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 ${ac^2 \neq b^2}$.

  • 1
    $\begingroup$ Why do a and c have the same prime factors? And why must some of the powers of $b^2$ must be odd? And you never used a at all, nor that it's not square. Those part I don't get. Your idea is correct. You do find which factors are even or odd. $\endgroup$ – fleablood Aug 12 '17 at 3:34
  • $\begingroup$ Thanks for the comments. You are correct in asking why do $a$ and $c$ have the same prime factors? I did not know if I could do that or not. Below "dxiv" pointed out that it is not a given and it is incorrect to do so. In the proof I gave I showed that some powers of $b^2$ are odd where in fact they must be all even for $b^2$ to be a square. Please note in the 2nd to last paragraph $a$ is used as shown by the factorization of $a c^2$ which are made of terms like $r_i^{\gamma_i+2\beta_i}$. $\endgroup$ – Rene Girard Aug 13 '17 at 1:25

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”.)

  • $\begingroup$ Thanks for the comments. I understand from your proof that ${k=gcd(b,c)}$. Is this correct? You indicate that ${q \mid p}$. Is this because ${c \mid b}$ and $b^2/c^2 = a$ where $a$ is integer? I am not familiar with the terminology " ${p^2/q^2}$ are in lowest terms ". Can you give some details on its meaning? $\endgroup$ – Rene Girard Aug 13 '17 at 2:18

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 $$ac^2 = b^2.$$


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.

  • $\begingroup$ P.S. The problem is of course equivalent to Prove that the square root of a positive integer is either an integer or irrational. In this case the square root of positive integer $a$ is $b/c$ which is obviously not irrational, so $b/c$ must be an integer, then it follows that $a = (b/c)^2$ is a perfect square. $\endgroup$ – dxiv Aug 12 '17 at 3:39
  • $\begingroup$ Thank you for the advice on the use of symbols for factorization exponents. The 2nd sentence of the 1st paragraph of your answer really help in improving my understanding of factorization when dealing with several integers. If I understand you correctly we have a set P that contains all the prime numbers that are used in the factorizations of $a,b,c$. Each of these factorization uses different subset of P. This enables writing the factorization of $a,b,c$ as you did above where some of the exponents are 0 if the corresponding prime number does not divide the integer being factored. $\endgroup$ – Rene Girard Aug 13 '17 at 1:53
  • $\begingroup$ @ReneGirard a set P that contains all the prime numbers that are used in the factorizations of a,b,c. Each of these factorization uses different subset of P. Right, that's precisely so. some of the exponents are 0 if the corresponding prime number does not divide the integer being factored Right again. $\endgroup$ – dxiv Aug 13 '17 at 1:58
  • $\begingroup$ Your reply made me understand better a proof of the Fund. Thm of Algebra that I am studying. This is great. Thanks again. $\endgroup$ – Rene Girard Aug 13 '17 at 2:23

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