The Polynomial t²(4x − n)² − 2ntx Does Not Always Admits a Perfect Square

1. Introduction

In his article : Partial Resolution of the Erdős-Straus, Sierpiński, and Generalized Erdős-Straus Conjectures Using New Analytical Formulas, in the page 13, Philemon Urbain MBALLA pose the following question: how can we prove that the polynomial $t^2{(4x-n)}^2-2ntx$ or $4t^2{(4x-n)}^2-8ntx$ always admits a perfect square for all $n\in \mathbb{N}$, with $n\ge 2$?

2. Definitions

Definition: we mean by a positive integer a natural number strictly greater than 1.

3. Principal Resulte

Théorème .The polynomial $t^2{(4x-n)}^2-2ntx$ does not always admits a perfect square with $n\ge 2$ and $x,t$ positive integers.

Proof. 

For $n=3$, the polynomial $t^2{(4x-n)}^2-2ntx$ become:

$t^2{(4x-3)}^2-6tx$

We need to have a positive integer m such that : $t^2{(4x-3)}^2-6tx=m^2$

In this expression if we show that x or t isn’t a postive integer then the theorem 3.1 is proved.

We have: $t^2{(4x-3)}^2-6tx=m^2$ ⇔ $t^2{(4x-3)}^2-m^2=6tx$

⇔$\left(t\right(4x-3)-m)\left(t\right(4x-3)+m)=6tx$

The two expression $t(4x-3)-m$ and $t(4x-3)+m$ have the same parity because it deffirencies by 2m, or the parity of it’s product is even(because it’s product is 6tx) then $t(4x-3)-m$ and $t(4x-3)+m$ are both even.

Let: $t(4x-3)-m=2r$ then: $2r+m=t(4x-3)$ and $r(r+m)=3xt$

So: $x={\displaystyle \frac{3r(r+m)}{4r(r+m)-6r+3m}}$

Remark: We need to find $(r,m)$ positive integeres such that x also a positive integer. Now We define the function: $x:t->x\left(t\right)$ with $t=r+m$ (then: $t\ge r+1$ because $m\ge 1$)

So: $x\left(t\right)={\displaystyle \frac{3rt}{t(4r+3)-9r}}$ Or $x\ge 0$, the denominator is positive, we studie x as a function of t with fixed r.

Proof that $x:t->x\left(t\right)$ is strictly decreasing:

If we calcule the derivative of x we find:

$x^{\prime }\left(t\right)={\displaystyle \frac{-27r^2}{{(t(4r+3)-9r)}^2}}$

Then $x^{\prime }\left(t\right)<0$ for all $t>{\displaystyle \frac{9r}{4r+3}}$, then x is strictly decreasing for all $t\ge r+1$.

And the maximum of the function x is atteint in the limit of left so when $t=r+1$ Then $x(t=r+1)={\displaystyle \frac{3r(r+1)}{4r^2-2r+3}}$ this value of x does not atteint 1: $x(t=r+1)={\displaystyle \frac{3r(r+1)}{4r^2-2r+3}}$$<1$ for all $r\ge 5$,

We will prove it by studiyng the function $g\left(r\right)={\displaystyle \frac{3r(r+1)}{4r^2-2r+3}}$ for all $r\ge 5$

The derivative of g is:$g^{\prime }\left(r\right)={\displaystyle \frac{9(-2r^2+2r+1)}{(4r^2-2r+3)}}$ this function is negative for all $r\ge {\displaystyle \frac{1+\sqrt{3}}{2}}$, So g atteint its maximum on the extrem of left mean on $r=5$

$g\left(5\right)=0.967<1$ mean $x(r+1)<1$ for all $r\ge 5$

So x is not an integer for all value $r\ge 5$.

Remaine the cases when $r=1,2,3,4$:

For every value of r we solve the equation $x=k\in {\mathbb{Z}}^{+}$ with $t\ge r+1$ this is equivalent to:

$t={\displaystyle \frac{9kr}{k(4r+3)-3r}}$, we need to have t positive integer and $t\ge r+1$:

For $r=1$, $t={\displaystyle \frac{9k}{7k-3}}$ any $k\ge 1$ integer does not give $t\ge 2$

Consider: $f\left(x\right)={\displaystyle \frac{9x}{7x-3}}$, the derivative of f is: $f^{\prime }\left(x\right)=-{\displaystyle \frac{27}{{(7x-3)}^2}}<0$ for all $x>{\displaystyle \frac{3}{7}}$, so f is strictly decreasing for all $x\ge 1$

Then the maximum of f is atteint in the limit of left when $x=1$ then $f\left(1\right)={\displaystyle \frac{9}{4}}=2.25$ and when $x\to +\infty$ then $f\left(x\right)\to {\displaystyle \frac{9}{7}}\approx 1.2857$

Since ${\displaystyle \frac{9}{4}}=2.25$ and the next integer less than $2.25$ is 2, and $2>{\displaystyle \frac{9}{7}}$, we need to check for $f\left(x\right)=2$:

When $f\left(x\right)=2$ then $x={\displaystyle \frac{6}{5}}\notin \mathbb{N}$

So no integer $k\ge 1$ gives $t={\displaystyle \frac{9k}{7k-3}}$ integer and $t\ge 2$

For $r=2$, $t={\displaystyle \frac{18k}{11k-6}}$ any $k\ge 1$ integer does not give $t\ge 3$

Consider: $f\left(x\right)={\displaystyle \frac{18x}{11x-6}}$, the derivative of f is: $f^{\prime }\left(x\right)=-{\displaystyle \frac{108}{{(11x-6)}^2}}<0$, so f is strictly decreasing for all $x\ge 1$

Then the maximum of f is atteint in the limit of left when $x=1$ then $f\left(1\right)={\displaystyle \frac{18}{5}}=3.6$ and when $x\to +\infty$ then $f\left(x\right)\to {\displaystyle \frac{18}{11}}\approx 1.636$

Possible integer values of $f\left(x\right)$ are 3 and 2 because $3<3.6$ and $2>{\displaystyle \frac{18}{11}}$

We study just the value of x when $f\left(x\right)=3$ (beacuse $t\ge 3$) then $x={\displaystyle \frac{6}{5}}\notin \mathbb{N}$

So no integer $k\ge 1$ gives $t={\displaystyle \frac{18k}{11k-6}}$ integer and $t\ge 3$

For $r=3$, $t={\displaystyle \frac{27k}{15k-9}}$ any $k\ge 1$ integer does not give $t\ge 4$

Consider: $f\left(x\right)={\displaystyle \frac{27x}{15x-9}}={\displaystyle \frac{9x}{5x-3}}$, the derivative of f is: $f^{\prime }\left(x\right)=-{\displaystyle \frac{27}{{(5x-3)}^2}}<0$, so f is strictly decreasing for all $x\ge 1$

Then the maximum of f is atteint in the limit of left when $x=1$ then $f\left(1\right)={\displaystyle \frac{9}{2}}=4.5$ and when $x\to +\infty$ then $f\left(x\right)\to {\displaystyle \frac{9}{5}}=1.8$

Possible integer values of $f\left(x\right)$ are 4 and 3 because $4<4.5$ and $3>1.8$

We study just the value of x when $f\left(x\right)=4$ (beacuse $t\ge 4$) then $x={\displaystyle \frac{12}{11}}\notin \mathbb{N}$

So no integer $k\ge 1$ gives $t={\displaystyle \frac{27k}{15k-9}}$ integer and $t\ge 4$

For $r=4$, $t={\displaystyle \frac{36k}{19k-12}}$ any $k\ge 1$ integer does not give $t\ge 5$

Consider: $f\left(x\right)={\displaystyle \frac{36x}{19x-12}}$, the derivative of f is: $f^{\prime }\left(x\right)=-{\displaystyle \frac{432}{{(19x-12)}^2}}<0$, so f is strictly decreasing for all $x\ge 1$

Then the maximum of f is atteint in the limit of left when $x=1$ then $f\left(1\right)={\displaystyle \frac{36}{7}}\approx 5.142857$ and when $x\to +\infty$ then $f\left(x\right)\to {\displaystyle \frac{36}{19}}\approx 1.8947$

Possible integer values of $f\left(x\right)$ are 5 and 4 because $5<5.142857$ and $4>1.8947$

We study just the value of x when $f\left(x\right)=5$ (beacuse $t\ge 5$) then $x={\displaystyle \frac{60}{59}}\notin \mathbb{N}$

So no integer $k\ge 1$ gives $t={\displaystyle \frac{36k}{19k-12}}$ integer and $t\ge 5$

So no couple $(r,m)$ of positive integers give x as a positive integer.

So no couple (x,t) of positive integers exists such that $t^2{(4x-3)}^2-6tx$ to be a perfect square. □

4. Conclusions

We have prove that the polynomial $t^2{(4x-n)}^2-2ntx$ does not always admits a perfect square with $n\ge 2$ and $x,t$ positive integers.