- #1
evinda
Gold Member
MHB
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Hey! (Smirk)
I am looking at the following exercise:
Prove that the diophantine equation $ x^4-2y^2=1 $ has only two solutions.
It is: $$(x−1)(x+1)(x^2+1)=2y^2 $$
If $x=2k \Rightarrow 2y^2=(2k-1)(2k+1)(4k^2+1), \text{ that is not possible, because } 2 \nmid (2k-1)(2k+1)(4k^2+1)$
So,it is $x=2k+1$.
Replacing this,we get:
$$4k(k+1)(2k^2+2k+1)=y^2$$
So, $k,k+1,2k^2+2k+1$ must all be a square.
$$ k=a^2 $$
$$k+1=a^2+1=b^2 $$
$$b^2−a^2=1⇒a=0,b= \pm 1 \Rightarrow k=0 $$
Therefore, $x=1$.I found only the solution: $(1,0)$ .. So,have I done something wrong? (Thinking) (Thinking)
I am looking at the following exercise:
Prove that the diophantine equation $ x^4-2y^2=1 $ has only two solutions.
It is: $$(x−1)(x+1)(x^2+1)=2y^2 $$
If $x=2k \Rightarrow 2y^2=(2k-1)(2k+1)(4k^2+1), \text{ that is not possible, because } 2 \nmid (2k-1)(2k+1)(4k^2+1)$
So,it is $x=2k+1$.
Replacing this,we get:
$$4k(k+1)(2k^2+2k+1)=y^2$$
So, $k,k+1,2k^2+2k+1$ must all be a square.
$$ k=a^2 $$
$$k+1=a^2+1=b^2 $$
$$b^2−a^2=1⇒a=0,b= \pm 1 \Rightarrow k=0 $$
Therefore, $x=1$.I found only the solution: $(1,0)$ .. So,have I done something wrong? (Thinking) (Thinking)