1. The problem statement, all variables and given/known data 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. 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 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?