I think this problem should be post in High School Math. (Even if it's 2001 Moldova MO Grade 10 P1.) $x=\frac{1}{5},y=\frac{2}{5},z=\frac{3}{5},t=\frac{4}{5}$

Can you prove that it's the only solution?

Our goal is to make them into squares notice that: x^2 + y^2/4 - xy is a square 3y^2/4 + z^2/3 - yz is a square 2z^2/3 + 3t^2/8 -zt is a square 5t^2/8 -t + 2/5 is a square Sum them up, we have L.H.S ≥ R.H.S When L.H.S = R.H.S, we have all the squares is 0. Hence, x = 1/5 y = 2/5 z = 3/5 t = 4/5 is the only solution.