Solving X^4+131=3y^4: No Integer Solutions

[dtl42]

This is a problem from Terence Tao's book, and I cannot locate the solution anywhere, so I thought I'd post it here.


1. Homework Statement

X^4+131=3y^4, Show that the equation has no solutions when x and y are integers.

Homework Equations

None? Not sure.

The Attempt at a Solution

I thought of using modular arithmetic, but I can't really work it correctly.

 

[Xevarion]

Okay, let's do some mods.

Mod (powers of) 2: 4th powers are 1 or 0 mod 16. 131 is 3 mod 16. So the LHS can be 3 or 4 mod 16 and the RHS can be 0 or 3 mod 16. Therefore we must have x even and y odd.

Mod 3: 4th powers are 1 or 0 mod 3. 131 is 2 mod 3. So the LHS can be 2 or 0 mod 3, and the RHS is 0 mod 3. So x cannot be divisible by 3.

Mod 5: 4th powers are 1 or 0 mod 5. 131 is 1 mod 5. So the LHS can be 1 or 2 mod 5, and the RHS can be 0 or 3 mod 5. Uh-oh! Looks like we're done!

Comments on strategy: When you're trying to solve a Diophantine equation, using mods is usually the place to start. If only one power appears, you might be able to guess what mod to use by Fermat's Little Theorem (in fact, in this case, this tells you to try 5, and that's the one that works). Powers of 2 and 3 are also good choices because there tend to be few possible residues mod those numbers. Using the given numbers can be useful too -- for example, it may also be possible to show that there are no solutions to this by looking mod 131. But I'm too lazy to compute all of the fourth powers mod 131...

 

[tacman]

I also tried factoring and using the difference of squares, but that didn't seem to lead anywhere. I'm not sure how to approach this problem.


After spending some time on this problem, I also could not find a solution using modular arithmetic or factoring. However, I did come across a proof by contradiction that shows there are no integer solutions to this equation.

First, assume that there exists an integer solution to the equation, where x and y are both integers. This means that x^4 + 131 = 3y^4 has a solution in integers.

Next, consider the equation modulo 3. This gives us x^4 + 2 ≡ 0 (mod 3). Since 3 is a prime number, we can apply Fermat's Little Theorem, which states that for any integer a and prime number p, a^p ≡ a (mod p). In this case, we have a = x and p = 4, so x^4 ≡ x (mod 3). Therefore, our equation becomes x + 2 ≡ 0 (mod 3).

This implies that x ≡ 1 (mod 3) or x ≡ 2 (mod 3). Substituting these values back into our original equation, we get (1)^4 + 131 = 3y^4 or (2)^4 + 131 = 3y^4. However, neither of these equations have integer solutions, as 132 is not a perfect fourth power and 147 is not divisible by 3.

Therefore, our assumption that there exists an integer solution to the equation is false. This means that the equation x^4 + 131 = 3y^4 has no solutions in integers.