 2$, $|B| \geq 2$, $|A| = n = |B|$, and $f:A \rightarrow B$ is a surjection. For the base case, let $n = 2$.   There are two elements in both $A$ and $B$. Due to surjection, every element $b \in B$ must be mapped to, through $f$, by at least one element $a \in A$. If each of the two elements in $B$ were mapped to by the same element in $A$, the definition of function would be violated. Therefore, they are mapped to by unique elements in $A$. Thus, for $f(p), f(q) \in B$, if $f(p) = f(q)$, it must be true that $p = q$ so $f$ is injective.  Now assume that the surjection implies an injection for $n \geq 2$. We must show this to be true for $|A| = n + 1 = |B|$. Since it is true for $|A| = n = |B|$, the $n + 1$ case represents the addition of one new element to both $A$ and $B$. The new element in $B$ cannot be mapped to any other element in $A$ except for the new one. If mapped to by an old one, the definition of function would be violated. It must be mapped to by something since $f$ is surjective, hence it must be the new element. Finally, the new element in $A$ cannot be mapped to an old element in $B$ because it is unique and the previous $B$ was shown to be injective.  $$\blacksquare$$  This is a very wordy and awkward proof in my opinion. I have been out of proofs for a long time. I would like to see one that is more clear or seek validation if there isn't. I know that I have only completed half of the proof and have yet to go the other way.
77252
I have difficulty in neatly writing down a proof for the following from Terence Tao's Analysis I, where $m{+\!+}$ is the :     Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m{+\!+})$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m\leq n$; this is know as the principle of backwards induction. (Hint: apply induction to the variable $n$.)   First of all, I am unsure about what the base case should look like. For the induction step, I understand that if we suppose inductively that $P(n)$ is true, that then for a natural number $a$ s.t. $a{+\!+}=n$ it holds that $P(a)$ is true, and then for a natural number $b$ s.t. $b{+\!+}=a$ it holds that $P(b)$ is true etc. Hence for all natural numbers $m\leq n$, $P(m)$ is true.  Could anyone please tell me what the base case should look like, and whether there is a neater way of writing down the induction step?
45015
Let $f \in \mathbb{Z}[x]$ be non constant. How to prove that exists infinitely many primes such $f$ have root in $\mathbb{Z/_{(p)}}$. I spent much time, but with no benefits.