<s>[INST] Prove that the equation $\ 5x^4 + x − 3 = 0\ $ has no rational solutions. I'm locked at $\ x\left(5x^3 + 1\right) = 3$. Not too sure where to go from there but I'm getting the feeling it's really really obvious..
 [/INST] Let us suppose  $x = a / b$ is a rational solution, where a, b are integers, mutually prime. Then: 
$$
5(a/b)^4 + a/b -3 = 0
$$
and after multiplying by $b^4$:
$$
5a^4 + ab^3 -3b^4=0
$$
or equivalent
$$
a(5a^3 + b^3) = 3b^4 (*)
$$
The last equality implies that $a$ divides $3b^4$. Because $a$ and $b$ are coprime, then $a$ must divide $3$, so $a \in \{-1, 1, -3, 3\}$.
Case 1
If $a = 1$ then the last equality becomes:
$$
5 + b^3 = 3b^4
$$ 
or equivalent
$$
5 = b^3(3b - 1)
$$
Therefore $b^3$ divides $5$, but the only cube numbers to divide $5$ are $-1$ and $1$ so $b \in \{-1, 1\}$. Both values for $b$ don't satisfy the last equality, so this case doesn't lead to a solution.
Case 2 
If $a=3$ then $(*)$ becomes:
$$
3(135 + b^3)=3b^4
$$
or equivalent
$$
135 = b^3(b-1)
$$
The last equality is not possible because $b^3(b-1)=b^2b(b-1)$ and $b(b-1)$ is always an even number, therefore $b^3(b-1)$ is even, while $135$ is odd number.
The other cases ($a=-1$ and $a=-3$) are pretty similar to the above ones.
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