If $a + b \sqrt{c} = d + e \sqrt{f}$ can we equate the numbers?

Question

Let $a, b, d, \text{ and } e$ be integers; Let $c \text{ and } f$ be positive integers greater than or equal to 2.

If the following is true:

$a + b \sqrt{c} = d + e \sqrt{f}$
$b \neq 0, e\neq0$
$c \text{ and } f$ are square-free.

Then is it true that $a=d, b=e,\text{ and } c=f $?

Edit: I added the condition that $c$ and $f$ are at least 2.

Please do not delete the question after receiving an answer. — lone student, Apr 07 '21 at 05:27
"Has no repeated prime factors" is a bit strange. Do you mean "are square-free"? — Brian Moehring, Apr 07 '21 at 05:28
@Brian Moehring, Yes, I'm not sure what the right term is but if the number under the square root has repeated prime factors, for example something like $\sqrt{12} = \sqrt{2^2 \cdot 3}$, we can simplify it to be $2 \sqrt{3}$. Edit: After looking up the definition for square-free integer, I think I meant square-free. — s114, Apr 07 '21 at 05:35
Definition of square-free: https://en.wikipedia.org/wiki/Square-free_integer — PM 2Ring, Apr 07 '21 at 05:39
@Empy2 you are right. I added the condition that $c$ and $f$ are at least 2. — s114, Apr 07 '21 at 06:08
By the linked Lemma it is true more generally if $,c,d,cd,$ are all nonsquares. — Bill Dubuque, Apr 07 '21 at 07:18

score 2 · Accepted Answer · answered Apr 07 '21 at 05:34

Yes.

Rearrange and square to get $$ e^2f=(a-d+b\sqrt c)^2=(a-d)^2+2(a-d)b\sqrt c+b^2c$$ and then the contradiction $$\sqrt c=\frac{e^2f-(a-d)^2-b^2}{2(a-d)b}\in \Bbb Q $$ unless we are not allowed to divide by $a-d$. We conclude that $a=d$. Thne $b\sqrt c=e\sqrt f$, so $$ \sqrt{cf}=\frac{ef}b\in\Bbb Q$$ which means $cf$ is a perfect square, and as $c,f$ are square-free, this makes $c=f$. As they are also $\ne 0$, then the last eequality, $b=e$, is immediate.

If $a + b \sqrt{c} = d + e \sqrt{f}$ can we equate the numbers?

1 Answers1