Show that the equation $x^4 + 3 y^4 = 131$ has no integer solutions.

Question

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. $\square$

Answer 1

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.

Answer 2

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.

Answer 3

$$x^4 + 3 y^4 = 131$$

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 $$x^4 + 3 y^4 = 131$$

Answer 4

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