For example,$ f (2) = 2$, and $f(3) = 6$.   I don't know what kind of technique I should use.   By putting $n = 1, 2, 3, \cdots$ it's even hard to know if $f(2) = 2$ for ALL integer $n$.
123401
Prove that $2730$ divides $n^{13} - n$ for all integers $n$.   What I attempted is breaking $2730$ into $2, 3, 5$, and $7, 13$. Thus if I prove each prime factor divides by $n^{13} - n$ for all integers $n$, it will prove the original problem.   I already proved $2, 3$, and $5$ work. Struggling with how to do $91$?  Other solutions are welcome, but they should be associated with mod and solutions of this kind, because my answer should pertain to what was learned in class.  (Edit: Fixed the 91 problem, being slow today...In any case, I've got this problem now, so no answers are necessary. Posting my own soon.)
14402
Let $x=0\,\, f(x+y)=f(0+y)$ every $y \in \mathbb{Z}$ So we get $f(0)+f(y)=f(y)$ then $f(0)=0$  $0=f(0)=f(x(-x))=f(x)+f(-x)=0$ Let $f(1)=k$ for some $k \in \mathbb{Z}$ $f(n)=f(1+1+1+...+1) (n \text{ many } 1)$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=f(1)+f(1)+...+f(1) (n \text{ many } f(1))$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=n\cdot k$  Thus $f: \mathbb{Z} \to \mathbb{Z}$ a function such that $f(n)=nk$ I found this but I must also show that there is no other function which satisfies $f(x+y)=f(x)+f(y)$ and I am stuck can somebody help me.
87297
Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?  Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.  Answer is $n(n+1)+an+b$
68529
Problems:Determine the number of functions $$f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$$ satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd.   My Attempt: Each integer in domain has $4$ choices and therefore the total number of functions is $f$ is $4^{1999}.$ Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be $2*4^{1998}.$ I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct?  Edit $1$: I tried to proceed further in the following manner:   For a given mapping $f$ let the number of integers assigned to $2000,2001,2002$ amd $2003$ be $p,q,r$ and $s$ respectively, where $p,q,r,s\in \mathbb{Z}^+$. Therefore the sum in question can be written as $f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.$ Clearly the sum will be odd iff $q+s$  is odd. Also note that $p+q+r+s=1999\Rightarrow p+r$ is even. On the other hand if we let $q+s$ as even then $p+r$ must be even. Now the number of values of $p,q,r$ and $s$ for which the above two lemmas hold is the same and therefore if we let $f$ to be the function which maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $q+s$ is even and $g$ to be the function  hich maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $p+q$ is even we will get a bijection.(Maybe!)
8071
I have tried to solve it for the case when $n=1,2,3,4$ which i found that there is only one group for $n=1$, one group for $n=2,$ one group for$ n=3 $ and two groups for $ n=4$, then my question come, if i were given a number $n$ which is larger, how do i know how many different groups are there?  (I treat two isomorphic groups as the same group in this question.)
105867
A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
4484
What's the smallest positive integer which is divisible by all integers $1, 2, \dots, n$?  Is there a simple way to represent the answer?  Call it $f(n)$ here.   Clearly factorial ($n!$) satisfies the condition of divisibility.  But once you get to $f(4)$ (which is $12$), factorial ($24$) is too large, because it includes as factors both $4$ and $2$, which are redundant.  Multiplying all the prime numbers up to $n$ gives you another estimate, which is this time too low, because you need to include primes multiple times when they're repeated in the factorization of the inputs.  So the answer is somewhere between factorial, and the product of primes.  Is there a simple answer?
123714
Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
95028
What can be said about$\ f^{-1}(f(S))$ and$\ S$, where$\ f(S):=\{f(x):x \in S\}$?   I feel as if$\ f^{-1}(f(S))$ will be smaller than or equal in size (number of elements) to$\ S$.  For example, if$\ f(x)=x$, then$\ f(\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$and$\ f^{-1}(f\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$ (same size).  But if$\ f(x)=x^2$, then$\ f(\{-1, 0, 1, 2\})=\{0, 1, 4\}$ and$\ f^{-1}(f\{0, 1, 4\})=\{0, 1, 2\}$, which is a little bit smaller than$\ S$.   What can be said about$\ f(f^{-1}(U))$ and$\ U$, where$\ f^{-1}(U):=\{x \in X:f(x) \in U\}$?   The same size as each other? Always??
36333
For the recurrence relation:  $f(0)=1$  $f(1)=1$  $f(2)=2$  $f(2n)=f(n)+f(n+1)+n\ \ \ (\forall n&gt;1)$  $f(2n+1)=f(n)+f(n-1)+1\ \ \ (\forall n\ge 1)$  (the first numbers of the sequence are: 1, 1, 2, 3, 7, 4, 13, 6, 15, 11, 22, 12, 25, 18, 28)  Given some number $S$, I would like to find $n$ such that $f(n) = S$, using an approach that runs in $O(\log n)$ time. Any ideas or tips how to solve it? I tried substitution so far but didn't get far.
58309
I am asked to find all positive integers $n$ such that $\phi(n)=6$, and to prove that I have found all solutions.  The way I am tackling this is by constructing all combinations of prime powers such that when passed to Euler's function will yield $6$.  For example, I know that $\phi(7)=6$, $\phi(3^2)=6$, and $\phi(2)=1$. Therefore, the numbers should be $7$, $7\cdot2=14$, $3^2=9$, and $3^2\cdot2=18$.  I believe that there cannot possibly be any others because of the way the $\phi$ function is defined. What do you guys think?
8402
I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.  My attempt at the program was as follows:   A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.   LONG and TEDIOUS way: $$(8k)^2 = 0 (mod 8)$$ $$(8k+1)^2 = 1 (mod 8)$$ $$(8k+2)^2 = 4 (mod 8)$$ $$(8k+3)^2 = 1 (mod 8)$$ $$(8k+4)^2 = 0 (mod 8)$$ $$(8k+5)^2 = 1 (mod 8)$$ $$(8k+6)^2 = 4 (mod 8)$$ $$(8k+7)^2 = 1 (mod 8)$$  That is, using three of these modulo there is no way to arrive at  $$a^2 + b^2 + c^2 = 7 (mod 8)$$
10141
Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
35358
Say we have two integers $n$ and $k$, such that $0 &lt; k &lt; n$ and $gcd(n, k) = 1$.  Why is it that if we take every integer $x\in\{0,\dots, n-1\}$ and compute $k\cdot x \ (mod \ n)$, we get as remainders all numbers from $0$ to $n-1$, with no repeats?  In other words, why is $\ f:\{0,\dots,n-1\}\rightarrow\{0,\dots,n-1\}$, $\ f(x)=k\cdot x\ (mod\ n)$ a bijection for $k,n\in\mathbb{N},\ gcd(k,n)=1$?
123555
An arithmetic function is defined as follows:$f(1)=1$,  $f(2k)=k$ and $f(2k+1)=f(k)+f(k+1)$. When (for which $k$) is $f(k)$ even?  While it is obvious that $f(4n-1)=f(4n)=2n$, therefore $f(k)$ is even for $k=4n$ and $k=4n-1$, which is possible to prove using induction (isn't it?), I don't know how to prove the converse.  I tried to solve the recurrence or make use of constant difference between certain elements of the sequence $\ldots$ What am I missing?  Thank you.
67789
I've been working with the Fermat numbers recently, but this problem has really tripped me up. If the Fermat theorem is set as $f_a=2^{2^a}+1$, then how can we say that for an integer $b&lt;a$, the $\gcd(f_b,f_a)=1$?
22403
Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. Then it's very easy to see that $f(2)=2$ and since $f(2)=2$, we can see that $f(4)=4$ and $f(5)=5$ because $5=2^2+1^2$. To see $f(3)=3$, note that $$25=f(5^2)=f(3)^2+f(4)^2 \implies f(3)=3$$ One can even see that $f(6)=6$, $f(7)=7$ and so on. Is it generally true that $f(n)=n$ for all $n$?  Edit. If anyone wondering how $f(7)=7$ here is the way. Note that $f$ satisfies $f(n^2)=f(n)^2$. One can see that $$25^{2}=24^2+7^2 \implies 25^{2}=f(24)^{2}+f(7)^{2}$$ We have to show $f(24)=24$. For this note that $$26^2 = 24^2+10^2 \implies f(26)^{2} = f(24)^{2}+f(10)^{2}$$ But $f(26)=26$ because $26=5^2+1^2$ and $f(10)=10$ since $10=3^{2}+1^{2}$. In short if $n=x^2+y^2$ for some $x,y$ and $f(1)=1$, then we can see that $f(n)=n$.