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## Topic review (newest first)

krassi_holmz
2005-12-05 16:29:08

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?

MathsIsFun
2005-12-05 07:45:41

#### 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

Ricky
2005-12-05 07:39:40

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.

krassi_holmz
2005-12-05 02:19:54

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?

Ricky
2005-12-05 00:39:40

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.

RickyOswaldIOW
2005-12-05 00:10:05

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

krassi_holmz
2005-12-04 23:22:31

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?