Do there exist quadratic trinomials $P, \ \ Q, \ \ R$ such that for every integers $x$ and $y$ an integer $z$ exists satisfying $P(x)+Q(y)=R(z)?$
Proposed by A. Golovanov
$ R(z)=az^{2}+bz+c $
$ 4aR(z)=(2az+b)^{2}+4c-b^{2} $.
in other words $ c_{1}x^{2}+c_{2}x+b_{1}y^{2}+b_{2}y+c=w^{2} $ where $ c_{i},b_{i},c $ are constants which are determined in function of the coefficients of the three polynomials.
if $ c_{1}<0 $ then for a very large $ x $ the number will be $ <0 $ false.
if $ c_{1}=0 $ then one of the polynomials will have degree $ <2 $.
if we put $ y+1 $ in the place of $ y $ then we obtain $ 2yb_{1}+b_{1}+b_{2}>x $ but if $ y $ is a constant and $ x $ tends to infinity then is false.
so such polynomials doesn't exist.
q.e.d.