Show that the equation $x^4 + 3 y^4 = 131$ has no integer solutions. I have seen the result demonstrated by considering the equation$\pmod{5}$, but was wondering if this proof, working in $\mathbb{Z}_3$, is also valid, since it seems to require less working:
$$\forall x \in \mathbb{Z}_3, x = \{0,1,2\} .$$
So, for any $x$, $x^4 = \{0,1\}$. Also, in $\mathbb{Z}_3$ we note that $3=0$, $3y^4=0$, for any $y$.
Since $131=2$, the equation $x^4 + 3 y^4 = 131$ has no solutions in $\mathbb{Z}_3$.
By definition, for any integers $x,y$
$$x^4 + 3 y^4 \equiv 131 \pmod{3} \iff x^4 + 3 y^4 -131=3k$$
for some integer $k$. Letting $k=0$, the result follows. $\square$
 A: What you did is correct, but I would not have used that language. For instance, when you say that $3=0$, I think that it would have been better to have written that $3\equiv0\pmod3$ and so on.
A: $$x^4 + 3 y^4 = 131$$
Note that remainder of the right side in dividing by $3$ is $2$.
Remainder of the left side in dividing by three is $r^4$ where $r=0, \pm 1$
Remainders do not match, so there is no  integral solution to $$x^4 + 3 y^4 = 131$$
A: I imagine if you've actually seen this done more than once via $\mod 5$ it is not because $\mod 5$ is easy, but because it is interesting.  
By Fermat's little Theorem $a^4 \equiv 0, 1\mod 5$ and determined by whether $5$ divides $a$ or not.  So $x^4 + 3y^4\equiv 0, 1, 3,4\mod 5$ and which value will uniquely determine which of $x,y$ are divisible by $5$ and which are not.  So $x^4 + 3y^4 \equiv 1 \mod 5\iff 5\not \mid x$ and $5\mid y$ (and as $3*5^4 > 131$ $y = 0$ and $x^4 = 131$). 
That we can do that is certainly interesting and informative.  Far more so then the final result we are trying to prove (As $3*5^4 > 131$ $y = 0$ and $x^4 = 131$ so there is no solution).  
But doing it $\mod 3$ is certainly easier and more intuitive.  But there is nothing particularly interesting or informative about $x^4 +3y^4 \equiv x^4 \equiv 0, 1\not \equiv 131 \mod 3$... at least nothing that can't be demonstrated by other problems.
....
More interesting would be to find all integer solutions to $x^4 + 3y^4 = 1956$.
$\mod 2$ we get $x$ are both even or odd but as $16\not \mid 1956$ they are both odd.  Not very useful.
$\mod 3$ we get $x \equiv 0 \mod 3$.  Somewhat useful, maybe we can try $\mod 9$ and get that $3y^4 \equiv 3 \mod 9$ so $3\not \mid y$.  but that's not that useful.
So far we have $x,y$ are both odd. $3|x$ and $3\not \mid  y$.  That's... not much.
But $\mod 5$ we get $x^4 + 3y^4 \equiv 1 \mod 5$ so $x^4 \equiv 1\mod 5$ and $y^4 \equiv 0 \mod 5$.  So $5\not \mid x$ and $5\mid y$.  And as $3*5^4 =  1875 < 1956 < 3*10^4 = 30000$ we have $x^4 = 1956;y=0$ of $x^4 = 1956-1875=81; |y| = 5$.  $x^4 = 1956$ is not possible.  $x^4 = 81$ means $|x| =3$.
so $(\pm 3, \pm 5)$ are the only solutions.
A: we use quadratic residue to solve this problem 
  for all $x$ belongs to integers $x^4 \equiv 0,1(mod3)$
   $131 \equiv 2(mod3)$ [LHS $\equiv2(mod3)$] and $3x^4 \equiv 0 (mod3)$ 
   and $x^4\equiv0,1 (mod3)$ for both cases[$x^4\equiv0 (mod3)$ and$x^4\equiv1 (mod3)$] RHS cannot be $2(mod3)$ .
  therefore this equation has no solution
