Such a unique question! Ma Zhao Yu is definitely one of my favorite Singaporean problem proposers. Checking a natural solution of $P(x,y,z) = (x-y)^2 + (y-z)^2$ at $(x,y,z) = (n, n+m, n+2m)$ gives us $2m^2$. Picking any odd $m$ bounds $r \leq 2$. Now we'll show that $m^2$ will always divide $P(n, n+m, n+2m)$. Note that, if there exists two points in $\mathbb{R}^3$ such that the value of $P$ at those two points have a different sign, we can consider a path between them that avoids the line $x = y = z$. Some point on the path must attain the value $0$, which contradicts the question's condition. Hence, the polynomial is always positive (which we'll assume WLOG) or negative outside the line. Now consider the integer polynomial $Q(t) = P(n, n+t, n+2t)$ for any fixed $n$. $Q$ has exactly $1$ root at $t = 0$, and is positive elsewhere. Hence the root is of multiplicity $\ge 2$, so $x^2 \mid Q$. Thus, $m^2 \mid Q(m) = P(n, n+m, n+2m)$.

The answer as claimed above is $r = 2$. We shall show that $P(x,y,z)$ can be written as a sum of terms with two of $x-y,y-z,z-x$ as factors (counting multiplicity) which will imply the answer. We call these factors good. Write $P(x,y,z)$ as $(x-y)P_1(x,y,z) + (y-z)P_2(y,z) + P_3(z)$ then $a = b = c \implies P(a,b,c) = 0$ tells us that $P_3(z) = 0$. Now by implicit function theorem, the solutions to $P(a,b,c) = 0$ at the points $a=b=c$ should form a $2$-dimensional surface and not a line, unless $dP/dx = dP/dy = dP/dz = 0$. Then $dP/dx = 0$ tells us that $P_1(x,y,z) = 0$ for all $x = y = z$ and then we can write it in a similar form and write it as a sum of terms with a good factors. Considering $dP/dy = 0$ will similarly tell us that $P_2(y,z)$ should have a good factor as desired.

Consider a polynomial $P(x,y,z)$ in three variables with integer coefficients such that for any real numbers $a,b,c,$ $$P(a,b,c)=0 \Leftrightarrow a=b=c.$$Find the largest integer $r$ such that for all such polynomials $P(x,y,z)$ and integers $m,n,$ $$m^r\mid P(n,n+m,n+2m).$$ Proposed by Ma Zhao Yu