(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that the equation 3x^{2}+ 2 = y^{2}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.

2. Relevant equations

3. 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 thebold 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 thebold equationto 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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# Rational solutions of equation 3x² + 2 = y²

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