integral solution of n such that [nC10][(n+1)C10] is a perfect square?

Question

Number of integral solution of n such that [nC10][(n+1)C10] is a perfect square? n>10 :options: A) 0 B) 1 C) 2 D) 3

if there is any then give the values of n

Answer 1

$$C_{10}^nC_{10}^{n+1}=C_{10}^n\cdot\frac{(n+1)!}{10!(n-9)!}=C_{10}^n\cdot\frac{(n+1)n!}{10!(n-10)!(n-9)}=\left(C_{10}^n\right)^2\left(1+\frac{10}{n-9}\right)$$

$\dfrac{10}{n-9}$ is an integer when $n-9=\pm1$, $\pm2$, $\pm5$ or $\pm 10$.

$ 1+\dfrac{10}{n-9}$ is not a perfect square in all cases.

Answer 2

Following CY Aries’s method:

We wish to know when $(n+1)/(n-9)$ is a square, equivalently when $(n+1)(n-9)$ is a square. Taking $k=n-4$, we are looking for solutions to $k^2 - 25 = m^2$, or $(k-m)(k+m)=25,$ where $k\ge 6$. The only factorization of $25$ that yields such solutions is $1\cdot 25$, which results in $k=13, m=12$. This corresponds to $n=17$ as the unique solution greater than $9$.

Answer 3

$$C(n,10)C(n+1,10)=\frac{n!}{10!(n-10)!}.\frac{(n+1)!}{10!(n-9)!}=C^2(10,n)\frac{n+1}{n-9}$$

Let $\frac{n+1}{n-9}= (\frac {p}{q})^2$ where gcd(p,q)=1 and $p,q \in Z$ and $p>q$

Now, $n+1=kp^2$ and $n-9=kq^2$ where $k \in Z$

Subtracting eqn 2 from 1, we get $10=k(p^2-q^2)$

$$1.2.5=k(p+q)(p-q)$$

Now, $p+q, p-q$ can be either both odd or both even

So the only solution possible is when $k=2, p+q=5 ,p-q=1$

So, we get $k=2, p=3, q=2 \Rightarrow n=17$

So, the only solution possible is n=17