Prove that for all $n$, $121\mid n^2+3n+5$.   I thought proving $n^2 +3n+5\pmod{121}\equiv 0$ had no solutions would be a good start, so I completed the square of $n^2+3n+5($, and arrived at $(n+62)^2 \pmod{121} \equiv 3718$. Replacing $n+62$ by $x$, I lastly arrived at $x^2 \equiv 88\pmod{121}$. Now, Im stuck. Any hints would be appreciated.
123320
Question:  Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.
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.)
41147
Problem :   Solve in $\mathbb N$ :   $8n+9\equiv 0\pmod{1163}$  The mathematics give me : $n=435+1163k$ , $k\in\mathbb N$   $1163$ is a prime number   $8n\equiv -9\pmod{1163}$   $8n\equiv 1154\pmod{1163}$   I don't know any ideas to complete my work ?   I have to see your hints
10938
Let $p&gt;3$ be a prime. Prove that $24 \mid p^2-1$.   I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.
111645
Prove by induction that $3\mid n^3 - n$.   I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:  Proof by induction:  Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;  Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.  Inductive Step:     $p(n+1) = (n+1)^3 - (n+1)$      $= n^3 + 3n^2 + 3n + 1 - n - 1$      $= n^3 + 3n^2 + 2n$      $= n^3 + 3n^2 + 2n - (n^3 - n)$      $= 3n^2 + 3n$   By induction, $p(n+1)$ is true if:      $3\mid n$ and $3\mid m$ and $3 | n - m$     to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.  My professor wanted:     $= (n^3 -n) + 3n^2 + (2n + n)$
11556
$n$ is a natural number. I’ve tried to turn this problem to this: Prove that: $n(n+1)(n+2)(n+3)+1 $is always a perfect square. So: $$n(n+1)(n+2)(n+3)+1 =x^{2}$$ And i’ve expanded the whole equation but it didn’t help me a lot.
3537
I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly.   $$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36$$
36054
Prove that if $n$ is a natural number and $(3n+1)$ &amp; $(4n+1)$ are both perfect squares, then $56$ will divide $n$.   Clearly we have to show that $7$ and $8$ both will divide $n$.  I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.  - so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.  - so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?  Thanks for the help.
4839
We can prove that there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:     Is it true that for every pair of primes $p&lt;q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p&lt;q$ and an integer $n&gt;p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n&gt;6$ then $2^n-1\nmid 5^n-1$?   Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &amp;3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &amp;7^{4}-1\mid 151^{4}-1. \end{array}  Now I prove there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1.$  Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.  If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again.
59103
Find n such that $n^5 - n$ is divisible by $120$. (*Here $n$ belongs to the natural number set)  Seems a trivial problem, but the answer mentioned in the text book seemed incomplete to me so I'm here just to ensure whether approach to this problem is correct or not.  So my approach is as follows: We know that $n^5 - n$ can be written as $n^5-n = n(n-1)(n+1)(n^2 + 1)$  Now since it's product of three consecutive natural numbers so it's divisible by $3! = 6$. Now using Fermat's Little theorem we can prove that $n^5= n \mod5$ hence $n^5 - n$ is divisible by $5$ as well. So $n^5 - n$ is divisible by $30$. Now to ensure their is another factor of $4$ in the term we need to ensure that $n^5 - n$ is divisible by $8$. I found that all odd natural numbers and multiples of $8$ makes $n^5 - n$ divisible by $8$, hence $n^5 - n$ is divisible by $120$. But in the text book they have only mentioned odd natural numbers as answer, which seemed incomplete to me.  Thanking everyone in advance:P
13989
I'm not sure I've got this right. When proving $a^n \mid b^n \Rightarrow a \mid b$, can we do this indirectly? In short,  "Suppose $a$ does not divide $b$, this implies that $a^n$ does not divide $b^n$. But $a^n \mid b^n$, hence $a$ divides $b$."  How about $n^n \mid m^m \Rightarrow n \mid m$? Can we do this the same way?
30883
If $a$ is an odd integer, prove that $24 \mid a(a^2-1)$.  I proved in my class that $a^2=8k+1$ for some $k$, provided $a$ was odd.  So $$a(a^2-1)=(2k-1)(8k+1-1)=8k(2k-1)$$ but I can't pull a $24$ out of it.
7687
I came across the following in the book Handbook of Mathematical Induction: $$ 19\mid (2^{2^n}+3^{2^n}+5^{2^n}),\quad n\in\mathbb{Z^+}\tag{1} $$ Apparently, this problem is not so bad if you think about it in a somewhat "outside of the box" sort of way, but it is supposed to be quite challenging otherwise.   Question: How might one prove $(1)$ using induction or otherwise? The "outside of the box" answer has been provided but with details oranged out so as not to spoil it for anyone who may come up with that solution individually.
106970
Prove that if $n\in\mathbb Z$, then $n^2$ is of the form $3q$ or $3q+1$ for some $q\in\mathbb Z$  I would like to show that 3q+2 is = 3q+1 thus $n^2$ can be of the form of 3q or 3q+1.  Case one   $(3k)^2=(3k)(3k)=9k^2=3(3k^2)$ and is still of the form $3q$  When $q=3k^2$  $3q$  Case two  $(3k+1)^2= (3k+1)(3k+1)= 9k^2+6k+1$  $(9k^2+6k)+1 =3(3k^2+2k)+1$ this is of the form $3q+1$ when $q=3k^2+2k$  $3q+1$  Case three  $(3k+2)^2= (3k+2)(3k+2) =9k^2+12k+4$  $(9k^2+12k+4) = 3(3k^2+4k+1)+1$  This is of the form $3q+1$ when $q=3k^2+4k+1$  $3q+1$  using a direct proof with cases we see that when $n$ is of the form $3k$ it's in the form $3q$ after squaring. Also when n is in the form $3k+1$, $n$ squared is still in the form $3q+1$ after squaring. Lastly we saw that when $n$ was in the form $3k+2$ we could simplify to the form $3q+1$.
57124
Prove that for any positive integer $n$, $$ 2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1 $$  the middle part can be expressed as a sum. of $1/\sqrt{x}$ from 1 to $n$, which leads to a $Hn^{0.5}$. Im not sure how that helps.  I tried differentiating both end functions and found they are both increasing.. But it doesn't seem to help much... help pls
55250
Prove that $ n^3 + 5n  $ is divisible by 6 for all $ n \in \textbf{N} $.  I provide my proof below.
109360
I need to show that it doesn't exist any $k \in \mathbb{Z}$ that satisfies $41 \mid (k^5-2)$.  Can I apply Fermat's Little Theorem to prove that in anyway? How? The only thing that came close when I was thinking was that $$41 \mid (k^5-1)(k^5+1)(k^{10}+1)(k^{20}+1)$$ (because $41\mid (k^{40}-1)$ by Fermat's Little Theorem), but from that I couldn't conclude anything. Thanks.