I want to prove for all integers $a$ that if $3\mid a$ then $3\mid a^2$.  Suppose that $a \in \mathbb{Z}$ and $3\mid a$. Then $a = 3k$ for any integer k.  Hence $$a^2 = (3k)(3k) = 9k^2 = 3(3k^2)$$  Since $k$ is a integer, $3k^2$ is an integer. Thus $a^2 = 3b$ for some integer $b$.  Therefore $3\mid a^2$.  Is this proof correct?
122429
I just need to know that if $ a $ divides $ b $, where $ a $ and $ b $ are integers, does $ a^{2} $ divide $ b^{2} $?
19168
Let $a$ be a nonzero number and $m$ and $n$ be integers. Prove the following equality: $a^{m+n}=a^{m}a^{n}$  I'm not really sure what direction to go in. I'm not sure if I need to show for $n$ positive and negative separately or is there an easier way. Can you use induction on integers?  My attempt: 1) Base case $m=0$. Prove that $a^{m+n}=a^m a^n$. Is true.  2) Assume the result holds for $m$.  So I want to prove it holds for $m+1$.  So I know that $a^{m+n}=a^ma^n$. So does this imply that: $a^{m+n+1}= a^{m+1+n}= a^{m+1}a^n$? Am I going in the right direction? I'm not sure what to do next...
55143
The problem's quite clear. Prove that $$\gcd(a^2, b^2) = \gcd(a, b)^2$$  This is easy to understand intuitively and using the Fundamental Theorem of Arithmetic would be easy but I want to prove it by using the Divisibility axioms and the GCD. I have attempted it and am mentioning my results.  Here $d = \gcd(a, b)$.   $$d | ax + by \implies d^2 | a^2x^2 + 2axby + b^2y^2 \implies d^2 | a^2x^2 + b^2y^2 \implies \gcd(a^2, b^2) | d^2$$  I feel that this is not necessary. Please help.  Also, the proof can be generalized to $$\gcd(a^n, b^n) = \gcd(a, b)^n$$  Any help would be appreciated.
56685
Suppose $p$ is a prime and $a$ and $k$ are positive integers. Prove that, if $p \mid a^k$, then $p^k\mid a^k$ also.  I have already proven that if $a,b,n\in\mathbb{N}$ and if $a^n\mid b^n$, then $a\mid b$. I tried using induction:  $k=1 \Rightarrow p\mid a^1 \Rightarrow p^1\mid a^1$. Assume the statement holds for some $k$. Then if $p\mid a^{k+1}\Rightarrow p\mid a^ka$. Since $p$ is prime, $p\mid a$ or $p\mid a^k$. If $p\mid a$, we are done. If $p\mid a^k$ then by assumption, $p^k\mid a^k$ and so by the lemma, $p\mid a$ and so $p^{k+1}\mid a^{k+1}$.  Does this work?
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?
99342
$a \mid b$ and $b \mid a$ if and only if $a= \pm b$.  How do I go about this proof? Step by step explanation please!
27107
Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$.   I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions.  This is what I have so far: Let $m = nq + r$, then $a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1$. Since $0\le r\lt n$, $r = 0$ then $a^m - 1$ = $a^{nq} - 1$ so $a^m$ = $a^{nq}$ . Thus $m = nq$ hence $n\mid m$.
85098
I need help!  I need to prove that for any $2 \le n,k$ positive integers $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$  Thanks!
11458
Prove, by contradiction, that, if $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.    Proposition: If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.  Hypothesis: $n$ is a positive integer such that $n^3 − n − 6 = 0$.  Conclusion: For every positive integer $m$ with $m \not = n$, $m^3  − m − 6 \not = 0$.    My workings  A (Hypothesis): $n$ is a positive integer such that $n^3 − n − 6 = 0$.  A1 ($\neg B$): There exists a positive integer $m$ with $m \not = n$ such that $m^3  − m − 6 = 0$.  A2: $n^3 - n = 6$  $\implies n(n^2 - 1) = 6$ where $n \in \mathbb{Z}^+$  A3: $m^3 - m = 6$  $\implies m(m^2 - 1) = 6$ where $m \in \mathbb{Z}^+$ and $m \not = n$.  A4: $n(n^2 - 1) = m(m^2 - 1)$ where $m \not = n$.  $\implies \dfrac{n(n^2 - 1)}{m(m^2 - 1)} = 1$ where $m \not = n$.  Since $m,n \in \mathbb{Z}^+$, $(n^2 - 1) \ge 0$ and $(m^2 - 1) \ge 0$. Therefore, $n(n^2 - 1) = m(m^2 - 1)$ iff $n = m$.   Contradiction.  $Q.E.D.$    I would greatly appreciate it if people could please take the time to provide feedback on the correctness of my proof. Given that $m \not = n$, did I construct my proof correctly? Or is it incorrect to structure it this way if $m \not = n$?
103641
Given that $p$ is a prime and $p\mid a^n$, prove that $p^n\mid a^n$.   I know that the fundamental theorem of arithmetic states that any positive integer can be represented as a product of primes but how do I apply this to the proof?
24840
Let $S = \{ax + by: x,y \in \mathbb{Z}\}$  and let $e &gt; 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$  I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.  I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this.
17489
Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$  So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$ Is this right so far? Where do I go from here? Step by step explanation please!
113901
Prove the following: If $a \mid bc$, then $a  \mid \gcd(a, b)c$.   I tried to set $\gcd(a, b)$ to $b$ and used the fundamental theorem of arithmetic to prove that it is divisible by $a$, but I can't prove that $a \mid bc$, if and only if $a\mid b$ and $a\mid c$. Please help. Thanks.
29081
I want to prove by induction that $a^n \mid b^n$ implies that $a \mid b$ holds for all integers $n\geq 1$.  Clearly for $n=1$ this is true, since if $a \mid b$, then $a \mid b$.  Suppose this is true for some $n = k$. Then $a^k \mid b^k$, so $a|b$.  $a^k \mid b^k$ means there exists some integer $m$ such that $b^k = ma^k$, and $a \mid b$ means there exists an integer $r$ such that $b=ra$.  Then we can multiply both sides of the above equality by any constant, namely $b$.   Then $b \cdot b^k = bma^k$  so $b^{k+1} = (ra)ma^k = rma^{k+1}$  which means that $a^{k+1} \mid b^{k+1}$ completing the induction step, and thus proving that for all integers $n \geq1$ the original statement holds.  Are there any logical holes in the proof?
103912
Here is the statement: Let $a,b \in \mathbb{Z}$ positive integers such that $a^2=b^4+b^3+b^2+b+1.$ Prove $b=3.$  I've tried is the following: Let $\Sigma=b^4+b^3+b^2+b+1$. If $a\equiv 0\mod 3$, then  $a^2\equiv 0 \mod 3$. If $a\equiv 1\mod 3$ or $a\equiv 2\mod 3$, then $a^2\equiv 1\mod 3$. In the other hand, $b\equiv 0\mod 3$ or $b\equiv 2\mod 3$, then $\Sigma\equiv 1\mod 3$ and if $b\equiv 1\mod 3$, $\Sigma\equiv 1\mod 3$. Thus $a=3n$ or $a=3n+2$ for some $n\in\mathbb{Z}$, and $b=3m$ or $b=3m+2$ for some $m\in\mathbb{Z}$. In a similar way, I have prove $a$ cannot be even.  If $a=3n$, for some $n\in \mathbb{Z}$ and $b=3m$, for some $n\in \mathbb{Z}$, we arrive at the following equation $(3n-1)(3n+1)=3m(3&#179;m&#179;+3&#178;m&#178;+3m~1).$ So, $a\equiv 0\mod 3$ or $b\equiv 0\mod 3$ but not both at same time.  I think I should use the Legendre symbol. In this case $(\frac{\Sigma}{3})=0 \text{ or }1$.
2699
Prove or disprove: For every integer a, if a is not congruent to 0 (mod 3), the a^2 is congruent to 1 (mod 3)  SO this is for abstract algebra and I am really struggling with this. Here are some of the definitions and theorems I think would apply.  -an integer a divides and integer b if there is an integer q such that b=aq -Let a and b be integers, with a>0. Then there exist unique integers q and r such that b= aq+r and 0&lt;=r  The question says hint: use 2 cases but I'm not really sure what that means. I'm guessing it wants me to look at it when a is congruent to 0 and when it's not? so i tried to do that but just ended up stuck. below is as far as i got with that attempt  Proof: Case 1 There should be some information about letting variables exist in certain number sets here  assume a is not congruent to 0 mod 3, then 3 does not divide a-0, therefore a-0 does not equal 3q for some q in the integers  I was trying to get this to wind down to thus 3 divides a^2-1 but even tho it seems true (I couldn't find a counter example) I couldn't figure out how to prove it with what I have.  Next i tried reversing it thinking I could contradict it or something.  I think this would be case 2? assume a is congruent to 0 mod 3, then 3 divides a-0 and a-0=3q for some q in the integers, then 3q=a since 0 is the additive identity, therefore 3 divides a,  and again stuck and my brain hurts a little..  Can anyone help me figure out how to work this.
51877
Does $a^n \mid b^n$ imply $a\mid b$? I think it does but haven't been able to prove it.  I don't know much number theory so an elementary answer would be great.