Rational solutions of equation 3x² + 2 = y²

[I like Serena]

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?

 

[Office_Shredder]

I have an ugly proposal which might lead to some insight. Once you have b=0 mod 3, write it as b=3b'

3(a/(3b'))²+2=(c/d)²

1/3*(a/b')²+2=(c/d)²

clearing denominators again

(ad)²+6(b'd)²=3(b'c)²

taking mod 3 again we see that ad=0 mod 3, so a=0 mod 3 or d=0 mod 3. We know a can't be zero mod 3 because of the reduction, so d=0 mod 3 as well. Let's do the same thing again. d=3d'

1/3*(a/b')²+2=1/9*(a/d')². Clear denominators

3(ad')²+18(b'd')=(ab')²

Taking mod 3 and noting that a can't be divisible by 3, we get that b' is divisible by 3. I think this recurses, i.e. using 3|b' you can get that 3|d', then get that 3|b'', then 3|d'' etc. which is a contradiction because 3 only divides b and d finitely many times

 

[I like Serena]

I have an ugly proposal which might lead to some insight. Once you have b=0 mod 3, write it as b=3b'

3(a/(3b'))²+2=(c/d)²

1/3*(a/b')²+2=(c/d)²

clearing denominators again

(ad)²+6(b'd)²=3(b'c)²

taking mod 3 again we see that ad=0 mod 3, so a=0 mod 3 or d=0 mod 3. We know a can't be zero mod 3 because of the reduction, so d=0 mod 3 as well. Let's do the same thing again. d=3d'

1/3*(a/b')²+2=1/9*(a/d')². Clear denominators

3(ad')²+18(b'd')=(ab')²

Taking mod 3 and noting that a can't be divisible by 3, we get that b' is divisible by 3. I think this recurses, i.e. using 3|b' you can get that 3|d', then get that 3|b'', then 3|d'' etc. which is a contradiction because 3 only divides b and d finitely many times

Thanks!
That works.
I did see that there was a sort of recursive relationship, but I didn't realize the contradiction that it implies. :)

 

[Crazy Horse 11]

my answer is (x^2 + y^2) / -2 = 1

in other words both x^2 and y^2 = -1 ... see?

3x^2 + 2 = y^2
-3 + 2 = -1

P.j.S .