The number of elements in the set S = (r,k) : k Z and 36r+1 = 635rk^2-3 is:

Question

The number of elements in the set $S = \left\{(r,k) : k \in \mathbb{Z} 	ext{ and } \binom{36}{r+1} = \dfrac{6\binom{35}{r}}{k^2-3}ight\}$ is:

Accepted Answer

Using $\dbinom{36}{r+1} = \dbinom{35}{r}\cdot\dfrac{36}{r+1}$, the given equation becomes:
$$\binom{35}{r}\cdot\frac{36}{r+1} = \frac{6\binom{35}{r}}{k^2-3}$$

Since $\binom{35}{r} \ne 0$:
$$\frac{36}{r+1} = \frac{6}{k^2-3} \implies 36(k^2-3) = 6(r+1) \implies r = 6k^2 - 19$$

Since $0 \le r \le 35$: $0 \le 6k^2-19 \le 35 \implies \dfrac{19}{6}\le k^2 \le 9$, so $k^2 = 4$ or $k^2=9$.

- $k^2=4 \Rightarrow k=\pm2,\ r=6(4)-19=5$: pairs $(5,2),(5,-2)$
- $k^2=9 \Rightarrow k=\pm3,\ r=6(9)-19=35$: pairs $(35,3),(35,-3)$

Total elements $= 4$

Solutions · 1