there is no integer $x \geq 2$ such that: $a+b=(a-b)x^2$

Question

I was thinking about this problem for a while and I can't get an answer even that seems really easy:

Let $a,b$ be positive integers such that $\left\lceil\dfrac{a}{4}\right\rceil \leq b \leq \left\lfloor\dfrac{a}{2}\right\rfloor$. Then, there is no integer $x \geq 2$ such that the next equaction hold:

$$a+b=(a-b)x^2$$

I have tried a lot of things but I can't even prove it. I should say that this is no homework or anything, is just a problem I've been thinking for too much time.

Answer 1

Since $a, b$ are integers, $\left\lceil \dfrac{a}{4}\right\rceil\le b\le \left\lfloor \dfrac{a}{2}\right\rfloor$ is the same as $\dfrac{a}{4}\le b\le\dfrac{a}{2}$.

$x\ge 2\implies x^2\ge 4$ so $(a-b)x^2\ge 4(a-b)$ (notice $a-b$ is positive since $b< 2b\le a$).

If $a+b = (a-b)x^2$, then $a+b\ge 4(a-b)$, thus $5b\ge 3a$ and $b\ge \dfrac{3}{5} a\ge \dfrac{a}{2}$. Contradiction!

Answer 2

Rewriting the equation, we can write $b = \frac{x^2-1}{x^2+1}a$. Let $f(x) = \frac{x^2-1}{x^2+1} = 1 - \frac{2}{x^2+1}$. For $x \geq 2$, we have $3/5 \leq f(x) <1$. So, $b= f(x)a \geq \frac{3}{5}a$.

It suffices to prove that we have $\frac{3}{5}a > \frac{1}{2}a$ which is always true for $a>0$.

Answer 3

If $a,b$ are real numbers with $0<b \leq \dfrac{a}{2}$, then $\frac{a+b}{a-b}\leq 3$ (*).

Proof. Since $b \leq \frac{a}{2}<a$, we have $a-b>0$. Hence multiplying by $a-b$ we see (*) is equivalent to $4b\leq 2a$, which is equivalent to $b \leq \dfrac{a}{2}$. $\square$

Note: We didn't need the first inequality $\left\lceil\dfrac{a}{4}\right\rceil \leq b$ at all, as well as the fact of there being a perfect square or integers, any $(a+b)\leq t(a-b)$ will have $t \leq 3$, in your case $t=x^2$.