Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$

Question

Question:

Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$

Answer 1

Let's use a proof by contradiction.

Assume for the sake of contradiction that there exists a prime $p$ such that $p=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$. This implies that $\frac{a^2}{b^2}$ is an integer, and since $a,b$ are integers, so is $\frac{a}{b}$. Let $\frac{a}{b}=k$. We know that $p=k^2$. However, we have a contradiction because both $k$ and $k^2$ divide $p$ (unless $k=k^2$, in which case $p=1$, which is not prime). We're done! $\blacksquare$

Answer 2

Note that each fraction $\frac{a}{b}$ has an irreducible representation, that is, there exist $a_{1},b_{1}$ such that $\frac{a}{b} =\frac{a_{1}}{b_{1}}$ and $\gcd(a_{1},b_{1}) = 1$.

So if $p = (\frac{a}{b})^{2}$, then $p = (\frac{a_{1}}{b_{1}})^{2}=\frac{a_{1}^{2}}{b_{1}^{2}} \implies b_{1}^{2}p = a_1{}^{2} \implies p \mid a_{1}^{2} \implies p \mid a_{1}.$

Similarly $p \mid b_{1} \implies p \mid a_{1},b_{1} \implies (a_{1},b_{1}) \neq 1$, a contradiction.

Answer 3

Let $v_p(k)$ be the exponent of the largest power of $p$ that divides $k$ so that $p^{v_p(k)} | k$ and $p^{v_p(k)+1} \not\mid k$.

Then if $m, a, b \ge 1$ then $ m |v_p((a/b)^m)$.

This problem is the case $m = 2$.

Proof.

If $ v_p((a/b)^m) = n$ where $n \ge 1$, let $v_p(a) = u$ and $v_p(b) = v$ where $u, v \ge 0$.

Then $v_p(a/b) = u-v$ so $v_p((a/b)^m) = m(u-v)$ so $n = m(u-v) $.

Therefore $m | n$.

Answer 4

This depends on what you know about factorization.

Do you know every number has a unique prime factorization? If you know that then you know that if $p = \frac{a^2}{b^2}$ and $p$ is an integer, then $b^2$ divides evenly into $a^2$ and the primes that divide into $b$ also divide into $a$ but to higher power and the powers that divide into $b^2$ and $a^2$ are the same that divide into $b$ and $a$ but to exactly half the powers. So if $b^2$ divides evenly into $a^2$ then $b$ divides evenly into $a$ and $\frac ab$ is an integer. SO $p = \frac{a^2}{b^2} = (\frac ab)^2$ and so $p$ is a perfect square. .... but $p$ is prime. SO that's impossible.

But knowing the every number has a unique prime factorization is a pretty sophisticated assumption? Maybe you aren't allowed to assume it.

BUt can you assume that $\frac ab$ can be written in "lowest terms"? If $\frac ab$ is in lowest terms then $a$ and $b$ have no factors in common. Does that mean that $a^2$ and $b^2$ have no factors in common? That depends on what you know. But if $a^2$ and $b^2$ have no factors in common then $b^2$ can't divide evenly into $a^2$ (unless we assume $b = 1$) and we cant have $\frac {a^2}{b^2}= p$ an integer.

Is it possible that $\frac ab$ is not in "lowest terms"? If $a,b$ have a factor $k$ in common then $\frac ab = \frac {\frac ak}{\frac bk}$ and we can do this for all factors. Is there always a finite number of factors and we will eventually factor them all out until we get $\frac ab = \frac {a'}{b'}$ where $a', b'$ don't have factors in common? THe only other option would be if $a$ and $b$ both have an infinite number of factors in common and that's just plain nutty.

But do we know this stuff?

If $a,b$ have no factors in common than neither to $a^2, b^2$ so we can't have $\frac {a^2}{b^2} = p$ as in integer.... unless..... is it possible that somehow squaring $a$ and squaring $b$ add a new factor out of nowhere. That's kinda sorta what the ancient pythagorians wanted to sweep under the rug.

DO we know that's impossible? Do we know Euclid's Lemma? If we have $p = \frac {a^2}{b^2}$ then $pa^2 = b^2$ so $p$ divides evenly into $a^2$. But $p$ is prime and indivisible. Does that mean we can't split up $p$ among the two copies of $a$ in $a\times a = a^2$? Does that mean $p$ must divide evenly into $a$?

But if $p$ divides evenly into $a$ that would mean $p^2$ divides into $a^2$, wouldn't it? And if $pb^2 = a^2$ and $p^2$ divides $a^2$ the we can divide both sides by $p^2$ and get that $p$ divides evenly into $b^2$. But that means $p$ divides $b$ for the same reason and $a, b$ have $p$ as common factor....

SO.... what do we know about factorization? Which of those observations are legitimate? What can we conclude?