I have the following problem:
Prove that there exists some $x\in\mathbb{N}$ with $x>7$, such that the following equation has no solution $(k,n)$, where $k,n\in\mathbb{N}$ with $k>3$ and $n>4$:
$$k(k+1)(kx+3-x)=3n(nx+2-x)$$
I have no idea how to start on this problem, is there a smart way to choose the value of $x$? My professor did mention choosing a smart value of $x$, but I do not know what a smart way is to choose that smart value.
The equation is linear in $x$. Expanding and collecting terms yields $$\big(3n(n-1)-k(k+1)(k-1)\big)x=3k(k+1)-6.$$ Note that all terms are divisible by $6$, and we can express the fractions as binomial coefficients: $$\left(\binom{n}{2}-\binom{k+1}{3}\right)x=\binom{k+1}{2}-1.$$ From here you can use Pascal's rule to find a 'smart' value of $x$.