Integer solution

For every prime p, does there exists integers x1, x2 and x3 (0x1,x2,x3cp1/3 and c is some large constant) such that p122cp1/3f(x1,x2,x3)p12, where, f(x1,x2,x3)=x1+x2+x3+2(x1x2+x2x3+x3x1)+4x1x2x3.

Answer

I doubt that the lower bound p122cp1/3 holds for all p. Here is a proof for the weaker bound p12cp1/2.

First of all, the inequality p12cp1/2f(x1,x2,x3)p12 is essentially equivalent to
pO(p1/2)(2x1+1)(2x2+1)(2x3+1)p.

Let us take y=p1/312. Then
p(2y+1)3<(p1/3+2)3<p+O(p2/3).

Define
z=((2y+1)3p4(2y+1))1/2.
Clearly, we have z=O(p1/6).

Now, let us take x1=y, x2=yz, x3=y+z so that
(2x1+1)(2x2+1)(2x3+1)=(2y+1)34z2(2y+1)
<(2y+1)34z2(2y+1)=p.
On the other hand, we have
(2x1+1)(2x2+1)(2x3+1)=(2y+1)34z2(2y+1)
>(2y+1)34(z+1)2(2y+1)=pO(p1/2).

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

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