# Solution of this Diophantine Equation

If $$x$$ and $$y$$ are prime numbers which satisfy $$x^2-2y^2=1$$, solve for $$x$$ and $$y$$.

My attempt:

$$x^2-2y^2=1$$

$$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$$

$$\implies (x+\sqrt{2}y)=1$$ and $$(x-\sqrt{2}y)=1$$

$$\implies x=1$$ and $$y=0$$

Clearly $$x$$ and $$y$$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?

\begin{align*}&x^2-2y^2=1\tag{1}\\\iff & x^2-1=(x+1)(x-1)=2y^2\end{align*}

Since $$2\mid (x+1)(x-1)$$, we conclude that both $$(x+1)$$ and $$(x-1)$$ have to be even, and hence $$4\mid 2y^2\implies 2\mid y^2\implies 2\mid y$$ and since $$y$$ is prime, $$\color{red}{y=2}$$. Can you end it now?

From (1), it follows immediately that $$x^2-2y^2=x^2-8=1$$. Thus, the only solution is $$\color{blue}{(3, 2)}$$.

The problem with your method is that for $$a,b\in\mathbb R$$

$$a·b=1\not\Rightarrow a=1\;\text{ and }\;b=1$$

In fact, this only works if $$a·b=0\implies a=0\;\text{ or }\;b=0$$

• +1 for the correct solution but you did not answer my question. What is wrong with my method? – MrAP Mar 28 at 18:33
• Can you put that in your answer. – MrAP Mar 28 at 18:35

The fault is that irrationals can also produce the product to $$1$$.
Consider $$x=3$$ and $$y=2$$ then we get, $$(3+\sqrt{2} \cdot 2)(3-\sqrt{2}\cdot 2)=1$$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $$a+b\sqrt{2}$$ in that step.

• This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there. – B. Goddard Mar 28 at 18:17
• "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?" – Mann Mar 28 at 18:18