Rational solutions of equation 3x² + 2 = y²

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In summary, the conversation discusses a problem from an introductory chapter in algebra, where the goal is to show that the equation 3x^2 + 2 = y^2 has no integer solutions. The first part is solved by calculating modulo 3, but the second part, showing that there are no rational solutions, is still being worked on. One possible approach is to assume that there are rational solutions and use a recursive method to show a contradiction. Another proposed solution is (x^2 + y^2) / -2 = 1, which implies that both x^2 and y^2 are equal to -1.
  • #1
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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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  • #2
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
 
  • #3
Office_Shredder said:
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. :)
 
  • #4
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 .
 
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1. What is a rational solution?

A rational solution is a solution to an equation that can be expressed as a ratio of two integers. In other words, it is a solution that can be written as a fraction.

2. How do you find rational solutions of an equation?

To find rational solutions of an equation, you can use the quadratic formula or factor the equation and solve for the variable. Once you have a solution, you can check if it can be written as a ratio of two integers.

3. Can an equation have more than one rational solution?

Yes, an equation can have more than one rational solution. For example, the equation 2x + 4 = 8 has two rational solutions: x = 2 and x = 4.

4. Are all rational solutions also integer solutions?

No, not all rational solutions are integer solutions. An integer solution is a solution that is a whole number, whereas a rational solution can be a fraction. For example, the equation x² = 4 has two rational solutions (x = 2 and x = -2) but only one integer solution (x = 2).

5. How do rational solutions differ from irrational solutions?

Rational solutions can be expressed as a ratio of two integers, while irrational solutions cannot be expressed as a ratio of two integers. Irrational solutions are often in the form of decimal numbers that do not terminate or repeat, such as pi or the square root of 2.

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