The problem is the following  Find all positive integers $n$ such that $3^{n-1} + 5^{n-1}$ divides $3^{n} + 5^{n}$.  And I attempted as the following: As  $3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n}$, $$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 5(3^{n-1} + 5^{n-1}) = -2\cdot 3 ^ {n-1}$$ And $$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 3(3^{n-1} + 5^{n-1}) = 2\cdot 5 ^ {n-1}$$ These expressions say that $$2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  p \in \mathbb{Z}^{+}$$ And $$2\cdot 5 ^ {n-1} = q(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  q \in \mathbb{Z}^{+}$$ By addition, it becomes clear that $q = 2 - p$. Hence, the last expression is equivalent to $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$. Solving for $p$, it is equal to $\frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$. Here, $p$ is odd only for $n = 1$, hence it is the only solution. Is this attempt a solution for the problem? And for the effort, Thanks!
120494
For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? I only got that $3^n+5^n=8k$ when $n$ is odd. How to solve this one?
87594
How to list all integer solutions of   $m^2-n^5 = m - n$  Here $m$ and $n$ are some positive integers.  Also, I want to know the name of this type equations (if name exist).   Regards Rosy
3368
Prove by mathematical  induction that:   $$\forall n \in \mathbb{N}: 3^{n} &gt; n^{3}$$  Step 1: Show that the statement is true for $n = 1$:  $$3^{1} &gt; 1^{3} \Rightarrow 3 &gt; 1$$  Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$  The general idea I had was to start with $(p+1)^{3}$ and during the process substitute in $3^{p}$ for $p^{3}$ as an inequality.  $$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 &lt; 3^{p} + 3p^{2} + 3p + 1$$  Now, if it can be shown that:  $$\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}$$  ...the proof is complete. This is because $3^{p+1} = 3 \cdot 3^{p}$ and one of those three have already been used.   We do this by mathematical induction. First, the base case of $n = 1$:  $$3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}$$  ..which turns out to be false.  What are some more productive approaches to this step?
29626
How do I prove that among any $5$ integers, you are able to find $3$ such that their sum is divisible by $3?$ I realize that this is a number theory question and we use modular arithmetic, but I'm unsure of where to begin with this specific situation.
122886
I think i know how to solve it but is that the best way? Is there a better way (using number theory). What i do is: knowing that  1st power last digit: 3 2nd power last digit: 9 3rd power last digit: 7 4rh power last digit: 1 5th power last digit: 3    $3^{347} = 3^{5\cdot69+2} = (3^5)^{69} \cdot3^2 = 3\cdot3^2=3^3=27 $ so the result is $7$.
102109
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem.
44709
I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:   Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.   And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
35496
Let N denote the number, and suppose $N=a^p \times b^q \times c^r \cdots$, where $a,b,c,\cdots$, are different prime numbers and $p,q,r,\cdots$ are positive integers.Then it is clear that each term of the product   $$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots$$  is a divisor of the given number,and that no other number is a divisor; then how could we use this to show/find that the number of ways in which this $N$ can be resolved into two factors which are prime to each other is $2^{n-1}$,where $n$ is the number of different prime factors in $N$?
70311
I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra  I know that $$a^n\equiv 1\pmod {a^n-1}$$  How do I proceed from there?
118073
let $x,y$ is positive integer,and such $$x^3-5x+10=2^y$$ find all $x,y$.  since $$x=1\Longrightarrow 1^3-5+10=6$$ can't $$x=2,2^3-5\cdot 2+10=8=2^3$$ so $x=2,y=3$  $$x=3,LHS=27-15+10=22$$ $$x=4,LHS=64-20+10=54$$ $$x=5,LHS=125-25+10=110$$ $$x=6,LHS=216-30+10=236$$ $$\cdots$$  I find $$(x,y)=(2,3)$$  I only find $x\le 7$ this solution.  maybe this have other solution.and This problem is from Mathematical olympiad problems Thank you
47264
Let $a$ and $b$ be positive integers such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5, \cdots $ Prove that $a = b$.  My way is as follows: Let $A=v_p(a), B=v_p(b)$ be the exact power of a prime $p$ dividing $a$ and $b$ respectively. Then by assumption, $(2n-1)A \leq 2nB, n=1,2,3,\cdots.$ Hence $A \leq (2n/(2n-1))B$. Taking limit as $n$ tends to infinite yields $A \leq B.$ Likewise, $2nB \leq (2n+1)A, n=1,2,3,\cdots,$ by a similar argument, we have $B \leq A.$ So that $A=B$ and we are done. But I am rather unsatisfied with the proof because it involves limit. I wanna ask if there is any other alternative proof, thank in advances.
68551
Use mathematical induction to show that:  $1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.  I am able to solve this problem, but the thing that confuses me on this is the Basis Step.  P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
23670
Prove that for all integers $n$, $n \geq 1$,   $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$   How would I go about proving this?
91210
Given a non-negative integer $p$. What is the number of solutions of $x_1+2x_2+3x_3 + \cdots + nx_n = p$, where the $x_i$'s are non-negative integers.   Can we answer this by using number of solutions of $x_1+x_2+x_3 + \cdots + x_m = q$ for any $m,q$?
52105
For any positive integer $n$ (using integer division only), let $P(n)$ denote the number of ways in which the integer $(n^2+n)/4$ can be expressed as a sum of exactly $n /2$ distinct elements of the set $\{1,2,3,\dots, n\}$.    What is $P(n)$ in terms of n?  Specifically, how exponential is it? Is this less than $2^{n/2}$?
69085
Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}.$   I'm currently trying to finish a task that requires me to use mathematical induction to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$. This is the first time that the divider is not a simple integer and I'm having trouble trying to prove it.   This is what I managed to do:     First step:      For $n=1$:      $2^3+1=9$      $3^2=9$      $L=P$      Second step: I assume that for some numbers $n&gt;=1$  $2^{3^n}+1$ is divisible by $3^{n+1}$      Third step: (induction hypothesis):   $2^{3^{n+1}}+1$ is divisible by $3^{n+1}$   I'm getting stuck here, trying to do some algebra and simplify the dividend:     $2^{3^{n+1}}+1 = 2^{3^n*3}+1$
24340
Show that there are infinitely many positive integers $N$ that cannot be written in the form $a^n+b^n+c^n$ where $a,b,c$ are positive integers and $n \geq 2$.   Note that in the question above, $n$ can vary and is not fixed.  I thought about using a modular arithmetic argument by finding a set of values which are not expressible when the exponent is $3n,3n+2,$ or $3n+4$ and then use the Chinese Remainder Theorem.  Note that for all expressions of the form $$a^{3n}+b^{3n}+c^{3n},$$ the numbers which are $4,5,6$ modulo $9$ are not expressible in these forms since each of the terms is $-1,0,1 \pmod{9}$. For expressions of the form $$a^{3n+2}+b^{3n+2}+c^{3n+2} = a^{3n} \cdot a^2+b^{3n} \cdot b^2+c^{3n} \cdot c^2,$$ note that $x^2 \equiv -1,0,1,4 \pmod{9}$ and that $x^3 \equiv -1,0,1 \pmod{9}$. But I couldn't find a set of numbers modulo $9$ which weren't expressible in this form.   For expressions of the form $$a^{3n+4}+b^{3n+4}+c^{3n+4},$$ note that $x^4 \equiv 0,1,3,9 \pmod{13}$ and $x^{3n} \equiv -1,0,1,8 \pmod{13}$. Therefore, $x^{3n+4} \equiv 0,1,3,7,8,9,10,11,12$, but I couldn't find a set of numbers modulo $13$ which weren't expressible in this form.  Can we use a modular arithmetic argument similar to this to solve the cases for $3n+2$ and $3n+4$ one or do we need to use a higher exponent such as $6n$ instead of $3n$?