As in the title. I have no idea how to deal with such equations, I'm completely new to this topic.
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Did you factor the polynomials on both sides ? – Peter Feb 18 '16 at 20:43
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It doesn't help very much. Or I've made a mistake. – Feb 18 '16 at 20:46
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1How does the factorization of your polynomials look like? – Roland Feb 18 '16 at 20:50
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HINT.- This problem is hard. I give here three solutions and I leave open the question that if these are the only ones.
Remark first that RHS is always even so $y$ must be even (because if not then RHS it would be odd).
Putting $$x(x-1)(x^2-x-1)=y(y^3+3y+1)$$ leads easily to the two solutions
$$(x,y)\in \color{red}{\{(0,0),(1,0)\}}$$ The problem is now a possible non-trivial solution
From the following table in $\mathbb F_7$
\begin{array}{|c|c|}\hline a&1&2&3&4&5&6&0\\ \hline a^2&1&4&2&2&4&1&0\\\hline a^3&1&1&6&1&6&6&0\\ \hline a^4&1&2&4&4&2&1&0\\\hline\end{array}
we get for $f(x)=x^4-2x^3+x$ and $g(y)=y^4+3y^2+y$
\begin{array}{|c|c|}\hline a&1&2&3&4&5&6&0\\ \hline f(a)&0&2&2&6&2&2&0\\\hline g(a) &5&2&6&0&5&3&0\\\hline\end{array}
It follows, modulo $7$,
$$f(x)=g(y)\iff (x,y)\in \{(2,2),(3,2),(5,2),(6,2)\}\cup \{(4,3),(0,0),(1,0),(1,4)\}$$
Among these “local” solutions, we have seen $(0,0)$ and $(1,0)$ correspond to “global” ones indeed.
Thus the candidate solutions are of the form
$x=2+7t\space\space\text {and}\space \space y=2+14s\\x=3+7t\space\space\text {and}\space\space y=2+14s\\x=5+7t\space\space\text {and}\space\space y=2+14s\\x=6+7t\space\space\text {and}\space\space y=2+14s\\x=4+7t\space\space\text {and}\space\space y=3+14s\\x=1+7t\space\space\text {and}\space\space y=4+14s$
I have just get the solution $(x,y)=\color{red}{(3,2)}$ of the form $(3+7t,2+14s)$ and I guess (after pay it some attention, of course) the three found solutions are the only ones. I invite you to finish the complete answer to this difficult question.
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