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

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The discussion revolves around the equation X^4 + 131 = 3y^4, which is posed as a problem from Terence Tao's book, seeking to demonstrate the absence of integer solutions. Initial attempts involve modular arithmetic, revealing that x must be even and y must be odd based on calculations mod 16. Further analysis shows that x cannot be divisible by 3 when considered mod 3. The exploration of mod 5 indicates a contradiction, suggesting that no integer solutions exist. The use of modular arithmetic is emphasized as a strategic approach for solving Diophantine equations.
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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.
 
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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...
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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