Find the fraction.

Question

Let us consider a fraction whose denominator is smaller than the square of the numerator by unity. If we add 2 to the numerator and the denominator, the fraction will exceed 1/3; now if we subtract 3 from the numerator and the denominator, the fraction remains positive but smaller than 1/10. Find the fraction.

My work:

Let $x/y$ be the fraction. Thus $y + 1 = x^2$. Also,$(x+2)/(y+2) > 1/3$ and $0<(x-3)/(y-3) <1/10$.

Substituting the first equality in the second inequality, $0<(x-3)/(x-2)(x+2)<1/10$

How do I proceed after this?

Answer 1

HINT From $\frac {x+2}{y+2} \gt \frac 1 3$ you get $\frac {x+2}{x^2+1} \gt \frac 1 3$ and from here $x^2 - 3x -5 \lt 0, x \in \mathbb{N}$

Answer 2

You have an answer already. To add some detail to it, for $x \in \mathbb Z$ $$\frac{x+2}{x^2+1} > \frac13 \implies x^2-3x-5 < 0 \implies \left(x - \frac32 \right)^2 < \frac{29}4 \implies x \leqslant 4 $$

Similarly, $\displaystyle 0 < \frac{x-3}{x^2-4} < \frac1{10} \implies $

Case $x^2 > 4$: gives from $0 < \dfrac{x-3}{x^2-4} \implies x > 3$

Case $x^2 < 4$: gives from $\dfrac{x-3}{x^2-4} < \dfrac1{10} \implies (x-5)^2 < -9 \tag{$\times$}$

So we are left with $x = 4$ as the only possibility.