If $a+\sqrt{b}=c+\sqrt{d}$, is it true that $a=c$ and $b=d$?

Question

Assume that $a,b,c,d$ are all positive integers. If $a+\sqrt{b}=c+\sqrt{d}$, is it true that $a=c$ and $b=d$?

I am grading some problems and I don't think this true, but all of a sudden I am doubting myself.

Answer 1

This is incorrect. For instance, let $a = 1$, $b = 4$, $c = 2$ and $d = 1$ from Mitra's example. Equality holds, but the corresponding values aren't. There are more than that counterexample.

Easiest counterexample: Let $a = d = 1$ and $b = c = 0$. Then, equality holds, but $a \neq c$ and $b \neq d$.

I am pretty sure that you can find another counterexample. ;)

Answer 2

a=2 b=4 c=3 d=1 is a counterexample ?

Answer 3

As you can see, there are many counterexamples. I won't add any, except to point out that we have problems when both $b$ and $d$ are perfect squares.

So what we can say is this: provided at least one of $b, d$ is not a perfect square, then yes, $$a + \sqrt b = c + \sqrt d \iff a = c\;\;\text {and} \;\;b = d$$

Answer 4

If $a+\sqrt{b}=c+\sqrt{d}$, then $(a-c)+\sqrt{b}=\sqrt{d}$, so it clearly suffices to study when is it possible to have $$a+\sqrt{b}=\sqrt{d}.$$ If $b,d$ are arbitrary positive real numbers, then given any $a$ and $b$ such that $a+\sqrt{b}\geq 0$, then there is a $d\geq 0$, namely $d=(a+\sqrt{b})^2$, such that $\sqrt{d}=a+\sqrt{b}$. So I will assume from now on that the OP meant $a\in\mathbb{Q}$ and $b,d\in \mathbb{Q}^{\geq 0}$.

Suppose $a+\sqrt{b}=\sqrt{d}$ holds. Then, by squaring both sides we obtain $$a^2+2a\sqrt{b} + b = d,$$ If $a=0$, then $b=d$. Otherwise, if $a\neq 0$, then
$$\sqrt{b}=\frac{d-a^2-b}{2a}\in \mathbb{Q}.$$ In particular, $b$ must be a perfect square, and $a+\sqrt{b}\geq 0$.

Conversely, if $b$ is a perfect square, say $b=B^2$ for some $B\geq 0$, so that $B=\sqrt{b}\in \mathbb{Q}$, and $a+\sqrt{b}=a+B\geq 0$, then one can take $d=(a+B)^2$, so that $$a+\sqrt{b}=a+B=\sqrt{d}.$$ Hence, $a+\sqrt{b}=\sqrt{d}$ has a solution if and only if $b$ is a perfect square, $b=B^2$ for some $B\geq 0$, and $a+B\geq 0$, or $a=0$ and $b=d$.

Thus, $a+\sqrt{b}=c+\sqrt{d}$ has a solution if and only if $b$ is a perfect square, $b=B^2$ for some $B\geq 0$, and $(a-c)+B\geq 0$ (so that $d=((a-c)+B)^2$ is also a perfect square), or $a=c$ and $b=d$.

Example: In the example by David Mitra in the comments, $$1+\sqrt{4}=2+\sqrt{1}$$ we have $b=4=2^2$, so $B=2$, and $(1-2)+2=1\geq 0$.