Are there infinite Pythagorean triples where the square root of one of the elements itself is a whole number?

[Unity+]

Let's take a look at the following picture:

 

Of course, a^2+b^2 = c^2 has infinite primitive(I think) Pythogorean triples where a,b, and c are whole numbers, proven by Euclid. However, are there infinite primitive Pythagorean triples where one of the elements, when square rooted, is also a whole number?

 

I am asking this because, for example, let us take the primitive Pythagorean triple (3,4, 5). Notice how the square root of 4 is 2, obviously. Let us apply this to a geometric visual.

 

This means that it takes a real whole number sided square to get two squares that have irrational sides. In interesting phenomena, I say.

[imatfaal]

To get two sides being perfect squares you are basically looking for solutions of the diophantine equations

[latex]x^4+y^4=z^2[/latex]

or

[latex]x^4+y^2=z^4[/latex]

Which I don't think have solutions. This problem is connect to the question of whether the area of a pythagorean triangle made from integer sides ever has a perfect square area - it doesn't.

When I was checking up my answer I came across this on mathworld - BTW seriously cool site

In 1643, Fermat challenged Mersenne to find a Pythagorean triplet whose hypotenuse and sum of the legs were squares. Fermat found the smallest such solution:

,,

 

He worked that out by hand!!!!

http://mathworld.wolfram.com/PythagoreanTriple.html

You have also gotta love the idea of Fermat and Mersenne setting each other challenges - bit too superhuman-difficulty for ordinary mortals.

[mathematic]

There are an infinite number of integer Pythagorean triples.

Let m > n > 0 be any integers. [Latex]a = 2mn, b = m^2 - n^2, c = m^2 + n^2[/Latex].

Edited April 16, 2014 by mathematic