Pythagorean triples,consecutive terms of an arithmetic progression

Question

I am looking at the exercise: Find all the positive Pythagorean triples that are consecutive terms of an arithmetic progression. $$$$ So, according to the solution that I saw in my notes, we want to find $x,y,z>0$ such that $x^2+y^2=z^2$ and $x+z=2y$. $$$$

How did we find the relation $x+z=2y$? Do we conclude to this, from the arithmetic progression? If yes, how?

Answer 1

An arithmetic progression is defined as the sequence of terms,

$$a,a+d,a+2d,a+3d,\cdots$$

where $a$ is the first term, and $d$ is the "common difference". Here we have a arithmetic progression of three terms, $x,y,z$

By definition, $y-x=d$ and $z-y=d$(difference of any two consecutive terms is $d$)

Therefore, we get $$y-x=z-y\implies x+z=2y$$

Note: Any arithmetic progression has the nice property that,

$$t_n = \frac{t_{n-k}+t_{n+k}}{2}$$

where $t_n$ is the nth term. To put that formula into english, the term exactly between two given terms is the average of the two terms. This is the special case, $n=2,k=1$

Presumably this is where the expression "arithmetic mean" gets its name.

Answer 2

Well, an interesting question and here is my attempt to answer it.

Assuming the three terms of required triple as:

       `a - d, a, a + d`,  upon applying Pythagorean theorem to them, we get:

(a + d)^2 = (a - d)^2 + a^2 or upon simplifying and rearranging things, we get

       a = 4d and providing only integer values to d, we get the first set as:

      when  d = 1, a = 4, and hence the triple is (3, 4, 5) and for all higher integer values of d, such as d = 2, d = 3 and so on, we can see that those are multiples of the primitive triple, namely (6, 8, 10), (9, 12, 15) etc. 

         However, when d = 1/60, we can have an interesting triple, which is in an A.P., viz. (1/20 ,1/15, 1/12) and lot more such '**fraction-based**' triples do exist, as long as d can be a fraction with a denominator - which is a multiple of 4.

Hope, this throws some light on it.