Using the fundamental theorem of arithmetics

Question

I’ve got a problem that I’m not quite sure how to solve. I can see the reasoning behind the problem, but I’m not sure how to apply the theorem.

Suppose that $a$, $b$, and $c$ are integers, and $a^2 = b \cdot c^2. $ Then, $b$ must be a square, e.g. there is some integer $m$ so that $m^2 = b$.

We’re supposed to show this using the fundamental theorem of arithmetic. It might just be because it’s rather late here, but I don’t quite see how the fact that these numbers can be factored into primes helps.

Is there a provision that $a \neq 0$? Because if there isn't, you can just say $0^2 = 3 \times 0^2$. — Robert Soupe, Nov 07 '19 at 08:21
Does this answer your question? Show that $a^n \mid b^n$ implies $a \mid b$ — Bill Dubuque, Feb 28 '23 at 02:10

Bernard · Answer 1 · 2020-08-31T15:38:03.773

2

You have to show that, for any prime $p$ in the factorisation of $b$, the exponent of $p$ (what is called the $p$-valuation of $b$) is even.

Indeed the equality $a^2=bc^2$ translates, at the $p$-valuations level, as $$v_p(a^2)=2v_p(a)=v_p(bc^2)=v_p(b)+2v_p(c),\enspace\text{whence}\quad v_p(b)=2\bigl(v_p(a)-v_p(c)\bigr).$$

edited Aug 31 '20 at 15:38

answered Nov 05 '19 at 21:37

Bernard

179,256

score 1 · Answer 2 · answered Nov 05 '19 at 21:30

1

Hint. Factor $a$, $b$ and $c$ into primes (in the unique way guaranteed by the Fundamental Theorem of Arithmetic). Then you want to show that each prime that occurs in $b$ does so an even number of times.

answered Nov 05 '19 at 21:30

Ethan Bolker

103,433

score 1 · Answer 3 · answered Nov 05 '19 at 21:30

If you consider the prime factorization of $a$, you can see that the prime factorization of $a^2$ will have each prime appearing an even number of times.

Similarly the prime factorization of $c^2$ will have each prime appearing an even number of times.

Then the prime factorization of $b$ must involve the primes in the prime factorization of $a^2$ that do not appear in the prime factorization of $c^2$. Each prime still appears an even number of times, so $b$ is also a square.

score 0 · Answer 4 · answered Nov 05 '19 at 21:31

0

Write $a=\prod_ip_i^{A_i},\,c=\prod_ip_i^{C_i}$, where the $p_i$ run over all prime factors of $ac$. Then $b=\prod_ip_i^{2(A_i-C_i)}$, so you can take $m=\prod_ip_i^{A_i-C_i}$. Since $c^2|a^2$, each $A_i\ge C_i$, so this $m$ is an integer.

answered Nov 05 '19 at 21:31

J.G.

118,053

score 0 · Accepted Answer · answered Nov 05 '19 at 21:32

Let $p$ be a prime factor of $b$. Then $p$ is a prime factor of $a$ and therefore, if you express $a^2$ as a product of powers of distinct primes, $p$ will appear there with an even power. But either $p\nmid c$ or $p$ appears in the expression of $c^2$ as a product of powers of distinct primes also with an even power. But then, since $a^2=bc^2$, $p$ appears in the expression of $b$ as a product of powers of distinct primes with an even power too. Since this occurs for every prime factor of $b$, $b$ is a perfect square.

Using the fundamental theorem of arithmetics

5 Answers5