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

In summary, the existence of n in the given integer expression can be proved using the rational root theorem, which states that if a polynomial equation with integer coefficients has a rational root, then that root must be a factor of the constant term divided by the leading coefficient. Proving the existence of n is important as it allows us to determine the values of x, y, and z that satisfy the equation, which can be useful in solving real-world problems and understanding mathematical concepts. However, the existence of n cannot be proved for all integer values of x, y, and z as there may be cases where the given expression does not have a rational root. Additionally, there are limitations to using the rational root theorem, as it can only be applied
  • #1
anemone
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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.
 
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  • #2
anemone said:
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:

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 $-\dfrac{x}{2}$.
 
  • #3
anemone said:
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

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
 

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

How do you prove the existence of n for the given integer expression?

The existence of n can be proved by using the rational root theorem. This theorem states that if a polynomial equation with integer coefficients has a rational root, then that root must be a factor of the constant term divided by the leading coefficient. By plugging in the given integer expression into the polynomial equation, we can determine the possible rational roots and check if they satisfy the equation.

What is the importance of proving the existence of n in this expression?

Proving the existence of n in this expression is important as it allows us to determine the values of x, y, and z that satisfy the equation. This can help in solving real-world problems, as well as in understanding the properties and relationships of different mathematical concepts.

Can the existence of n be proved for all integer values of x, y, and z?

No, the existence of n cannot be proved for all integer values of x, y, and z. There may be cases where the given expression does not have a rational root, which means that n does not exist for those values of x, y, and z. In such cases, different methods may be needed to find the values of n that satisfy the equation.

Are there any limitations to using the rational root theorem to prove the existence of n?

Yes, there are limitations to using the rational root theorem. This theorem can only be applied to polynomial equations with integer coefficients. In addition, it can only determine the possible rational roots, not the exact value of n. Other methods may be needed to find the exact value of n for the given expression.

Can the existence of n be proved for non-integer values of x, y, and z?

No, the existence of n can only be proved for integer values of x, y, and z. This is because the given expression is defined for integer values, and the rational root theorem can only be applied to polynomial equations with integer coefficients. For non-integer values, different methods may be needed to determine the existence of n.

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