n=x^3+y^3+z^3-3xyz=u(u^2-3v).

The property u is a multiple of 3, which is equivalent to that u 2-3v is a multiple of 3.

We have identities.

k^3+k^3+(k+ 1)^3-3k^2*(k+ 1)=3k+ 1，

k^3+k^3+(k- 1)^3-3k^2*(k- 1)=3k- 1，

k^3+(k- 1)^3+(k+ 1)^3-3k(k- 1)(k+ 1)=9k，

Therefore, integers that are not multiples of 3 or multiples of 9 (for example, 1, 5,2014) have the property p.

Essentially, an integer that is a multiple of 3 but not a multiple of 9 (for example, 20 13) does not have the property p.

The following is the initial answer process.

(1)n= 1, then x+y+z= 1, x 2+y 2+z 2-xy-yz-zx =1.

It is easy to know that xy+yz+zx=0,

X=y=0, z= 1 satisfies the above equation, and 1 has the property p.

(2)n=5, represented by property 2, u 2-3v =1,u = 5, v = 8,

Substitute z=5-x-y into the above formula to get xy+(x+y) (5-x-y) = 8, ① xy.

(3/4)(x+y)^2-5(x+y)+8<； =0,

8/3 & lt； = x+y & lt； =4,

So x+y=3 or 4.

When x+y=3, ① becomes xy=2, with integer solution x = 1, y = z = 2, and 5 has property p.

(3)n = 20 13 = 3 * 1 1 * 6 1，

U, u 2-3v, one is a multiple of 3 and the other is not a multiple of 3, which is impossible. 20 13 has no attribute p.

(4)n=20 14=2* 19*53。

(i)u^2-3v= 1,u=20 14.v= 1352065,②

Imitation, (x+y) 2/4+(x+y) (2014-x-y) > =1352065,

(3/4)(x+y)^2-20 14(x+y)+ 1352065<； =0,

1342<= x+y & lt； =4030/3,

So x+y= 1342 or 1343.

When x+y= 1342, ② becomes xy=45024 1,

X = Y = 67 1, Z = 672, 20 14 has the property p.

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