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Homework Statement
Show that the equation 3x2 + 2 = y2 has no integer solutions by calculating modulo 3.
Proof that the equation has no rational solutions.
This is a problem from an introductory chapter in algebra that I'm teaching.
Homework Equations
The Attempt at a Solution
I've got the first part down, since the left side is always 2 mod 3, while the right side (a square) is always either 0 or 1 mod 3.However, I'm stuck on the second part.
I can write x=a/b and y=c/d, that can not be simplified, yielding:
3(a/b)² + 2 = (c/d)² ?
3(ad)² + 2(bd)² = (bc)² ?
(bd)² is either 0 or 1 mod 3.
If (bd)² = 1 mod 3, then we can check whether 3(ad)² + 2 = (bc)² mod 3.
Since this is effectively the same as the original equation for integer numbers, we already know this is not possible.
So (bd)² = 0 mod 3.
Consequently from the bold equation: (bc)² = 0 mod 3.
So b = 0 mod 3, or both c and d are 0 mod 3.
If both c and d are 0 mod 3, then the fraction c/d could have been simplified.
This violates the precondition.
So b = 0 mod 3.
For the bold equation to hold then a or d must be 0 mod 3.
If a = 0 mod 3, then a/b could have been simplified.
Again this violates the precondition.
So we have:
a ≠ 0 mod 3, b = 0 mod 3, c ≠ 0 mod 3, d = 0 mod 3.
As yet I still don't see the end of this.This is already quite a long line of reasoning and since it is from an introductory chapter in algebra I'm tentatively drawing the conclusion that I'm missing something.
Can someone give me a clue?
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