I'm learning Real Analysis by myself and I wanted to prove that if a prime $p$ divides $n^2$ where $n$ is an integer, then $p$ divides $n$ itself. I saw that proving this is the same as saying that the prime factors of $n^2$ are "the same" than those of $n$.

In this case, assume there exists a prime factor $a_q$ in $n^2$ which is not in $n$ (proof by contradiction). By Fundamental Theorem of Arithmetic, we can represent $n^2$ as $n^2= a_1 * a_2 * ... * a_x * a_q$, where every $a$ is prime and $n=k_1 * k_2 * ... * k_y$ (every $k$ is also prime). We know that $\frac{n^2}{n}=n$. So (by susbstituting):

$\frac{n^2}{n}= \frac{a_1*...*a_x*a_q}{k_1*...*k_y}=n \rightarrow \frac{a_1*...*a_x}{k_1*...*k_y}=\frac{n}{a_q}$

Because $\frac{n}{a_q}$ it's not an integer (initial statement), there's at least one $a_s$ in the left side of the equation that is not divisible by any $k$. Now we have two factors of $n^2$ that don't divide $n$. We can repeat this process,

$\frac{a_1*...*a_x*a_s}{k_1*...*k_y}=\frac{n}{a_q} \rightarrow \frac{a_1*...*a_x}{k_1*...*k_y}=\frac{n}{a_q*a_s}$,

for each prime factor of $n^2$ which means that there's no factor $a_i$ in $n^2$ divisible by any factor $k_j$ of $n$. But we knew that $\frac{n^2}{n}=n$, hence we have a contradiction ($n^2$ is not divisible by $n$). So there's no prime factor $p$ in $n^2$ that does not divide $n$ itself $\rightarrow$ every prime factor $p$ in $n^2$ also divides $n$.

I'm obviously not an expert in demonstrations so I want to know if this is a valid argument. I'm also aware of Euclid's Lemma, but for this case, ignore it.

  • 1
    $\begingroup$ As in the answer from Alex R., an important part of the Fundamental Theorem of Arithmetic is that for $1<n\in \Bbb Z^+,$ the finite sequence $p_1,...,p_m$ such that $n=\prod_{j=1}^mp_j$ is $unique,$ up to a re-arrangement of the order of the terms $p_1,...,p_m$. $\endgroup$ – DanielWainfleet Jan 5 '19 at 1:59

A shorter way would be to write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (where each $\alpha_i>0$) by the Fundamental Theorem of Arithmetic. Then $n^2= p_1^{2\alpha_1}\cdots p_k^{2\alpha_k}$. The only primes that divide $n^2$ are $p_1,\cdots,p_k$ and these clearly also divide $n$.

  • $\begingroup$ Woah, thanks! I'm just a beginner, so this helps me a lot. $\endgroup$ – Data Space Jan 4 '19 at 18:40
  • $\begingroup$ @DataSpace Unfortunately this answer is wrong - your proof is not correct. The claim in the first sentence of the 3rd paragraph is not true, e.g. $p/p^2$ is not an integer, but but there is no prime in the numerator that is not divisible by $p$. $\endgroup$ – Bill Dubuque Jan 4 '19 at 19:06
  • $\begingroup$ @BillDubuque If $\frac{p}{p^2} = \frac{n}{a_q}$ then $a_q = n*p$ (simple algebra), this contradicts the idea that every $a$ is prime. So your counterexample is not valid.Sorry for the delay :s $\endgroup$ – Data Space Jan 4 '19 at 22:36
  • $\begingroup$ @BillDubuque Nevertheless, it is true that this proof is not clear at all and contains argumentative"holes". It is, of course, a better idea to just take Alex's proof. $\endgroup$ – Data Space Jan 4 '19 at 23:05
  • $\begingroup$ @DataSpace But I didn't claim it equals $n/a_q$. The point is that your argument doesn't handle such multiple factors that may occur in the (reducible) fraction. $\endgroup$ – Bill Dubuque Jan 5 '19 at 0:02

Here is a different approach. Unique prime factorization of $n,$ for $1<n\in \Bbb N,$ follows from the Lemma that if $a,b,c\in \Bbb Z$ and $\frac {ab}{c}\in \Bbb Z$ while $\gcd (a,c)=1$ then $\frac {b}{c}\in \Bbb Z.$ The Lemma follows from something called Bezout's Identity, although it is implicit in Euclid's algorithm for computing a $\gcd$ : For any $a,b\in \Bbb Z$ (not both $0$) there exist $x,y \in \Bbb Z$ such that $ax+by=\gcd (a,b).$

Let $p$ be prime and let $n\in \Bbb Z$ such that $p|n^2.$ Let $m=\gcd (p,n).$ Since $m$ divides the prime $p$ and $m\ge 1,$ we have $m=1$ or $m=p.$

Now there exist $x,y\in \Bbb Z$ such that $px+ny=m,$ so $\frac {(m-px)^2}{p}=\frac {n^2y^2}{p}=\frac {n^2}{p}\cdot y^2\in \Bbb Z.$ But if $m=1$ then $\frac {(m-px)^2}{p}=\frac {1-2px+p^2x^2}{p}=\frac {1}{p}-2x+px^2\in \Bbb Z,$ implying $\frac {1}{p}\in \Bbb Z,$ which is absurd.

So, since we can't have $m=1,$ we have $m=p$. And since $m=\gcd (p,n)|n,$ we have therefore $p|n.$

  • $\begingroup$ As matter of style, we could also say that $p|n^2\implies$ $ p|n^2y^2=$ $(m-px)^2=$ $=m^2+p(-2mx+px^2)\implies$ $ p|m^2,$ to show that $m\ne 1$. $\endgroup$ – DanielWainfleet Jan 5 '19 at 2:55
  • $\begingroup$ <3 I really liked this proof, thanks for such logical piece. $\endgroup$ – Data Space Jan 5 '19 at 2:57
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    $\begingroup$ @DataSpace If you're going to permit Bezout then it's easy: $$\,p\mid nn,\ p\nmid n\,\Rightarrow\, kn\!+\!jp = 1\,\Rightarrow\, p\mid knn\!+\!jpn = (kn\!+\!jp)n = n\qquad\qquad$$ Or, equivalently said in terms of gcds $\ \ p\mid nn,pn\,\Rightarrow\, p\mid (nn,pn)=(n,p)n = n.\ $ This is essentially a special-case of the Bezout-based proof of EL = Euclid's Lemma (see here for various proofs of EL) $\endgroup$ – Bill Dubuque Jan 5 '19 at 3:39

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