"Classic.."

[grayfalcon89]

Say you have three sides of a right triangle that form an arithmetic sequence. The only possible one is 3-4-5. Prove this.

[matt grime]

I can't prove it; it is false: there are infinitely many counter examples. Perhaps you omitted some other condition on the primitivity of the triangle (6,8,10 or 9,12,15 etc are the counter examples).

[the tree]

I have the feeling that the OP meant the only possible proportion is 3:4:5.

[matt grime]

That makes most sense, but I did also give the method of proving the correct version, albeit without stating it: the form of **primitive** pythagorean triples is a well known one and can easily be found.

 

the vey first hit on google for primitive pythagorean triples gives the answer, for instance.

[BigMoosie]

My proof.

 

It has been shown that all pythagorean triples can be formed by this formula where a and b are any integers and the third value is the hypotenuse.

 

[math]a^2 - b^2, 2 a b, a^2 + b^2[/math]

 

So our arithmetic series could be defined by either of the following.

 

(1) [math](a^2 + b^2) - 2ab = 2ab - (a^2-b^2)[/math]

 

(2) [math](a^2 + b^2) - (a^2-b^2) = (a^2 - b^2) - 2ab[/math]

 

Simplifying (1):

 

[math]2a^2 = 4ab^2[/math]

 

[math]a=2b[/math]

 

Substituting a back into our original formula:

 

[math]3b^2, 4b^2, 5b^2[/math]

 

The values are in the 3,4,5 ratio, therefore true for any a or b in the first form.

 

Simplifying (2):

 

[math]3b^2 + 2ab - a^2 = 0[/math]

 

[math]b = \frac{-2a \pm \sqrt{4a^2 + 12a^2}}{b} = \frac{-a \pm 2a}{3}[/math]

 

Taking the first root is trivial as [math]b=-a[/math] and the first value will be zero (not technically a pythagorean triple).

 

The second root:

 

[math]a = 3b[/math]

 

Substituting a into the original formula:

 

[math]8b^2, 6b^2, 10b^2[/math]

 

In the ratio of 3,4,5.

Since it is true for any a or b it is true for any pythagorean triple.