I have done this several times and do not find that x is an integer. I get an imaginary number. Crazy as it seems, here's what I got all 3 times I did this:. That is definitely NOT an integer!
There appears to be a mistake in the LaTeX, so that the equation is supposed to be If we let z=√x, then this is
By Descartes' rule of signs, this will have one positive real root. The rational root theorem says it will be one of 1, 5, 7, or 35. It is actually z=7. The corresponding value of x is x = z² = 7² = 49.
A graphing calculator finds the solution to the original equation easily. (I find it useful to put it in the form f(x)=0.)