yes exist. the idia hers is conseder $x^{2}+y^2+z^2+u^2+v^2=xyzuv-65$ like a equation in such variable and observing that if $(x,y,z,v;u)$ soulition then $(yzuv-x,x,y,z,u,v)$ soulition

First notice that there is a solution in positive integers given by $ (x,y,z,u,v) = (1,2,3,4,5)$ Suppose we have a solution in positive integers, since the equation is symmetric we can reorder so that $ x$ is less than or equal to $ y,z,u$ and $ v$. For fixed $ y,z,u,v$, the equation is a quadratic in $ x$ and the sum of its roots is $ yzuv$ Since we have one root $ x_1 = x$ already we can find the second $ x_2 = yzuv - x$ $ x_2$ is also an integer and it is larger than $ x$ provided at least one of the other integers is at least 2. Therefore given a solution in positive integers we can always find another with a larger smallest number. This means that the size of the smallest element is unbounded and there are solutions where all elements are greater than 1998.

Can someone explain why the size of the smallest element is unbounded? I get that if $x=1$ is a solution, then $yzuv-1=2*3*4*5-1=119$ must also be possible for $x$. But how does this lead to the conclusion that the size of the smallest element is unbounded? And how do we know, given a smallest, element, there must exist integers $y, z, u, v$ such that the equation is satisfied?

It's because the quadratic PhilG made is w.r.t the smallest of $x,y,z,u,v$, and the second root of the quadratic is clearly greater than the root we currently have. This can be applied every single time we have a solution to get a solution with a strictly higher minimum element, and the result follows.