The equation with integer solution: $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$

Question

I am trying to solve this equation:

Given $x,y\in \text {Z}$: Solve : $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$

Obviously, when $x=0$, it is easy to find that $y=0,y=1$ and $y=0$ then $x=0, x=-1$.

My question is how to find another solutions.

My try is factorization, but the power of $x,y$ is too big, i guess there is relation between $x,y$ so i tried to divide polynomial $x^4+x-y^4+y-2x^2y^2(x^4-y^4+1)^3$ by $xy$ or $xy-1$ (base on CAS gives solutions $x,y$ and $xy=0,xy=1$) but i failed.

I am bad at inequalities so i can't find another approach.

Every help is precious to me,thank you for reading.

Answer 1

Write $a=x+y$, $b=x-y$, $c=x^2+y^2$ then the equation becomes $$2a(bc+1)=(c-ab)(c+ab)(abc+1)^3.$$ Now if $|a|>1$ and $|bc|>1$ then by the fundamental theorem of arithmetic either $a$ or $bc+1$ must contain three copies of all the prime factors of $abc+1$. However $a$ is coprime with $abc+1$, so $bc+1$ must contain three copies of all the prime factors of $abc+1$, an impossibility. Hence there are no integer solutions other than the trivial cases.

★ Note that

$$ \begin{align} c-ab&=x^2+y^2-(x^2-y^2)=2y^2\\ c+ab&=x^2+y^2+(x^2-y^2)=2x^2\\ abc&=x^4-y^4\\ \end{align} $$

Answer 2

Given $x,y\in\mathbb Z\;,\;$ solve the following equation :

$x^4+x-y^4+y=2x^2y^2\left(x^4-y^4+1\right)^3\;.$

We can rearrange the terms in the following way :

$x+y-1=2x^2y^2\left(x^4-y^4+1\right)^3-\left(x^4-y^4+1\right)\;,$

$x+y-1=\left(x^4-y^4+1\right)\left[2x^2y^2\left(x^4-y^4+1\right)^2-1\right].\;\;\color{blue}{(*)}$

There are four possible cases :

$1)\;x=0\;,\;$

$2)\;x\ne0\;\land\;y=0\;,$

$3)\;x\ne0\;\land\;y\ne0\;\land\;x=y\;,$

$4)\;x\ne0\;\land\;y\ne0\;\land\;x\ne y\;.$

First case : $\;x=0\;.$

If $\;x=0\;,\;$ from the equation $(*)\;,\;$ we get that

$y-1=y^4-1\;,$

$y=y^4\;,$

$y\left(1-y^3\right)=0\;,$

$y=0\;\lor\;y=1\;.$

Hence, we have obtained the following two solutions :

$\begin{cases}x=0\\y=0\end{cases}\;\lor\;\begin{cases}x=0\\y=1\end{cases}\;.$

Second case : $\;x\ne0\;\land\;y=0\;.$

If $\;x\ne0\;\land\;y=0\;,\;$ from the equation $(*)\;,\;$ we get that

$x-1=-x^4-1\;,$

$x=-x^4\;,$

$x\left(1+x^3\right)=0\;,$

and, since $\;x\ne0\;,$ it follows that

$1+x^3=0\;,$

$x=-1\;.$

Hence, we have obtained the following solution :

$\begin{cases}x=-1\\y=0\end{cases}.$

Third case : $\;x\ne0\;\land\;y\ne0\;\land\;x=y\;.$

If $\;x\ne0\;\land\;y\ne0\;\land\;x=y\;,\;$ from $\;(*)\;,\;$ we get that

$2x-1=2x^4-1\;,$

$2x=2x^4\;,$

$2x\left(1-x^3\right)=0\;,$

and, since $\;x\ne0\;,\;$ it follows that

$1-x^3=0\;,$

$x=1\;.$

Hence, we have obtained the following solution :

$\begin{cases}x=1\\y=1\end{cases}.$

Fourth case : $\;x\ne0\;\land\;y\ne0\;\land\;x\ne y\;.$

If $\;x\ne0\;\land\;y\ne0\;\land\;x\ne y\;,\;$ we get that

$\begin{align} \left|x^4-y^4+1\right|&=\left|(x+y)(x-y)\left(x^2+y^2\right)+1\right|\geqslant\\ &\geqslant\left|(x+y)(x-y)\left(x^2+y^2\right)\right|-1\geqslant\\ &\geqslant2\left|x+y\right|-1=\\ &=2\big|(x+y-1)+1\big|-1\geqslant\\ &\geqslant2\left|x+y-1\right|-3\;. \end{align}$

Since $\;2x^2y^2\left(x^4-y^4+1\right)^2-1\;$ is an odd number, it results that $\;\left|2x^2y^2\left(x^4-y^4+1\right)^2-1\right|\geqslant1\;.$

From the equation $\;(*)\;,\;$ it follows that

$\begin{align} \big|x+y-1\big|&=\left|x^4-y^4+1\right|\left|2x^2y^2\left(x^4-y^4+1\right)^2-1\right|\geqslant\\ &\geqslant\left|x^4-y^4+1\right|\geqslant2\left|x+y-1\right|-3\;, \end{align}$ hence ,

$\left|x+y-1\right|\leqslant3\;.$

Now, we are going to prove that $\;x\ne-y\;.$

If $\;x=-y\;,\;$ from the equation $\;(*)\;,\;$ it would follow that

$-1=2x^4-1\;,\;$ but it is a contradiction since $\;x\ne0\;.$

Hence, it results that $\;x\ne-y\;.$

So , $\;x^2\ne y^2\;$ because $\;x\ne y\;$ and $\;x\ne-y\;.$

Consequently,

$\begin{align} \left|x^4-y^4+1\right|&=\left|\left(x^2-y^2\right)\left(x^2+y^2\right)+1\right|\geqslant\\ &\geqslant\left|\left(x^2-y^2\right)\left(x^2+y^2\right)\right|-1\geqslant\\ &\geqslant1\cdot5-1=4\;,\\ &\text{indeed }\;x^2\ne y^2\;\text{ and }\;x^2\ne0\;,\;y^2\ne0\;. \end{align}$

Moreover, from the equation $\;(*)\;,\;$ it follows that

$\begin{align} 3\geqslant\big|x+y-1\big|&=\left|x^4-y^4+1\right|\left|2x^2y^2\left(x^4-y^4+1\right)^2-1\right|\geqslant\\ &\geqslant4\cdot1=4\;, \end{align}$

which is a contradiction and it means that there is not any solution in this fourth case.

Hence, all the solutions of the equation are the following ones :

$\begin{cases}x=0\\y=0\end{cases}\;\lor\;\begin{cases}x=0\\y=1\end{cases}\;\lor\;\begin{cases}x=-1\\y=0\end{cases}\;\lor\;\begin{cases}x=1\\y=1\end{cases}.$