<s>[INST] How to find all the integer pairs $(x,y)$ satisfying the equation $x^{17}+6 x^{14}+2 x^{5}=y !+2 $ We are going to solve the equation
$$x^{17}+6 x^{14}+2 x^{5}=y !+2 \tag*{(*)}$$
by the  Fermat Little Theorem:
$$x^3\equiv x \quad  \pmod 3 \quad \forall x\in Z.$$
Assume that $y \geqslant 3$.
Applying the  Fermat Little Theorem to  $(*)$  in modulo $3$ yields
$$x^{5} \equiv x^3\cdot  x ^2\equiv x^3 \equiv x$$
$$x^{17} \equiv\left(x^{5}\right)^{3} \cdot x^2  \equiv x^{3}\cdot x^2\equiv x\\y !=0 $$
Expressing the equation $(*)$ in modulo $3$ yields
$$ x+0+ 2x\equiv 2 \quad(\bmod 3) \\ 0\equiv 2 \quad(\bmod 3) $$
which is a contradiction.
$$\boxed{ \textrm{There is no such integer pairs for }y\geqslant 3.}$$

Now let’s check for the integer pairs
$(x,y)$ for $y=0,1,2$.
When $y=2$,
$$0 \equiv x+0+2x\equiv 2!+2 \equiv 1   \pmod 3,  $$
which is a contradiction.
$\therefore$ there is no such integer solution.

When $y=0,1$, $(*)$ becomes
$$f(x)=x^{17}+6 x^{14}+2 x^{5}-3=0$$
because $f(-1)=0$ but $f(1) \neq 0$ and $f(\pm 3) \neq 0$
$\therefore x=-1$ is the ONLY integer solution.

Now we can conclude that the integer pairs $(x, y)$ satisfying $(*)$ are only $(-1,0)$ and $(-1,1)$.
My Question
Is there simpler solutions?
 [/INST] You could also note that $y!+2 \pmod 8 \in \{2,4\}$ for any $y \ge 2$, which is even, so $x$ must be even. But then the RHS would be a multiple of $x^5$ and thus as $2|x^5$, a multiple of $32$. So then the RHS $x^{17}+6x^{14}+2^{x^5}$ must be $0 \pmod{8}$ $\not \in \{2,4\}$....
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