# 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. $\square$

• That is correct: there are no solutions mod $3$, and therefore no solutions. Commented May 9, 2018 at 7:36
• This is correct, but given that there are only even powers with positive coefficients I would have done this as follows: $x^4\le 131$ so $x^4$ must be one of $0^4,1^4,2^4,3^4$. Similarly $y^4$ must be one of $0^4,1^4,2^4$. Checking that the resulting 12 combinations won't give any solutions is very fast. True, modulo three has even less cases to test, but selecting the right modulus may take some time. Experience helps, of course. Not much difference between the two approaches, so make your pick. Yours would be superior to mine if the equation were $$x^4+3y^4=1333333331$$ instead :-) Commented May 9, 2018 at 8:04
• Are you sure about the $\bmod 5$ ? That proof doesn't work well.
– user65203
Commented May 9, 2018 at 8:41
• Does 5 work. $(0,1,2,-2,-1)^4 \equiv (0,1)\mod 5$ and $3(0,1,2,-2,-1)^4 \equiv 0,3$ so $x^4 + 3y^4 \equiv 0, 1, 3, 4 \mod 5$ and .... $131$ is $1 \mod 5$ so $x^4 \equiv 1\mod 5$ and $y\equiv 0 \mod 5$. $5^4 > 131$ so $y = 0$ and $x^4 = 131$. But $131$ is prime. So impossible. It works. And not too badly. But certainly not as well as mod 3. 5 certainly would not have been my first guess. (Not sure my first guess, mod 4, would have been all the good either.) Commented Jun 13, 2018 at 16:00

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.

• Thanks for the response. So, $\pmod{5}$ is more interesting purely by virtue of Fermat’s Little Theorem, and that the method of applying that theorem has a greater number of applications? Commented Jun 14, 2018 at 5:07
• Eh.... maybe. It's pretty subjective. All I can say is to solve $\mod 3$ as you did is easier. And everyone I know would probably find it more intuitive. However I was kind of impressed that $x^4 + 3y^4 \equiv 1 \mod 5 \iff 5|x, 5\not\mid x$ and $\equiv 3 \mod 5 \iff 5|y, 5\not \mid x$ and $x^4 +3y^4\equiv 4\mod 5 \iff 5\not \mid x; 5\not \mid y$ and $x^4 + 3y^4 \equiv 0 \iff 5|x, 5|y$. It surprised me a little. Because $\mod 5$ is harder and less intuitive, I am guessing the reason you've seen it more is only because somebody finds it interesting. Commented Jun 14, 2018 at 5:18

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.

• Thanks. Yeah, I was pretty careless typing this up and the notation is not good. Could you offer an insight as to why most proofs seem to work $\pmod{5}$, though? My immediate instinct was $\pmod{3}$, based on the coefficient of the y value. Would 5 be selected by considering 131? Commented May 9, 2018 at 8:16
• @MoedPolBollo I have no idea. I would have tried $\pmod3$ first too. Commented May 9, 2018 at 8:20
• Well, this is the first time I've ever seen $x^4 + 3y^4 = 131$ being asked much less proven, so I imagine that one person somewhere made up a prove with mod 5 and it's been recycled and circulated. I imagine if you asked anyone to prove it without given them any reference I image 95% of all people would try powers of 2 or mod 3 first. If "most" proofs are mod 5, I'd imagine it's because so very few people have bothered to prove it. (It's not a very generalizable or useful result). Commented Jun 13, 2018 at 4:36
• @fleablood Shouldn't this be a comment to the original question? Commented Jun 13, 2018 at 6:11
• It was an answer to the "Would 5 be selected by considering 131?" in his comment to you. Commented Jun 13, 2018 at 15:52

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

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

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