Trying to show that an equation in two variables implies that the variables are the same

Question

Let $a ,b$ be non-zero integers such that $\frac{a^2 + \sqrt{2}}{a} = \frac{b^2 + \sqrt{2}}{b}$. Prove that $a=b$.

My proof:

1. Assume that $a \neq b$. Then, $a - b \neq 0$, implying that one can divide by $a-b$.
2. The given expression simplifies like so

$$ a^2b +b\sqrt2=ab^2+a\sqrt2 \\ ab(a-b) = \sqrt2(a-b) \\ ab = \sqrt2 $$

In the third step here, we have divided by $a-b$.

3. The conclusion $ab = \sqrt2$ is absurd. The product of two non-zero integers cannot be $\sqrt2$.

$\implies$ the initial assumption is false. $a = b \text{ QED.}$

I think something is wrong with my proof.

1. Have I taken into account during the assumption that $a$ and $b$ are integers?

Answer 1

Indeed, as the comments indicated, you actually did use the fact that $a, b$ are integers in your proof, since the product of two integers cannot equal $\sqrt{2}$.

To illustrate, note that if $a = \sqrt{2}$ and $b = 1$, the equality should hold, and in fact it does, since $$\frac{(\sqrt{2})^2 + \sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2} + 1)}{\sqrt{2}} = 1 + \sqrt{2} = \frac{1^2 + \sqrt{2}}{1}.$$ So if the condition that $a, b$ are integers is relaxed, it is possible for $a \ne b$.

Answer 2

Great job. here is a more calculus approach.

Let $f(x) = x + \frac{\sqrt2}{x}$ which is an odd function, it is positive when $x$ is positive and negative when $x$ is negative.

$f'(x) =1-\frac{\sqrt2}{x^2}$, it increases on $(\sqrt{2}, \infty)$.

We just have to check that $f(1) \le f(\sqrt{2})$ which is true since $f(1)=f(\sqrt2)=1+\sqrt2$