Prove that if integer $a > 0$ is not a square , then $ a \neq \frac{b^2}{c^2} $ for non-zero integers b,c

Question

Prove that if integer $a > 0$ is not a square , then $ a \neq \frac{b^2}{c^2} $ for non-zero integers b,c.

I would like to know if the proposed proof below is valid.

Assume that $ac^2 = b^2$.

With $a > 0$ we have $ac^2 > 0$. $b^2$ is a square and can written with the following factorization:

$b^2 = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \ldots p_n^{\alpha_n}$

where ${\alpha_1,\alpha_2,\ldots,\alpha_n}$ are all even integers and ${p_1,p_2,\ldots,p_n}$ are prime numbers.

Now $c^2$ and $a$ can be written with the following factorizations:

$c^2 = r_1^{2\beta_1}\cdot r_2^{2\beta_2} \ldots r_n^{2\beta_n}$

$a = r_1^{\gamma_1}\cdot r_2^{\gamma_2} \ldots r_n^{\gamma_n}$

where ${r_1,r_2,\ldots,r_n}$ are prime numbers and, ${\beta_1,\beta_2,\ldots,\beta_n}$ and ${\gamma_1,\gamma_2,\ldots,\gamma_n}$ are positive integers.

Then for any prime $s$ in the factorization of $b^2$ we have that ${s\mid r_1^{\gamma_1+2\beta_1}\cdot r_2^{\gamma_1+2\beta_2} \ldots r_n^{\gamma_n+2\beta_n}}$ which implies there is a unique $r_i$ such that ${s\mid r_i^{\gamma_i+2\beta_i}}$. It also implies $s\mid r_i$ and $s= r_i$ since $s$ and $r_i$ are both prime. (Believe last implication is correct but needs confirmation).

By the uniqueness of the factorization of $b^2$, $s$ has the power $t$ so that ${s^t = r_i^{\gamma_i+2\beta_i}}$ and $t = \gamma_i+2\beta_i$. $t$ is odd if $\gamma_i$ is odd but $t$ is even if $\gamma_i $ is even. So some of the powers in

${b^2 = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \ldots p_n^{\alpha_n}}$

are odd. This is a contradiction and ${ac^2 \neq b^2}$.

Answer 1

Almost correct, But don't start your proof with $$ac^2 = b^2$$

instead, start with; assume $$a = \frac{b^2}{c^2},$$ now we get $$ac^2 = b^2.$$

Answer 2

Yes, your proof looks correct.

I think proof by contrapositive might also work here, too: if $a=b^2/c^2$ is an integer, suppose $b^2/c^2 = (kp)^2/(kq)^2$, where $p/q$ and hence $p^2/q^2$ are in lowest terms. Then $q^2=1$ because $a$ is an integer, and hence $a = b^2$ is the square of an integer.

(You would need a similar kind of factorization argument if you wanted to prove in more detail the part which says “$p/q$ and hence $p^2/q^2$ are in lowest terms”.)

Answer 3

The intuition is right, but the formalization of the proof has some holes.

$c^2$ and $a$ can be written $\,\dots\,$ where $\,\dots\,$ ${\beta_1,\beta_2,\ldots,\beta_n}$ and ${\gamma_1,\gamma_2,\ldots,\gamma_n}$ are positive integers

This assumes that $a$ and $c$ have the same prime factors, which is not a given.

Also, as just a side advice, it helps with making proofs more readable (to others, and even yourself) to use consistent, easy to follow notation. The posted proof uses:

$b^2 = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \ldots p_n^{\alpha_n}$

$c^2 = r_1^{2\beta_1}\cdot r_2^{2\beta_2} \ldots r_n^{2\beta_n}$

Using $\alpha_i$ for twice the exponent of prime factors of $b$ vs. $\beta_i$ for exponents of prime factors of $c$ doesn't make it any easier to follow.

Below is my alternative writeup of what is essentially the same proof...

Assume that $a = b^2/c^2$ which is equivalent to $ac^2=b^2$. Let $p_1, p_2, \dots, p_n$ be the prime factors that divide either of $a,b,c$ so that $a = p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}\,$, $b = p_1^{\beta_1}p_2^{\beta_2}\dots p_n^{\beta_n}\,$, $c = p_1^{\gamma_1}p_2^{\gamma_2}\dots p_n^{\gamma_n}\,$, where $\alpha_i, \beta_i,\gamma_i$ are non-negative integers (some can be $0\,$, for example $\alpha_k=0$ if $p_k$ does not divide $a$).

Then, by the unique factorization theorem (also known as FTA), the powers of each prime on the two sides of the equality $ac^2=b^2$ must be equal, therefore:

$$ \alpha_i+2\gamma_i = 2\beta_i $$

It follows that $\alpha_i = 2(\beta_i-\gamma_i)\,$ is even, but in that case each prime factor of $a$ occurs at an even power, therefore $a$ is a perfect square, in fact $a=\big(p_1^{\beta_1-\gamma_1}p_2^{\beta_2-\gamma_2} \dots p_n^{\beta_n-\gamma_n}\big)^2$ which proves the contrapositive of the problem statement, and so the statement itself.