MHB Prove Existence of $n$ for Integer $\sqrt{n^3+xn^2+yn+z}$

[anemone]

Prove that for any integers $x,\,y,\,z$, there exists a positive integers $n$ such that $\sqrt{n^3+xn^2+yn+z}$ is not an integer.

 

[anemone]

Prove that for any integers $x,\,y,\,z$, there exists a positive integers $n$ such that $\sqrt{n^3+xn^2+yn+z}$ is not an integer.

Solution of other:

Spoiler

Let $f(n)=n^3+xn^2+yn+z$. Find an integer $a$ large enough that $8a^3$ exceeds both $(x^2-4y)a^2-xa-z+\dfrac{1}{4}$ and $(4y-x^2)a^2-xa+z-\dfrac{1}{4}$, i.e.

$\begin{align*}f(4a^2)&=(4a^2)^3+x(4a^2)^2+y(4a^2)+z\\&=64a^6+16a^4x+4a^2y+z\end{align*}$

$\begin{align*}\left(8a^3+xa-\dfrac{1}{2}\right)^2&=64a^6+16a^3\left(xa-\dfrac{1}{2}\right)+x^2a^2-xa+\dfrac{1}{4}\\&=64a^6+16a^4x-8a^3+x^2a^2-xa+\dfrac{1}{4}\end{align*}$

$\begin{align*}\left(8a^3+xa+\dfrac{1}{2}\right)^2&=64a^6+16a^3\left(xa+\dfrac{1}{2}\right)+x^2a^2+xa+\dfrac{1}{4}\\&=64a^6+16a^4x+8a^3+x^2a^2+xa+\dfrac{1}{4}\end{align*}$

and

$-8a^3+x^2a^2-xa+\dfrac{1}{4}<4a^2y+z$ which gives $(x^2-4y)a^2-xa-z+\dfrac{1}{4}<8a^3$

whereas

$8a^3+x^2a^2+xa+\dfrac{1}{4}>4a^2y+z$ which gives $8a^3>(4y-x^2)a^2-xa+z-\dfrac{1}{4}$

therefore we get

$\left(8a^3+xa-\dfrac{1}{2}\right)^2<f(4a^2)<\left(8a^3+xa+\dfrac{1}{2}\right)^2$

So if $f(4a^2)$ is a square, then $f(4a^2)=(8a^3+xa)^2$, i.e., $(x^2-4y)a^2=z$. This cannot hold for more than one value of $a$ unless $4y=x^2$ and $z=0$. But then $f(n)=n\left(n+\dfrac{x}{2}\right)^2$; this is not a square if $n$ is a non-square different from $-\dfrac{x}{2}$.

 

[kaliprasad]

Prove that for any integers $x,\,y,\,z$, there exists a positive integers $n$ such that $\sqrt{n^3+xn^2+yn+z}$ is not an integer.

this can be solved using modular arithmetic as below

Spoiler

If we can show that it cannot be for n = 1,2,3,4 then we are through
for n = 1 we have $f(1) = 1 + x + y+ z\cdots(1)$
for n= 2 we have $f(2) = 8 + 4x + 2y + z\cdots(2)$
$f(3) = 27 + 9x + 3y + +z\cdots(3)$
$f(4) = 64 + 16x + 4y + z\cdots(4)$
if f(1) and f(3) are perfect squares then from (1) and (3) we get $26+ 8x+ 2y \equiv 0 \pmod 4$ or $2y \equiv 2 \pmod 4\cdots(5)$
if f(2) and f(4) are perfect squares then from (2) and (4) we get $56+ 12x+ 2y \equiv 0 \pmod 4$ or $2y \equiv 0 \pmod 4\cdots(6)$
(5) and (6) cannot be satisfied at the same time so for n = 1 to 3 it cannot be a perfect square