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

  • Context: Graduate 
  • Thread starter Thread starter dtl42
  • Start date Start date
  • Tags Tags
    Integer
Click For Summary
SUMMARY

The equation X^4 + 131 = 3y^4 has been proven to have no integer solutions through modular arithmetic analysis. By examining the equation modulo 16, 3, and 5, it was established that x must be even and y must be odd, while also demonstrating that x cannot be divisible by 3. The analysis concluded that the fourth powers modulo 5 provide a definitive contradiction, confirming the absence of integer solutions.

PREREQUISITES
  • Understanding of Diophantine equations
  • Familiarity with modular arithmetic
  • Knowledge of Fermat's Little Theorem
  • Basic number theory concepts
NEXT STEPS
  • Study modular arithmetic applications in number theory
  • Explore Fermat's Little Theorem and its implications
  • Investigate other Diophantine equations and their solution strategies
  • Learn about residue classes and their significance in modular systems
USEFUL FOR

Mathematicians, students studying number theory, and anyone interested in solving Diophantine equations or applying modular arithmetic techniques.

dtl42
Messages
118
Reaction score
0
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.
 
Mathematics news on Phys.org
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...
 

Similar threads

Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
30K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K