## Task B-4.6.

Determine all natural numbers $x$ for which the equality

$$
3 \cdot\binom{2 x^{2}-10 x+16}{x^{2}-5 x+9}=2 \cdot\binom{2 x^{2}-10 x+17}{x^{2}-5 x+7}
$$

holds.
## First Solution.

Notice that the expressions in the binomial coefficients in the given equality are similar, and let's introduce the substitution $t=x^{2}-5 x+7$. Then we have $2 t=2 x^{2}-10 x+14$, so we get

$$
3 \cdot\binom{2 t+2}{t+2}=2 \cdot\binom{2 t+3}{t}
$$

Then we have

$$
3 \cdot \frac{(2 t+2)!}{t!(t+2)!}=2 \cdot \frac{(2 t+3)!}{t!(t+3)!}
$$

After simplification, we get $3=2 \cdot \frac{2 t+3}{t+3}$

from which it follows that $3(t+3)=2(2 t+3), t=3$.

Now from $x^{2}-5 x+7=3$ we get $x=1$ or $x=4$.