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..
 A: Explanation of user1442 's answer: If $A,B$ are integers and $A/B$ is in lowest terms and $A/B$ is a solution then [1] $5A^4+AB^3=3B^4$ is divisible by $A$, so $3$ is divisible by $A$, and [2] $3B^4-AB^3=5A^4$ is divisible by $B$, so $5$ is divisible by $B$.
A: Check whether$x=1,-1,3,-3,\frac{1}{5},\frac{-1}{5},\frac{3}{5},\frac{-3}{5}$ satisfies the equation or not.If they do not satisfy then there is no other rational root possible.
A: Given a polynomial $a_nx^n+\cdots+a_1x+a_0\in\Bbb C[X]$, every its possible rational roots are of the type $a/b$, where $a|a_0$ and $b|a_n$.
In your case $a_n=5$ and $a_0=-3$, thus all possible rational roots are $\pm1,\pm3,\pm\frac15,\pm\frac35$; just plug these values inside your polynomial and check they don't make it vanish.
A: $There\,\, is\,\,a\,\,general\,\, result$.
$Let\,\, f\, be\,\, a\,\, polynomial\,\,\, with \,integer\,\,coefficients\,\,say\,\,$ $f(x)=a_{n}x^n+a_{n-1}x^n+...+a_{0}$
$If\,f\,has\,rational\,roots\,then\,it\,is\,of\,the\,form\,\,\,\,p/q$. 
$where\,p\,divides\,\,a_{0}\,and\,q\, divides\,\, a_{n}$. 
A: 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.
A: Any rational solution $x=\frac{p}{q}$, where $p$ and $q$ are coprime integers, would have to satisfy $\frac{5 p^4}{q^4}+\frac{p}{q}-3=0$. Multiplying both sides by $q^4$ we get
$$
5 p^4+p q^3-3 q^4=0.
$$
If we look at this equation modulo two we get the following
$$
p+p q+q\equiv 0 \pmod{2}.
$$
Which is only true for $p\equiv q \equiv 0 \pmod{2}$, which contradicts the fact that $p$ and $q$ are coprime.
