This is an exercise in my abstract algebra reader, in the chapter about polynomial rings.      Let $n \in \Bbb N$. Let $p&gt;2$ a prime number. And let $n$ not divisble by $p-1$. Show that      $$1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}$$   Our teacher assistant gave a hint to solve this with induction. We have tried to solve this using induction, but after some hours tring, I don't believe anyomre that this is possible. At least, I don't think I'm able to do that with my current knowlegde about mathematics. If someone has an idea how to solve this with induction (or maybe an other way), then I'm all ears.
43674
Calculate for every positive integer $n$ and for every prime $p$ the expression  $$1^n +2^n + \cdots +(p-1)^n \mod p$$    I need your help for this. I don't know what to do, but I'll show you what I know.   Wilson's theorem The identity $X^p-X = \prod_{a \in \mathbb{F}_p-1}(X-a)$ $\exists a \in \mathbb{Z}, \ a^2 \equiv -1 \mod p \qquad \iff \qquad p \equiv 1 \mod 4$   Now can you please provide me a hint?
19148
Why is it that $\forall n \in N$, $2^n$ is not divisible by $3$?  I can prove it easily by induction, but I don't understand the intuition of why this is true. Could anyone please supply the intuition? [This has been answered.]  Moreover, I understand that there's also a proof of this fact that uses modular arithmetic. How does this proof go? [Also answered - it follows immediately from the definition of the modulus operation.]
63841
Prove or disprove: If $a \in \mathbb{Z}$, $a &gt; 0$, and $p$ is a prime such that $p^4 \mid a^3$, then $p^2 \mid a$.  I looked at the answers to the duplicate questions and I'm still confused. I can't even think of any counterexamples. Any tips on where to start?
37609
For all $n\ge1$, use mathematical induction to establish each other the following divisibility statements:   $$5\mid2^{4n}-1$$    I was wondering if someone could help me with the set up of this proof. I only have done proofs with mathematical induction when I am given a series of numbers and I can incorporate my induction hypothesis in for one of my terms in the series.   Step by step explanation please?
2770
Show that if $n&gt;2$, then $(n!)^2&gt;n^n$.     My work: I tried to apply induction. So, at the induction step, I need to prove, $n^n&gt;(n+1)^{n-1}$ Here, I tried to use induction again without any luck. I also took log of both sides, but I did not get anything. Please help!
115127
By using the principle of Mathematical Induction, prove that: $P(n)=n(n+1)(2n+1)$ is divisible by $6$.  My Attempt: Base Case: $n=1$ $$P(1)=1(1+1)(2\times 1+1)$$ $$=2\times 3$$ $$=6$$, Which is divisible by $6$. $P(1)$ is divisible by $6$  Induction Hypothesis: $(n=k)$ $P(k)=k(k+1)(2k+1)$  Now, how.should I move on?
68334
Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.   I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure how that would help.
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!
6856
Here is a problem which i have not been able to do for quite sometime.   For $n \geq 2$, show that $n \nmid 2^{n}-1$.   I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have $2^{p-1} \equiv 1 \ (\text{mod} \ p)$, which actually says that for $p \mid 2^{p}-2$. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?
25419
Prove by induction that $5^n - 1$ is divisible by $4$.   How should I use induction in this problem. Do you have any hints for solving this problem?  Thank you so much.
54889
Sequence $a_0,a_1,a_2,...$ satisfies that $a_0=2,a_1=1,a_{n+1}=a_n+a_{n-1}$  Prove that if $p$ is a prime divisor of $a_{2k}-2$,then $p$ is also a prime divisor of $a_{2k+1}-1$    If $x_{1,2}={1\pm\sqrt{5}\over 2}$ then $a_k = x_1^k+x_2^k$ and $a_{2k}-2(-1)^k = (x_1^k-x_2^k)^2$.
4426
Not sure if I'm doing this correctly but this is what I've done:  Assume that $n$ is composite and suppose $2^n-1$ is a prime for $n \gt 2$. Then, $2^n-1 = 2k$ for some $k \in \Bbb Z $, $\forall n$.  But this would be a contradiction since $n \gt 2$.  I'm not sure if this a correct proof.  EDIT: I realized that it's to prove $2^n - 1$ NOT $2^{n-1}$ so my attempt above is completely wrong.
94592
For any integer n greater than 1, show that $4^n+n^4$ is never a prime number.  My approach: I tried to use mathematical induction. But somehow couldn't manage to prove it. Is it to be done by induction or can it be done by another method?
72644
Prove that for $ n \geq 2$, n has at least one prime factor.  I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.
99210
Prove that for all  $n \in \mathbb{N}$: $$2^{n-1}(a^n + b^n) \ge (a + b)^n$$ I used induction (for $k = 1,2,...n-1$), and ended up with the following equation:  $$(a + b)(a + b)^{n-1} \le (a + b)(a^{n-1} + b^{n-1})2^{n-2}$$  Now, I have a feeling I'm on the right track, but I'm not totally sure. Could someone give me a pointer one where I should go from here? Any help would be greatly appreciated. Thanks!
91733
Let $p,q$ be primes, estimate the degree $[\Bbb Q(\sqrt[p]{2}\cdot\sqrt[q]{2}):\Bbb Q]$ and prove that the polynomial $X^q-2$ is irreducible in the ring $\Bbb Q(\sqrt[p]{2})[X]$  I found this problem on mathematical competition without any solution and it would be very interesting if we can find a proof for this.
89356
Let $R$ be a field. Show that the polynomial ring $R[x_{1},...,x_{n}]$ is not a PID if $n&gt;1$.    How do I show this?   So far, I've shown that $(2)+(x)$ wouldn't be principal in $\mathbb{Z}[x]$ so I was thinking of showing that $(x_1)+(x_2)$ would not be principal.