Is It Really A Symmetry?

We have $p = x^3 + y^3 + 1 - 3xy$ .

Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a+b+c) \times ( a^2 + b^2 + c^2 - ab - ab - bc )$ and substituting $a = x, b = y, c = 1$ , we see that

$p = (x+y+1)( x^2 + y^2 + 1 - x - y - xy )$

Since $x + y + 1 > 1$ and $p$ is a prime, hence in the above factorization of $p$ , we must have $(x^2 + y^2 + 1 - x - y - xy ) = 1$ .

This in turn implies that $(x-y)^2 + (x-1)^2 + (y-1)^2 = 2$ . Since these are perfect squares, we must have 2 of them equal to 1, and 1 of them equal to 0. This gives us solutions of $(x,y) = (2,2), (2,1), (1,2)$ .

We can check that the largest value of $p$ corresponds to $(x,y) = (2,2)$ , which gives us the value fo $p = 5$ .

Take 1 to LHS, And show that (Use the algebraic identity: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$ )

$p= (x+y+1)(x^{2} - x(1+y) + y^{2} -y +1 )$ .

Now since $p$ is prime either of the brackets on RHS ought to be one.

But since x & y are positive integers $x+y +1\neq 1$ .

Hence $x^{2} - x(1+y) + y^{2} -y$ = $0$

Since x & y are real $D\geq 0$ Hence ;

$x,y \epsilon [-0.12, 2.15]$ .

Hence $x,y = {1,2}$ .

Thus $p_{largest} = \boxed{5}$

Note : One must check weather for a value finded out of $y$ must give what is finded out of $x$ or vice-versa.