For every prime p, does there exists integers x1, x2 and x3 (0≤x1,x2,x3≤⌊cp1/3⌋ and c is some large constant) such that p−12−⌊2cp1/3⌋≤f(x1,x2,x3)≤p−12, where, f(x1,x2,x3)=x1+x2+x3+2(x1x2+x2x3+x3x1)+4x1x2x3.

**Answer**

I doubt that the lower bound p−12−2cp1/3 holds for all p. Here is a proof for the weaker bound p−12−cp1/2.

First of all, the inequality p−12−cp1/2≤f(x1,x2,x3)≤p−12 is essentially equivalent to

p−O(p1/2)≤(2x1+1)(2x2+1)(2x3+1)≤p.

Let us take y=⌈p1/3−12⌉. Then

p≤(2y+1)3<(p1/3+2)3<p+O(p2/3).

Define

z=((2y+1)3−p4(2y+1))1/2.

Clearly, we have z=O(p1/6).

Now, let us take x1=y, x2=y−⌈z⌉, x3=y+⌈z⌉ so that

(2x1+1)(2x2+1)(2x3+1)=(2y+1)3−4⌈z⌉2(2y+1)

<(2y+1)3−4z2(2y+1)=p.

On the other hand, we have

(2x1+1)(2x2+1)(2x3+1)=(2y+1)3−4⌈z⌉2(2y+1)

>(2y+1)3−4(z+1)2(2y+1)=p−O(p1/2).

**Attribution***Source : Link , Question Author : Shivaraj , Answer Author : Max Alekseyev*