Solution of: Let $k,n \in \mathbb{N}$ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $ Let $k,n \in \mathbb{N}$, $k,n < 101 $ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $.
How can I solve this and find $n$ and $k$ in an analytic way without trying some solutions?

solving the given equation $$n^2-n=2k^2-2k$$ for $n$ we get
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}$$
or
$$n=\frac{1}{2}-\sqrt{\frac{1}{4}+2k^2-2k}$$
now you Can solve the inequality
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}<101$$
writing the inequality in the form
$$\sqrt{\frac{1}{4}+2k^2-2k}<101-\frac{1}{2}$$
squaring this we get
$$2k^2-2k<101^2-101$$ and from here we get the interval
$$\frac{1}{2}\left(1-\sqrt{20201}\right)<k<\frac{1}{2}\left(1+\sqrt{20201}\right)$$
