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

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).

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

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.

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.