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

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In summary, the diophantine equation $ x^4-2y^2=1 $ has only two solutions: $(1,0)$ and $(-1,0)$. The equation can be rewritten as $(x-1)(x+1)(x^2+1)=2y^2$, and by considering possible values for $x$, it can be shown that only $x=1$ and $x=-1$ result in integer solutions for $y$. The proof involves showing that for any other value of $x$, the resulting factors cannot all be squares.
  • #1
evinda
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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)
 
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  • #2
Shouldn't (-1,0) be a solution as well?
 
  • #3
Deveno said:
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)
 
  • #4
"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...
 
  • #5
evinda said:
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 $k = -1$?
 
  • #6
Deveno said:
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 $k = -1$?

Taschee said:
"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)
 
  • #7
evinda said:
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
 
  • #8
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.
 
  • #9
evinda said:
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;
 

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

1. What is meant by "only one solution" in scientific terms?

In science, when we say "only one solution", it means that there is only one correct answer or explanation for a particular problem or question. This solution must be supported by evidence and must be able to consistently explain the observed phenomena.

2. Can there be multiple solutions to a scientific problem?

Yes, there can be multiple solutions to a scientific problem. However, only one solution will be considered the most accurate and reliable based on the evidence and data collected. Other solutions may also be proposed, but they may not be as well-supported or accepted by the scientific community.

3. How do scientists determine if there is only one solution to a problem?

Scientists use the scientific method to determine if there is only one solution to a problem. This involves making observations, formulating a hypothesis, conducting experiments, and analyzing data. Through this process, scientists can determine if their hypothesis is the only solution that can explain the observed phenomena.

4. Are there any exceptions to the idea of there being only one solution in science?

Yes, there can be exceptions to the idea of there being only one solution in science. In some cases, multiple solutions may be equally valid and supported by evidence. This can happen when there are different ways to interpret the data or when new evidence is discovered that challenges the previously accepted solution.

5. Why is it important to have only one solution in science?

Having only one solution in science is important because it allows for consistency and accuracy in understanding and explaining natural phenomena. It also helps to avoid confusion and conflicting information. Additionally, having one well-supported solution allows for further research and advancement in the field.

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