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• Hi Peter, Not 100% on topic but close enough for a response from me. Chances are number fiddlers, hypothesizers and number enthusiasts make this discovery
Message 1 of 3 , Jul 14, 2009
Hi Peter,

Not 100% on topic but close enough for a response from me.

Chances are number fiddlers, hypothesizers and number enthusiasts make this

discovery every day. I pounced upon it messing around with x^n+y^n = z^n

in my yout back in the late 50's. At that time I assumed your result were the
only solutions.

What I should have done was look more closely at the numbers at that time.

n x y z

1 3 4 5

2 5 12 13

3 7 24 25

4 9 40 41

5 11 60 61

....

Indeed, 3^2+4^2=5^2, 5^2 + 12^2 = 13^2, etc

Notice that x is the set of odd numbers. Then notice that

n x

1 3 = 4-1

2 5 = 9-4

3 7 = 16-9

4 9 = 25-16

...

Yep looks like x is the difference between 2 squares. Now

n z

1 5 = 4+1

2 13 = 9+4

3 25 = 16+9

4 41 = 25+16

...

Yep looks like z is the sum of 2 squares. Hmmmm....Difference and sum of 2 squares?

Ok x = a^2-b^2 for some a and b. Now notice that when

n is 1 x = 3 = 4 - 1 and z = 5 = 4+1

n is 2 x = 5 = 9 - 4 and z = 13 = 9+4

n is 3 x = 7 = 16 -9 and z = 25 = 16+9

...

Sure Looks like a pattern to me. Yep, kooks like x = a^2-b^2 and z = a^2 + b^2.

We are almost there. We now need to find y. Lets go ahead and assume this for

x and z.

y^2 = z^2 - x^2 = (a^2+b^2) - (a^2-b^2) =

(a^4+2a^2*b^2+b^4) - (a^4-2a^2*b^2+b^4) = 4a^2*b^2

so y = 2ab and the picture show now is

if x^2+y^2=z^2 then

x = a^2-b^2

y = 2ab

z = a^2+b^2

Lets try it out. Let a =7,b=2 then

45^2 + 28^2 = 53^2

It works!

So multiply it out for the general case.

(a^2-b^2)^2 + (2ab) = (a^2+b^2)^2

(You can do this)

Now we have proved the general case for all a>0,b.

Heuristics Rock,

El cino

From: plesala@...
Date: Tue, 14 Jul 2009 06:09:33 +0200

Hi,

I figured out that there is a formula for sets of integers for which Pythagoras theorem is true.

3 4 5

5 12 13

7 24 25

The formula is

2n + 1 2n(n + 1) 2n(n + 1) + 1

The question I am having difficulty with is whether or not there are sets of integers that do not belong to the set which is defined by the above relation; i.e are there some integers a, b and c for which

a^2 + b^2 = c^2 that are connected by the relationship that if a is defined as

a = 2n + 1, then b and c have to be

b = 2n(n + 1) and c = 2n( n + 1) + 1, respectively?

Thank you.

Peter.

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