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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Can anybody solve the system

|x^2+y^2=a^2

|x^2-y^2=b^2,

where x, y, a and b are integers?

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

What is the | for? Is it a mathematical symbol?

*Last edited by rickyoswaldiow (2005-12-04 01:10:15)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

a, b, x, and y = 0 is a solution, though uninteresting.

Other than that, I can only come up with the restrictions:

0 ≤ y² ≤ x² ≤ a²

0 ≤ b² ≤ x² ≤ a²

By using the fact that squares have to be positive, so x² and y² must be positive, and thus b² has to be less than or equal to x², since y² is at least 0, and at most, x².

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Let write the system as:

|x^2+y^2=a^2

|b^2+y^2=x^2

Thus we get 2 pythagorean triples (Sorry if the syntax is incorrect. I don't know English well)

x=u^2-v^2

y=2uv

a=u^2+v^2

and

b=w^2-z^2

y=2wz

x=w^2+z^2.

To solve the system is enough to solve:

|2uv=2wz

|u^2-v^2=w^2+z^2

<=>

|uv=wz

|u^2=w^2+z^2+v^2

what to do further?

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

What you're basically looking for is two pythagorean triplets that go:

b, y, x, and then y, x, a.

I don't believe any such triplets exist, although I can't prove it.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

rickyoswaldiow wrote:

What is the | for? Is it a mathematical symbol?

Just being used to "group" the two equations together - *more typographic than mathematic*. If it was |x| that would mean absolute value.

BTW, visually I get this:

b² (<-y²->) x² (<-y²->) a²

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

More colouricaly, we search for 4 squares such the sum of first and second is equal to third and the sum of second and third is equal to fourth. So we can make a generalized question. Does the system:

|a1²+a2²=a3²

|a2²+a3²=a4²

|...

|a{N-2}²+a{N-1}²=aN²

have integer solutions?

I'm sure the upper system hasn't general solution.

Why?

*Last edited by krassi_holmz (2005-12-04 17:31:08)*

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