2
$\begingroup$

I think I need to use the Fundamental Theorem of Arithmetic but that only applies to $a > 1$ so I think I need to do it by cases.

Case 1) $a = 0$.

Case 2) $a = 1$.

Case 3) $a > 1$.

Case 4) $a < 0$.

Cases 1 and 2 are easy. I think Case 4 will follow from Case 3 but I struggle with Case 3.

Case 3) $a > 1$. Then by Fundamental Theorem of Arithmetic, $a$ is equal to a unique product of primes. Then, $a^2$ is a product of those primes squared. Then, because $p$ divides $a^2$, $a^2 = pk$ for some $k$ (an integer). So that product of primes squared equals $pk$. Now how do I argue that one of those primes must be $p$? Once we see that one of those primes is $p$ (i.e. $p$ is a prime factor of $a$), we can say that $p$ divides $a$.

So my questions are: How do I argue that one of those primes must be $p$? Can the proof be done without resorting to four cases?

$\endgroup$
  • 1
    $\begingroup$ How do you have primes defined? One of the most common definitions is that $p$ is a prime integer iff for any integers $x,y$ one has $p\mid xy$ implies $p\mid x$ or $p\mid y$, from which the question is immediately proven noting that $a^2=a\cdot a$. $\endgroup$ – JMoravitz Sep 10 '16 at 21:35
  • 1
    $\begingroup$ This is implied by my question here: math.stackexchange.com/questions/1464864/… $\endgroup$ – marty cohen Sep 10 '16 at 21:47
  • $\begingroup$ Often proofs can be done with a single, short line that addresses all possible cases. But such proofs can be very difficult for people who are not math professionals to understand. $\endgroup$ – Mr. Brooks Sep 12 '16 at 21:21
0
$\begingroup$

Using prime factorization of $a = p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m}$ where $p_j$'s are distinct primes, and $n_j$'s are natural numbers. Thus $p\mid a^2 = p_1^{2n_1}p_2^{2n_2}\cdots p_m^{2n_m}$. Since $p$ is a prime $p = p_k$, for some $1 \le k \le m$, and this implies $p \mid p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m} = a$

$\endgroup$
2
$\begingroup$

If $p$ does not divide $a$ then $\gcd(p,a)=1.$ By Euclidean Algorithm (or Bezout) there are integers $x$ and $y$ so that $px+ay = 1$. Multiply by $a$ to get $apx+a^2y = a.$ Since $p$ divides the left side, it must divide the right, contradiction.

$\endgroup$
  • $\begingroup$ Your sentence "if $p$ does not divide $a$ then $\text{gcd}(p,a) = 1$ "in the problem's context is a question in its own and needs further explanation. $\endgroup$ – DeepSea Sep 10 '16 at 21:45
2
$\begingroup$

Assume that $p \nmid a$. Then, $a=p \cdot q +r$ for some $r >0$, s.t. $r \neq 0 (modp) $.

Since, ${p \mid a^2 } \Rightarrow {a^2 \equiv 0 (mod p)}$

So, $a^2=(p \cdot q +r)^2=p^2 q^2+2pqr+r^2 \equiv r^2 (modp)\neq 0 (modp) \Rightarrow p \nmid a^2$ Contradiction.

$\endgroup$
0
$\begingroup$

This is a special case of my question (which I answered) here: If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?

In particular, I show that if $p^m | a^2$, then $p^{\lceil m/2 \rceil} | a $. This question is the case $m=1$.

$\endgroup$
0
$\begingroup$

Here is my understanding for what it's worth.

If $a$ is some integer then using the Fundamental Theorem of Arithmetic like you mentioned $a$ can be written as a product of prime numbers. Let's say $a=bcd$ where $b$ , $c$ and $d$ are prime numbers. From this we can see that $a^2=bbccdd$. If $p$ divides evenly into $a^2$ then $p$ must be equal to either one of $b$ , $c$ or $d$. It can't be a product of any of them as it is prime itself. Let's say $p=c$. We can see that $c$ is also a factor of $a$. Whichever factor of $a^2$ that equals $p$ also exists in $a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.