The number of ordered pairs (r,k) for which 6^35C_r = (k^2-3)^36C_r+1, where k is an integ…

Question

The number of ordered pairs $(r,k)$ for which $6\cdot{}^{35}C_r = (k^2-3)\cdot{}^{36}C_{r+1}$, where $k$ is an integer, is:

Accepted Answer

Using $^{36}C_{r+1} = {}^{35}C_r\cdot\dfrac{36}{r+1}$:

$$6\cdot{}^{35}C_r = (k^2-3)\cdot{}^{35}C_r\cdot\frac{36}{r+1} \implies \frac{6(r+1)}{36}=k^2-3 \implies k^2 = 3+\frac{r+1}{6}$$

For $r\in\{0,\ldots,35\}$ (valid range for both $^{35}C_r$ and $^{36}C_{r+1}$), integer $k$ requires $r=5$ ($k^2=4,\ k=\pm2$) or $r=35$ ($k^2=9,\ k=\pm3$).

$$\text{Total ordered pairs} = 4$$

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