A linear equation $c_1x_1 + \cdots + c_sx_s = 0$ is *partition regular* if for every partition of the natural numbers into colour classes $A_1, \ldots, A_r$, there is a solution to the equation in which the variables $x_i$ are from the same colour class (a "monochromatic solution"). Rado proved that an equation of this form is partition regular if and only if some subset of the $c_i$ sum to $0$.

We could instead ask whether we can always find monochromatic solutions where the $x_i$ satisfy some additional property, such as being square (this would correspond to the the equation $c_1x_1^2 + \cdots + c_sx_s^2 = 0$ being partition regular). If an equation is to be partition regular inside the squares then it must certainly be partition regular.

Question 1Is there a partition regular equation that does not remain partition regular if we insist that all variables must be square?

Presumably the answer is yes, although nobody I've asked has known of an example. If we allow multiple equations then we do lose something by restricting to the squares: the squares have no arithmetic progressions of length $4$, but we can write down a system of three equations in five variables whose solutions are precisely arithmetic progressions of length $4$ together with their common difference, and this system of equations is partition regular.

A lot of work on these questions has come from the additive combinatorics community, using density arguments and Fourier analysis (see this recent paper by Tim Browning and Sean Prendiville, the seminar version of which prompted this question). In this setting it is conventional to assume that $c_1 + \cdots + c_s = 0$ to ensure that we cannot find dense sets with no solutions through congruence considerations. But this assumption is not necessary for the Ramsey question.

Question 2Is there an equation which is partition regular in the squares but does not have its coefficients summing to $0$?

This is another very simple question that must have been looked at before, but again I've not found anybody who knows the answer.