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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?

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What about

\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 $\color{blue}{(3, 2)}$.


Addendum

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 $$a·b=0\implies a=0\;\text{ or }\;b=0$$

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

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    $\begingroup$ 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. $\endgroup$ – B. Goddard Mar 28 at 18:17
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    $\begingroup$ "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?" $\endgroup$ – Mann Mar 28 at 18:18
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    $\begingroup$ Please read the question. I answered exactly what has been asked. $\endgroup$ – Mann Mar 28 at 18:18
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    $\begingroup$ @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer? $\endgroup$ – Martin Argerami Mar 29 at 18:29

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