MHB Are Both Solutions to the Diophantine Equation $x^4 - 2y^2 = 1$ Valid?

[evinda]

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)

 

[Deveno]

Shouldn't (-1,0) be a solution as well?

 

[evinda]

Shouldn't (-1,0) be a solution as well?

Yes,but I haven't found this solution with the way that I did it. (Doh)
So,do I have to solve the exercise in an other way? (Thinking)

 

[Taschee]

"So, k,k+1,2k^2+2k+1 must all be a square."

I don't see an apparent reason for that.
Like Deveno said, (-1,0) is another solution, then with x=2k+1 follows k=-1,
and -1 is not a square...

 

[Deveno]

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.

I don't see how you draw this conclusion. Yes, they are all co-prime, but -1 divides any integer, and is not prime. What happens if $k = -1$?

 

[evinda]

I don't see how you draw this conclusion. Yes, they are all co-prime, but -1 divides any integer, and is not prime. What happens if $k = -1$?

"So, k,k+1,2k^2+2k+1 must all be a square."

I don't see an apparent reason for that.
Like Deveno said, (-1,0) is another solution, then with x=2k+1 follows k=-1,
and -1 is not a square...

Sice, $4k(k+1)(2k^2+2k+1)=y^2$

doesn't it mean that $\exists a,b,c \text{ such that } k=a^2, k+1=b^2 \text{ and } 2k^2+2k+1=c^2$ ? Or am I wrong? (Sweating)

 

[kaliprasad]

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)

you have found

$k , k+1, 2k^2 + 2k +1$ must all be square

you also can have
( alternatively)
$- k , -(k+1), 2k^2 + 2k +1$ must all be square which gives the second solution

 

[Deveno]

Put another way, we can factor $c^2 = a^2b^2$ two ways:

$c = a^2b^2 = (-a^2)(-b^2)$.

Prime factorization is only unique "up to units" and the integers have TWO units: 1 and -1. People often overlook the second unit.

 

[Taschee]

Sice, $4k(k+1)(2k^2+2k+1)=y^2$

doesn't it mean that $\exists a,b,c \text{ such that } k=a^2, k+1=b^2 \text{ and } 2k^2+2k+1=c^2$ ? Or am I wrong? (Sweating)

Not necessarily.
You need to prove, that k, k+1 and 2k^2+2k+1 are coprime:
k and k+1 are obviously coprime, if p|k, then p|2k^2 and p|2k,
so p \nmid 2k^2+2k+1;
If p|k+1, then p|2(k+1)^2 = (2k^2+2k+1) + (k+1) + k
Suppose p|2k^2+2k+1 , then you have p|k in contradiction to p|k+1

Then you still have the possibility that two of those terms are negativ squares,
i.e. k=-a^2, k+1=-b^2;