The Polynomial t²(4x−n)² −2xtn Is Always Admitting 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 $n\ge 2$? We show in this article that it is indeed the case.

2. Problem

We show that for every integer $n\ge 2$, there exist positive integers t and x such that

$$E=t^2{(4x-n)}^2-2xtn$$

is a perfect square.

Proof. 

We consider four cases depending on the residue of n modulo 4.

Case 1:$n=4k$ (so $n\equiv 0(mod4)$), with $k\ge 1$ because $n\ge 2$.

Let

$$x=k+1,t={\displaystyle \frac{k(k+1)}{2}}.$$

Then $4x-n=4(k+1)-4k=4$, and

$$E=t^2·4^2-2·(k+1)·t·\left(4k\right)=16t^2-8k(k+1)t.$$

Substituting $t={\displaystyle \frac{k(k+1)}{2}}$, we obtain

$$E=16{\left({\displaystyle \frac{k(k+1)}{2}}\right)}^2-8k(k+1)·{\displaystyle \frac{k(k+1)}{2}}=4k^2{(k+1)}^2-4k^2{(k+1)}^2=0.$$

Thus $E=0$, which is a perfect square. Moreover $x=k+1\ge 2$ and $t={\displaystyle \frac{k(k+1)}{2}}\ge 1$ are positive integers.

Case 2:$n=4k+2$ (so $n\equiv 2(mod4)$), with $k\ge 0$ (for $n=2$, we have $k=0$).

Let

$$x=k+1,t=(2k+1)(k+1).$$

Then $4x-n=4(k+1)-(4k+2)=2$, and

$$E=t^2·2^2-2·(k+1)·t·(4k+2)=4t^2-2(k+1)t·2(2k+1)=4t^2-4(2k+1)(k+1)t.$$

Substituting $t=(2k+1)(k+1)$, we get

$$E=4{(2k+1)}^2{(k+1)}^2-4(2k+1)(k+1)·(2k+1)(k+1)=4{(2k+1)}^2{(k+1)}^2-4{(2k+1)}^2{(k+1)}^2=0.$$

Hence $E=0$. For $k=0$ we have $x=1$, $t=1$; for $k\ge 1$ both are clearly positive.

Case 3:$n=4k+3$ (so $n\equiv 3(mod4)$), with $k\ge 0$ (for $n=3$, $k=0$).

Let

$$x=k+1,t=2(4k+3)(k+1).$$

Then $4x-n=4(k+1)-(4k+3)=1$, and

$$E=t^2·1^2-2·(k+1)·t·(4k+3)=t^2-2(k+1)t(4k+3).$$

Substituting $t=2(4k+3)(k+1)$, we obtain

$$E=4{(4k+3)}^2{(k+1)}^2-2(k+1)·2(4k+3)(k+1)·(4k+3)=4{(4k+3)}^2{(k+1)}^2-4{(4k+3)}^2{(k+1)}^2=0.$$

Thus $E=0$. For $k=0$, we have $x=1$, $t=6$, which are positive.

Case 4:$n=4k+1$ (so $n\equiv 1(mod4)$), with $k\ge 1$ because $n\ge 2$ (for $n=5$, $k=1$).

Let

$$x=k,t=2k(4k+1).$$

Then $4x-n=4k-(4k+1)=-1$, so ${(4x-n)}^2=1$, and

$$E=t^2·1-2·k·t·(4k+1)=t^2-2kt(4k+1).$$

Substituting $t=2k(4k+1)$, we get

$$E=4k^2{(4k+1)}^2-2k·2k(4k+1)·(4k+1)=4k^2{(4k+1)}^2-4k^2{(4k+1)}^2=0.$$

Hence $E=0$. For $k=1$, we have $x=1$, $t=10$, which are positive.

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3. Conclusions

In every case, we have constructed positive integers t and x such that $t^2{(4x-n)}^2-2xtn=0$, which is a perfect square. Therefore the statement holds for all $n\ge 2$.