MHB Why aren't there coprime integers?

[evinda]

Hey again! (Smile)

I am looking at the following exercise:

Why aren't there coprime integers $a,b>1$, such that $a^2 b^3=8100$?

That's what I have tried:

$b>1$,so it has a prime divisor $p$.

$p$ can be $2 , 3 \text{ or } 5$.

• $p=2$:
  
  $$b=2k, k \in \mathbb{Z}$$
  
  Then, $a^2 \cdot 2^3 \cdot k^3=2^2 \cdot 3^4 \cdot 5^2$ ,that is not possible,because the prime $2$ appears at the left side more times than at the right side.
  
• $p=5$
  
  $$b=5k, k \in \mathbb{Z}$$
  
  Then, $a^2 \cdot 5^3 \cdot k^3=2^2 \cdot 3^4 \cdot 5^2$, that is not possible,because the prime $5$ appears at the left side more times than at the right side.
  
• $p=3$
  
  $$ b=3k, k \in \mathbb{Z}$$
  
  Then, $a^2 \cdot 3^3 \cdot k^3=2^2 \cdot 3^4 \cdot 5^2$, that is also not possible,because, even if , $3$ is a factor of $k$,it would be $3^3 \cdot \text{ something }$,and so, it can't be that we have the same number of $3$ at both sides.

Can I say it like that or have I done something wrong? Also,don't I have to say which values can $k$ take? Because it can't take different values from powers of $2,3,5$.. (Thinking)

 

[Deveno]

I would approach it this way:

If $p|b^3$ then $p|b$, so that $p^3|b^3$ and thus $p^3|8100$.

Thus the only prime which can divide $b$ is $3$, as $8\not\mid 8100$ and $125\not\mid 8100$.

Since $b^3$ is a perfect cube, and $3^4 = 81$ is not, the only possible value for $b^3$ is $27$, or $b = 3$.

But then, we have that $3|a^2 = \dfrac{8100}{27} = 300$, that is: $3|a$, so that $a$ and $b$ are not co-prime.

Also, observe $300$ is NOT a perfect square.

 

[I like Serena]

Hey! :)

Can I say it like that or have I done something wrong? Also,don't I have to say which values can $k$ take? Because it can't take different values from powers of $2,3,5$.. (Thinking)

Yep. You can say it like that. It is perfect! ;)
Furthermore, for the line of reasoning it is irrelevant which values $k$ might take. So that is fine too.

 

[evinda]

Since $b^3$ is a perfect cube, and $3^4 = 81$ is not, the only possible value for $b^3$ is $27$, or $b = 3$.

But then, we have that $3|a^2 = \dfrac{8100}{27} = 300$, that is: $3|a$, so that $a$ and $b$ are not co-prime.

Also, observe $300$ is NOT a perfect square.

Could you explain it further to me? Why is the only possible value for $b, 3$ ?

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Hey! :)
Yep. You can say it like that. It is perfect! ;)
Furthermore, for the line of reasoning it is irrelevant which values $k$ might take. So that is fine too.

Great!Thanks a lot! (Clapping)

 

[Deveno]

Could you explain it further to me? Why is the only possible value for $b, 3$ ?

You already understand that the only prime divisors of $b$ are $2,3,5$.

If $b = 2k$, as you say, then $b^3 = 8k^3$. This means $8|b^3$ and $b^3|8100$, so $8|8100$, which is false.

The same kind of reasoning applies if $5|b$ (this is the same as what you wrote "without the $k$").

So out of $2,3,5$, the only prime which stands a chance of dividing $b$ is 3. This means $b$ is a power of 3.

The only powers of 3 that divide 8100 are: 1, 3, 27 and 81. Since $b > 1$, that leaves 3, 27 and 81.

Now if $b$ is a power of 3, so is $b^3$: if $b = 3^k$, then $b^3 = (3^k)^3 = 3^{3k}$.

But the highest power of 3 that divides 8100 is $3^4 = 81$.

This means that $3k \leq 4$, so $k = 0,1$. We have already ruled out $k = 0$ (since $b > 1$), so it HAS to be that $k = 1$.

And if $k = 1$, so that $b = 3$, then $b^3 = 27$, so:

$a^2 = \dfrac{8100}{27} = 300$, which is divisible by 3.

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I'd like to explain a bit about WHY I approach it this way: your method "counts" powers of each prime for possible prime factors of $b$. There is nothing wrong with this, because the possible powers of prime factors of:

$8100 = 2^2\cdot3^4\cdot5^2$

are all SMALL, if nothing else, we could try every single possibility.

This approach would be much more involved if we tried to use it for:

$n = 2^{352}\cdot5^{682}\cdot7^{22}$.

What I am trying to get across is if $\text{gcd}(a,b) = 1$, then if $p$ divides $b$, it doesn't divide $a$. So if we were trying to express:

$n = a^2b^3$, for example, then if 5 (for example) was a factor of $b$, then $b$ contains ALL the powers of 5 that occur in $n$.

This can only happen if 682 is divisible by 3, which is not the case. We can use the same reasoning to deduce that 2 is not a factor of $b$. This means that $b = 7^k$, and thus $b^3 = 7^{3k}$, and since ALL the powers of 7 in the factorization of $n$ occur in $b$, it HAS to be that $3k = 22$, which is impossible.

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I do NOT want to give the impression your approach is "wrong", because it is fine. I merely hope that you can see that a method which generalizes better is more useful than one tied to a "specific case".